# IBDP Maths AA: Topic: SL 4.6: Venn diagrams and tree diagrams : IB style Questions SL Paper 1

### Question

Andre will play in the semi-final of a tennis tournament.

If Andre wins the semi-final he will progress to the final. If Andre loses the semi-final,

he will not progress to the final.

If Andre wins the final, he will be the champion.

The probability that Andre will win the semi-final is p . If Andre wins the semi-final, then the

probability he will be the champion is 0.6.

1. Complete the values in the tree diagram. [1]

1. b.   The probability that Andre will not be the champion is 0.58.
2. c.  Find the value of p . [2]
3. d. Given that Andre did not become the champion, find the probability that he lost in the semi-final. [3]
Answer/Explanation

Ans:

(a)

(b)

p × 0.4+ (1-9)=0.58

OR

P × 0.6 = 1- 0.58

P= 0.7

(c)

$$\frac{0.3}{0.58}(\frac{1-0.7}{0.58})$$

OR

$$\frac{0.3}{0.3+0.7\times 0.4}$$

$$\frac{15}{29}$$(0.517,0.517241…,51.7%)

### Question

In a class of 30 students, 19 play tennis, 3 play both tennis and volleyball, and 6 do not play either sport.

The following Venn diagram shows the events “plays tennis” and “plays volleyball”.

The values t and v represent numbers of students.

1.           (i) Find the value of t .

(ii) Find the value of v . [4]

2. Find the probability that a randomly selected student from the class plays tennis or volleyball, but not both. [2]

Answer/Explanation

Ans:

(a)

(i)

valid approach to find t

eg  t+ 3 = 19, 19 – 3

t =16 (may be seen on Venn diagram)

(ii)

valid approach to find v

eg t+ 3 + v + 6 = 30, 30- 19 – 6

v = 5 (may be seen on Venn diagram)

(b)

valid approach

eg 16+  5  , 21 students, $$1-\frac{3+6}{30}$$

$$\frac{21}{30}(=\frac{7}{10})$$

## Question

Two boxes contain numbered cards as shown below.

Two cards are drawn at random, one from each box.

Copy and complete the table below to show all nine equally likely outcomes.

[2]
a.

Let S be the sum of the numbers on the two cards.

Find the probability of each value of S.

[2]
b.

Find the expected value of S.

[3]
c.

Anna plays a game where she wins $$\ 50$$ if S is even and loses $$\ 30$$ if S is odd.

Anna plays the game 36 times. Find the amount she expects to have at the end of the 36 games.

[3]
d.
Answer/Explanation

## Markscheme

A2     N2

[2 marks]

a.

$${\rm{P}}(12) = \frac{1}{9}$$ , $${\rm{P}}(13) = \frac{3}{9}$$ , $${\rm{P}}(14) = \frac{3}{9}$$ , $${\rm{P}}(15) = \frac{2}{9}$$     A2     N2

[2 marks]

b.

correct substitution into formula for $${\text{E}}(X)$$     A1

e.g. $${\rm{E}}(S) = 12 \times \frac{1}{9} + 13 \times \frac{3}{9} + 14 \times \frac{3}{9} + 15 \times \frac{2}{9}$$

$${\rm{E}}(S) = \frac{{123}}{9}$$     A2     N2

[3 marks]

c.

METHOD 1

correct expression for expected gain E(A) for 1 game     (A1)

e.g. $$\frac{4}{9} \times 50 – \frac{5}{9} \times 30$$

$${\rm{E}}(A) = \frac{{50}}{9}$$

amount at end = expected gain for 1 game $$\times 36$$     (M1)

= 200 (dollars)     A1     N2

METHOD 2

attempt to find expected number of wins and losses     (M1)

e.g. $$\frac{4}{9} \times 36$$ , $$\frac{5}{9} \times 36$$

attempt to find expected gain E(G)     (M1)

e.g. $$16 \times 50 – 30 \times 20$$

$${\text{E}}(G) = 200$$ (dollars)     A1     N2

[3 marks]

d.

## Question

Consider the events A and B, where $${\rm{P}}(A) = 0.5$$ , $${\rm{P}}(B) = 0.7$$ and $${\rm{P}}(A \cap B) = 0.3$$ .

The Venn diagram below shows the events A and B, and the probabilities p, q and r.

Write down the value of

(i)     p ;

(ii)    q ;

(iii)   r.

[3]
a(i), (ii) and (iii).

Find the value of $${\rm{P}}(A|B’)$$ .

[2]
b.

Hence, or otherwise, show that the events A and B are not independent.

[1]
c.
Answer/Explanation

## Markscheme

(i) $$p = 0.2$$     A1     N1

(ii) $$q = 0.4$$     A1     N1

(iii) $$r = 0.1$$     A1     N1

[3 marks]

a(i), (ii) and (iii).

$${\rm{P}}(A|B’) = \frac{2}{3}$$     A2     N2

Note: Award A1 for an unfinished answer such as $$\frac{{0.2}}{{0.3}}$$ .

[2 marks]

b.

valid reason     R1

e.g. $$\frac{2}{3} \ne 0.5$$ ,  $$0.35 \ne 0.3$$

thus, A and B are not independent     AG     N0

[1 mark]

c.

## Question

The diagram below shows the probabilities for events A and B , with $${\rm{P}}(A’) = p$$ .

Write down the value of p .

[1]
a.

Find $${\rm{P}}(B)$$ .

[3]
b.

Find $${\rm{P}}(A’|B)$$ .

[3]
c.
Answer/Explanation

## Markscheme

$$p = \frac{4}{5}$$     A1     N1

[1 mark]

a.

multiplying along the branches     (M1)

e.g. $$\frac{1}{5} \times \frac{1}{4}$$ , $$\frac{{12}}{{40}}$$

adding products of probabilities of two mutually exclusive paths     (M1)

e.g. $$\frac{1}{5} \times \frac{1}{4} + \frac{4}{5} \times \frac{3}{8}$$ , $$\frac{1}{{20}} + \frac{{12}}{{40}}$$

$${\rm{P}}(B) = \frac{{14}}{{40}}$$ $$\left( { = \frac{7}{{20}}} \right)$$     A1     N2

[3 marks]

b.

appropriate approach which must include $${A’}$$ (may be seen on diagram)     (M1)

e.g. $$\frac{{{\rm{P}}(A’ \cap B)}}{{{\rm{P}}(B)}}$$ (do not accept $$\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}$$ )

$${\rm{P}}(A’|B) = \frac{{\frac{4}{5} \times \frac{3}{8}}}{{\frac{7}{{20}}}}$$     (A1)

$${\rm{P}}(A’|B) = \frac{{12}}{{14}}$$ $$\left( { = \frac{6}{7}} \right)$$     A1     N2

[3 marks]

c.

## Question

In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn diagram below shows the events art and music. The values p , q , r and s represent numbers of students.

(i)     Write down the value of s .

(ii)    Find the value of q .

(iii)   Write down the value of p and of r .

[5]
a(i), (ii) and (iii).

(i)     A student is selected at random. Given that the student takes music, write down the probability the student takes art.

(ii)    Hence, show that taking music and taking art are not independent events.

[4]
b(i) and (ii).

Two students are selected at random, one after the other. Find the probability that the first student takes only music and the second student takes only art.

[4]
c.
Answer/Explanation

## Markscheme

(i) $$s = 1$$     A1     N1

(ii) evidence of appropriate approach     (M1)

e.g. $$21 – 16$$ , $$12 + 8 – q = 15$$

$$q = 5$$     A1     N2

(iii) $$p = 7$$ , $$r = 3$$     A1A1     N2

[5 marks]

a(i), (ii) and (iii).

(i) $${\rm{P(art|music)}} = \frac{5}{8}$$     A2     N2

(ii) METHOD 1

$${\rm{P(art)}} = \frac{{12}}{{16}}$$ $$\left( { = \frac{3}{4}} \right)$$     A1

evidence of correct reasoning     R1

e.g. $$\frac{3}{4} \ne \frac{5}{8}$$

the events are not independent     AG     N0

METHOD 2

$${\rm{P(art)}} \times {\rm{P(music)}} = \frac{{96}}{{256}}$$ $$\left( { = \frac{3}{8}} \right)$$     A1

evidence of correct reasoning     R1

e.g. $$\frac{{12}}{{16}} \times \frac{8}{{16}} \ne \frac{5}{{16}}$$

the events are not independent     AG     N0

[4 marks]

b(i) and (ii).

$${\text{P(first takes only music)}} = \frac{3}{{16}}$$ (seen anywhere)     A1

$${\text{P(second takes only art)}} = \frac{7}{{15}}$$ (seen anywhere)     A1

evidence of valid approach     (M1)

e.g. $$\frac{3}{{16}} \times \frac{7}{{15}}$$

$${\text{P(music and art)}} = \frac{{21}}{{240}}$$ $$\left( { = \frac{7}{{80}}} \right)$$     A1     N2

[4 marks]

c.

## Question

The Venn diagram below shows events A and B where $${\rm{P}}(A) = 0.3$$ , $${\rm{P}}(A \cup B) = 0.6$$ and $${\rm{P}}(A \cap B) = 0.1$$ . The values m , n , p and q are probabilities.

(i)     Write down the value of n .

(ii)    Find the value of m , of p , and of q .

[4]
a(i) and (ii).

Find $${\rm{P}}(B’)$$ .

[2]
b.
Answer/Explanation

## Markscheme

(i) $$n = 0.1$$     A1     N1

(ii) $$m = 0.2$$ , $$p = 0.3$$ , $$q = 0.4$$     A1A1A1     N3

[4 marks]

a(i) and (ii).

appropriate approach

e.g. $${\rm{P}}(B’) = 1 – P(B)$$ , $$m + q$$ , $$1 – (n + p)$$     (M1)

$${\rm{P}}(B’) = 0.6$$     A1     N2

[2 marks]

b.

## Question

Events A and B are such that $${\rm{P}}(A) = 0.3$$ , $${\rm{P}}(B) = 0.6$$ and $${\rm{P}}(A \cup B) = 0.7$$ .

The values q , r , s and t represent probabilities.

Write down the value of t .

[1]
a.

(i)     Show that $$r = 0.2$$ .

(ii)    Write down the value of q and of s .

[3]
b(i) and (ii).

(i)     Write down $${\rm{P}}(B’)$$ .

(ii)    Find $${\rm{P}}(A|B’)$$ .

[3]
c(i) and (ii).
Answer/Explanation

## Markscheme

$$t = 0.3$$     A1     N1

[1 mark]

a.

(i) correct values     A1

e.g. $$0.3 + 0.6 – 0.7$$ , $$0.9 – 0.7$$

$$r = 0.2$$     AG     N0

(ii) $$q = 0.1$$ , $$s = 0.4$$     A1A1     N2

[3 marks]

b(i) and (ii).

(i) $$0.4$$     A1     N1

(ii) $${\rm{P}}(A|B’) = \frac{1}{4}$$     A2     N2

[3 marks]

c(i) and (ii).

## Question

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

(i)     Copy and complete the following tree diagram.

(ii)    Find the probability that two white balls are chosen.

[5]
a(i) and (ii).

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$ .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Find the probability that the two balls are white.

[5]
b.

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$ .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Given that both balls are white, find the probability that they were chosen from bag A.

[4]
c.
Answer/Explanation

## Markscheme

(i)

$$\frac{4}{6},\frac{3}{6}{\rm{and}}\frac{3}{6}\left( {\frac{2}{3},\frac{1}{2}{\rm{and}}\frac{1}{2}} \right)$$    A1A1A1     N3

(ii) multiplying along the correct branches (may be seen on diagram)     (A1)

e.g. $$\frac{3}{7} \times \frac{2}{6}$$

$$\frac{6}{{42}}\left( { = \frac{1}{7}} \right)$$     A1     N2

[5 marks]

a(i) and (ii).

$${\rm{P(bag A) = }}\frac{2}{6}\left( { = \frac{1}{3}} \right)$$ , $${\rm{P(bag B) = }}\frac{4}{6}\left( { = \frac{2}{3}} \right)$$ (seen anywhere)     (A1)(A1)

appropriate approach    (M1)

e.g. $${\rm{P(}}WW \cap A) + {\rm{P}}(WW \cap B)$$

correct calculation     A1

e.g. $$\frac{1}{3} \times \frac{1}{7} + \frac{2}{3} \times \frac{2}{7}$$ , $$\frac{2}{{42}} + \frac{8}{{42}}$$

$${\rm{P}}(2W) = \frac{{60}}{{252}}\left( { = \frac{5}{{21}}} \right)$$    A1     N3

[5 marks]

b.

recognizing conditional probability     (M1)

e.g. $$\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}$$ , $${\rm{P}}(A|WW) = \frac{{{\rm{P}}(WW \cap A)}}{{{\rm{P}}(WW)}}$$

correct numerator     (A1)

e.g. $${\rm{P}}(A \cap WW) = \frac{6}{{42}} \times \frac{2}{6},\frac{1}{{21}}$$

correct denominator     (A1)

e.g. $$\frac{6}{{252}},\frac{5}{{21}}$$

probability $$\frac{{84}}{{420}}\left( { = \frac{1}{5}} \right)$$    A1     N3

[4 marks]

c.

## Question

Bill and Andrea play two games of tennis. The probability that Bill wins the first game is $$\frac{4}{5}$$.

If Bill wins the first game, the probability that he wins the second game is $$\frac{5}{6}$$.

If Bill loses the first game, the probability that he wins the second game is $$\frac{2}{3}$$.

Copy and complete the following tree diagram. (Do not write on this page.)

[3]
a.

Find the probability that Bill wins the first game and Andrea wins the second game.

[2]
b.

Find the probability that Bill wins at least one game.

[4]
c.

Given that Bill wins at least one game, find the probability that he wins both games.

[5]
d.
Answer/Explanation

## Markscheme

A1A1A1     N3

Note:   Award A1 for each correct bold probability.

[3 marks]

a.

multiplying along the branches (may be seen on diagram)     (M1)

eg   $$\frac{4}{5} \times \frac{1}{6}$$

$$\frac{4}{{30}}\left( {\frac{2}{{15}}} \right)$$     A1     N2

[2 marks]

b.

METHOD 1

multiplying along the branches (may be seen on diagram)     (M1)

eg   $$\frac{4}{5} \times \frac{5}{6},{\text{ }}\frac{4}{5} \times \frac{1}{6},{\text{ }}\frac{1}{5} \times \frac{2}{3}$$

adding their probabilities of three mutually exclusive paths     (M1)

eg   $$\frac{4}{5} \times \frac{5}{6} + \frac{4}{5} \times \frac{1}{6} + \frac{1}{5} \times \frac{2}{3},{\text{ }}\frac{4}{5} + \frac{1}{5} \times \frac{2}{3}$$

correct simplification     (A1)

eg   $$\frac{{20}}{{30}} + \frac{4}{{30}} + \frac{2}{{15}},{\text{ }}\frac{2}{3} + \frac{2}{{15}} + \frac{2}{{15}}$$

$$\frac{{28}}{{30}}{\text{ }}\left( { = \frac{{14}}{{15}}} \right)$$     A1     N3

METHOD 2

recognizing “Bill wins at least one” is complement of “Andrea wins 2”     (R1)

eg   finding P (Andrea wins 2)

P (Andrea wins both) $$= \frac{1}{5} \times \frac{1}{3}$$     (A1)

evidence of complement     (M1)

eg   $$1 – p,{\text{ }}1 – \frac{1}{{15}}$$

$$\frac{{14}}{{15}}$$     A1     N3

[4 marks]

c.

P (B wins both) $$\frac{4}{5} \times \frac{5}{6}{\text{ }}\left( { = \frac{2}{3}} \right)$$     A1

evidence of recognizing conditional probability     (R1)

eg   $${\text{P}}(A\left| B \right.),{\text{ P (Bill wins both}}\left| {{\text{Bill wins at least one), tree diagram}}} \right.$$

correct substitution     (A2)

eg   $$\frac{{\frac{4}{5} \times \frac{5}{6}}}{{\frac{{14}}{{15}}}}$$

$$\frac{{20}}{{28}}{\text{ }}\left( { = \frac{5}{7}} \right)$$     A1     N3

[5 marks]

d.

## Question

In a class of 21 students, 12 own a laptop, 10 own a tablet, and 3 own neither.

The following Venn diagram shows the events “own a laptop” and “own a tablet”.

The values $$p$$, $$q$$, $$r$$ and $$s$$ represent numbers of students.

A student is selected at random from the class.

Two students are randomly selected from the class. Let $$L$$ be the event a “student owns a laptop”.

(i)     Write down the value of $$p$$.

(ii)     Find the value of $$q$$.

(iii)     Write down the value of $$r$$ and of $$s$$.

[5]
a.

(i)     Write down the probability that this student owns a laptop.

(ii)     Find the probability that this student owns a laptop or a tablet but not both.

[4]
b.

(i)     Copy and complete the following tree diagram. (Do not write on this page.)

(ii)     Write down the probability that the second student owns a laptop given that the first owns a laptop.

[4]
c.
Answer/Explanation

## Markscheme

(i)     $$p = 3$$     A1     N1

(ii)     valid approach     (M1)

eg$$\,\,\,\,\,$$$$(12 + 10 + 3) – 21,{\text{ }}22 – 18$$

$$q = 4$$    A1     N2

(iii)     $$r = 8,{\text{ }}s = 6$$     A1A1     N2

a.

(i)     $$\frac{{12}}{{21}}{\text{ }}\left( { = \frac{4}{7}} \right)$$     A2     N2

(ii)     valid approach     (M1)

eg$$\,\,\,\,\,$$$$8 + 6,{\text{ }}r + s$$

$$\frac{{14}}{{21}}{\text{ }}\left( { = \frac{2}{3}} \right)$$    A1     N2

b.

(i)          A1A1A1     N3

(ii)     $$\frac{{11}}{{20}}$$     A1     N1

[4 marks]

c.

## Question

In a group of 20 girls, 13 take history and 8 take economics. Three girls take both history and economics, as shown in the following Venn diagram. The values $$p$$ and $$q$$ represent numbers of girls.

Find the value of $$p$$;

[2]
a.i.

Find the value of $$q$$.

[2]
a.ii.

A girl is selected at random. Find the probability that she takes economics but not history.

[2]
b.
Answer/Explanation

## Markscheme

valid approach     (M1)

eg$$\,\,\,\,\,$$$$p + 3 = 13,{\text{ }}13 – 3$$

$$p = 10$$     A1     N2

[2 marks]

a.i.

valid approach     (M1)

eg$$\,\,\,\,\,$$$$p + 3 + 5 + q = 20,{\text{ }}10 – 10 – 8$$

$$q = 2$$     A1     N2

[2 marks]

a.ii.

valid approach     (M1)

eg$$\,\,\,\,\,$$$$20 – p – q – 3,{\text{ }}1 – \frac{{15}}{{20}},{\text{ }}n(E \cap H’) = 5$$

$$\frac{5}{{20}}\,\,\,\left( {\frac{1}{4}} \right)$$     A1     N2

[2 marks]

b.

## Question

A bag contains 5 green balls and 3 white balls. Two balls are selected at random without replacement.

Complete the following tree diagram.

[3]
a.

Find the probability that exactly one of the selected balls is green.

[3]
b.
Answer/Explanation

## Markscheme

correct probabilities

A1A1A1     N3

Note:     Award A1 for each correct bold answer.

[3 marks]

a.

multiplying along branches     (M1)

eg$$\,\,\,\,\,$$$$\frac{5}{8} \times \frac{3}{7},{\text{ }}\frac{3}{8} \times \frac{5}{7},{\text{ }}\frac{{15}}{{56}}$$

adding probabilities of correct mutually exclusive paths     (A1)

eg$$\,\,\,\,\,$$$$\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7},{\text{ }}\frac{{15}}{{56}} + \frac{{15}}{{56}}$$

$$\frac{{30}}{{56}}{\text{ }}\left( { = \frac{{15}}{{28}}} \right)$$     A1     N2

[3 marks]

b.

## Question

Pablo drives to work. The probability that he leaves home before 07:00 is $$\frac{3}{4}$$.

If he leaves home before 07:00 the probability he will be late for work is $$\frac{1}{8}$$.

If he leaves home at 07:00 or later the probability he will be late for work is $$\frac{5}{8}$$.

Copy and complete the following tree diagram.

[3]
a.

Find the probability that Pablo leaves home before 07:00 and is late for work.

[2]
b.

Find the probability that Pablo is late for work.

[3]
c.

Given that Pablo is late for work, find the probability that he left home before 07:00.

[3]
d.

Two days next week Pablo will drive to work. Find the probability that he will be late at least once.

[3]
e.
Answer/Explanation

## Markscheme

A1A1A1 N3

Note: Award A1 for each bold fraction.

[3 marks]

a.

multiplying along correct branches      (A1)
eg  $$\frac{3}{4} \times \frac{1}{8}$$

P(leaves before 07:00 ∩ late) = $$\frac{3}{32}$$    A1 N2

[2 marks]

b.

multiplying along other “late” branch      (M1)
eg  $$\frac{1}{4} \times \frac{5}{8}$$

adding probabilities of two mutually exclusive late paths      (A1)
eg  $$\left( {\frac{3}{4} \times \frac{1}{8}} \right) + \left( {\frac{1}{4} \times \frac{5}{8}} \right),\,\,\frac{3}{{32}} + \frac{5}{{32}}$$

$${\text{P}}\left( L \right) = \frac{8}{{32}}\,\,\left( { = \frac{1}{4}} \right)$$    A1 N2

[3 marks]

c.

recognizing conditional probability (seen anywhere)      (M1)
eg  $${\text{P}}\left( {A|B} \right),\,\,{\text{P}}\left( {{\text{before 7}}|{\text{late}}} \right)$$

correct substitution of their values into formula      (A1)
eg $$\frac{{\frac{3}{{32}}}}{{\frac{1}{4}}}$$

$${\text{P}}\left( {{\text{left before 07:00}}|{\text{late}}} \right) = \frac{3}{8}$$    A1 N2

[3 marks]

d.

valid approach      (M1)
eg  1 − P(not late twice), P(late once) + P(late twice)

correct working      (A1)
eg  $$1 – \left( {\frac{3}{4} \times \frac{3}{4}} \right),\,\,2 \times \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4}$$

$$\frac{7}{{16}}$$    A1 N2

[3 marks]

e.

## Question

A four-sided die has three blue faces and one red face. The die is rolled.

Let B be the event a blue face lands down, and R be the event a red face lands down.

Write down

(i)     P(B);
(ii)    P(R).

[2]
a.

If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where p, s, t are probabilities.

Find the value of p, of s and of t.

[2]
b.

Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face lands down, he scores 1 and rolls one more time. Let X be the total score obtained.

(i)     Show that $${\text{P}}(X = 3) = \frac{3}{{16}}$$ .
(ii)    Find $${\text{P}}(X = 2)$$ .

[3]
c.

(i)     Construct a probability distribution table for X.

(ii)    Calculate the expected value of X.

[5]
d.

If the total score is 3, Guiseppi wins $$\ 10$$. If the total score is 2, Guiseppi gets nothing.

Guiseppi plays the game twice. Find the probability that he wins exactly $$\ 10$$.

[4]
e.
Answer/Explanation

## Markscheme

(i) P(B) $$= \frac{3}{4}$$     A1     N1

(ii) P(R) $$= \frac{1}{4}$$     A1     N1

[2 marks]

a.

$$p = \frac{3}{4}$$     A1     N1

$$s = \frac{1}{4}$$, $$t = \frac{3}{4}$$     A1     N1

[2 marks]

b.

(i) $${\text{P}}(X = 3)$$

$$= {\text{P (getting 1 and 2)}} = \frac{1}{4} \times \frac{3}{4}$$     A1

$$= \frac{3}{{16}}$$     AG     N0

(ii) $${\text{P}}(X = 2) = \frac{1}{4} \times \frac{1}{4} + \frac{3}{4}{\text{ }}\left( {{\text{or }}1 – \frac{3}{{16}}} \right)$$     (A1)

$$= \frac{{13}}{{16}}$$      A1     N2

[3 marks]

c.

(i)

A2      N2

(ii) evidence of using $${\text{E}}(X) = \sum{x{\text{P}}(X = x)}$$     (M1)
$${\text{E}}(X) = 2\left( {\frac{{13}}{{16}}} \right) + 3\left( {\frac{3}{{16}}} \right)$$    (A1)
$$= \frac{{35}}{{16}}{\text{ }}\left( { = 2\frac{3}{{16}}} \right)$$     A1     N2

[5 marks]

d.

win $$\ 10 \Rightarrow$$ scores 3 one time, 2 other time     (M1)

$${\text{P}}(3) \times {\text{P}}(2) = \frac{{13}}{{16}} \times \frac{3}{{16}}$$ (seen anywhere)     A1

evidence of recognising there are different ways of winning $$\ 10$$     (M1)
e.g. $${\text{P}}(3) \times {\text{P}}(2) + {\text{P}}(2) \times {\text{P}}(3)$$ , $$2\left( {\frac{{13}}{{16}} \times \frac{3}{{16}}} \right)$$ , $$\frac{{36}}{{256}} + \frac{3}{{256}} + \frac{{36}}{{256}} + \frac{3}{{256}}$$
$${\text{P(win }}\ 10) = \frac{{78}}{{256}}{\text{ }}\left( { = \frac{{39}}{{128}}} \right)$$     A1     N3

[4 marks]

e.

## Question

There are 20 students in a classroom. Each student plays only one sport. The table below gives their sport and gender.

One student is selected at random.

(i)     Calculate the probability that the student is a male or is a tennis player.

(ii)    Given that the student selected is female, calculate the probability that the student does not play football.

[4]
a(i) and (ii).

Two students are selected at random. Calculate the probability that neither student plays football.

[3]
b.
Answer/Explanation

## Markscheme

(i) correct calculation     (A1)

e.g. $$\frac{9}{{20}} + \frac{5}{{20}} – \frac{2}{{20}}$$ , $$\frac{{4 + 2 + 3 + 3}}{{20}}$$

$${\text{P(male or tennis)}} = \frac{{12}}{{20}}$$     A1     N2

(ii) correct calculation     (A1)

e.g. $$\frac{6}{{20}} \div \frac{{11}}{{20}}$$ , $$\frac{{3 + 3}}{{11}}$$

$${\text{P(not football|female)}} = \frac{6}{{11}}$$     A1     N2

[4 marks]

a(i) and (ii).

$${\text{P(first not football)}} = \frac{{11}}{{20}}$$ , $${\text{P(second not football)}} = \frac{{10}}{{19}}$$     A1

$${\text{P(neither football)}} = \frac{{11}}{{20}} \times \frac{{10}}{{19}}$$     A1

$${\text{P(neither football)}} = \frac{{110}}{{380}}$$     A1     N1

[3 marks]

b.

## Question

Let A and B be independent events, where $${\text{P}}(A) = 0.6$$ and $${\text{P}}(B) = x$$ .

Write down an expression for $${\text{P}}(A \cap B)$$ .

[1]
a.

Given that $${\text{P}}(A \cup B) = 0.8$$ ,

(i)     find x ;

(ii)    find $${\text{P}}(A \cap B)$$ .

[4]
b(i) and (ii).

Hence, explain why A and B are not mutually exclusive.

[1]
c.
Answer/Explanation

## Markscheme

$${\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B)( = 0.6x)$$    A1     N1

[1 mark]

a.

(i) evidence of using $${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)$$     (M1)

correct substitution     A1

e.g. $$0.8 = 0.6 + x – 0.6x$$ , $$0.2 = 0.4x$$

$$x = 0.5$$    A1     N2

(ii) $${\text{P}}(A \cap B) = 0.3$$     A1     N1

[4 marks]

b(i) and (ii).

valid reason, with reference to $${\text{P(}}A \cap B)$$     R1     N1

e.g. $${\text{P(}}A \cap B) \ne 0$$

[1 mark]

c.

## Question

Consider the events A and B, where $${\rm{P}}(A) = 0.5$$ , $${\rm{P}}(B) = 0.7$$ and $${\rm{P}}(A \cap B) = 0.3$$ .

The Venn diagram below shows the events A and B, and the probabilities p, q and r.

Write down the value of

(i)     p ;

(ii)    q ;

(iii)   r.

[3]
a(i), (ii) and (iii).

Find the value of $${\rm{P}}(A|B’)$$ .

[2]
b.

Hence, or otherwise, show that the events A and B are not independent.

[1]
c.
Answer/Explanation

## Markscheme

(i) $$p = 0.2$$     A1     N1

(ii) $$q = 0.4$$     A1     N1

(iii) $$r = 0.1$$     A1     N1

[3 marks]

a(i), (ii) and (iii).

$${\rm{P}}(A|B’) = \frac{2}{3}$$     A2     N2

Note: Award A1 for an unfinished answer such as $$\frac{{0.2}}{{0.3}}$$ .

[2 marks]

b.

valid reason     R1

e.g. $$\frac{2}{3} \ne 0.5$$ ,  $$0.35 \ne 0.3$$

thus, A and B are not independent     AG     N0

[1 mark]

c.

## Question

José travels to school on a bus. On any day, the probability that José will miss the bus is $$\frac{1}{3}$$ .

If he misses his bus, the probability that he will be late for school is $$\frac{7}{8}$$ .

If he does not miss his bus, the probability that he will be late is $$\frac{3}{8}$$ .

Let E be the event “he misses his bus” and F the event “he is late for school”.

The information above is shown on the following tree diagram.

Find

(i)     $${\rm{P}}(E \cap F)$$ ;

(ii)    $${\rm{P}}(F)$$ .

[4]
a(i) and (ii).

Find the probability that

(i)     José misses his bus and is not late for school;

(ii)    José missed his bus, given that he is late for school.

[5]
b(i) and (ii).

The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Copy and complete the probability distribution table.

[3]
c.

The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Find the expected cost for José for both days.

[2]
d.
Answer/Explanation

## Markscheme

(i) $$\frac{7}{{24}}$$     A1     N1

(ii) evidence of multiplying along the branches     (M1)

e.g. $$\frac{2}{3} \times \frac{5}{8}$$ , $$\frac{1}{3} \times \frac{7}{8}$$

adding probabilities of two mutually exclusive paths     (M1)

e.g. $$\left( {\frac{1}{3} \times \frac{7}{8}} \right) + \left( {\frac{2}{3} \times \frac{3}{8}} \right)$$ , $$\left( {\frac{1}{3} \times \frac{1}{8}} \right) + \left( {\frac{2}{3} \times \frac{5}{8}} \right)$$

$${\rm{P}}(F) = \frac{{13}}{{24}}$$     A1     N2

[4 marks]

a(i) and (ii).

(i) $$\frac{1}{3} \times \frac{1}{8}$$     (A1)

$$\frac{1}{{24}}$$     A1

(ii) recognizing this is $${\rm{P}}(E|F)$$     (M1)

e.g. $$\frac{7}{{24}} \div \frac{{13}}{{24}}$$

$$\frac{{168}}{{312}}$$ $$\left( { = \frac{7}{{13}}} \right)$$     A2     N3

[5 marks]

b(i) and (ii).

A2A1     N3

[3 marks]

c.

correct substitution into $${\rm{E}}(X)$$ formula     (M1)

e.g. $$0 \times \frac{1}{9} + 3 \times \frac{4}{9} + 6 \times \frac{4}{9}$$ , $$\frac{{12}}{9} + \frac{{24}}{9}$$

$${\rm{E}}(X) = 4$$ (euros)     A1     N2

[2 marks]

d.

## Question

A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a second marble.

Write down the probability that the first marble Anna selects is red.

[1]
a.

Find the probability that Anna selects two red marbles.

[2]
b.

Find the probability that one marble is red and one marble is blue.

[3]
c.
Answer/Explanation

## Markscheme

Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.

$${\rm{P}}(R) = \frac{6}{8}\left( { = \frac{3}{4}} \right)$$     A1     N1

[1 mark]

a.

attempt to find $${\rm{P(Red)}} \times {\rm{P(Red)}}$$     (M1)

e.g. $${\rm{P(}}R{\rm{)}} \times {\rm{P(}}R{\rm{)}}$$ , $$\frac{3}{4} \times \frac{3}{4}$$ , $$\frac{6}{8} \times \frac{5}{7}$$

$${\rm{P}}(2R) = \frac{{36}}{{64}}\left( { = \frac{9}{{16}}} \right)$$     A1     N2

[2 marks]

b.

METHOD 1

attempt to find $${\rm{P(Red)}} \times {\rm{P(Blue)}}$$     (M1)

e.g. $${\rm{P(}}R{\rm{)}} \times {\rm{P(}}B{\rm{)}}$$ , $$\frac{6}{8} \times \frac{2}{8}$$ , $$\frac{6}{8} \times \frac{2}{7}$$

recognizing two ways to get one red, one blue     (M1)

e.g. $${\rm{P}}(RB) + {\rm{P}}(BR)$$ , $$2\left( {\frac{{12}}{{64}}} \right)$$ , $$\frac{6}{8} \times \frac{2}{7} + \frac{2}{8} \times \frac{6}{7}$$

$${\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)$$     A1     N2

[3 marks]

METHOD 2

recognizing that $${\rm{P}}(1R,1B)$$ is $$1 – {\rm{P}}(2B) – {\rm{P}}(2R)$$     (M1)

attempt to find $${\rm{P}}(2R)$$ and $${\rm{P}}(2B)$$     (M1)

e.g. $${\rm{P}}(2R) = \frac{3}{4} \times \frac{3}{4}$$ , $$\frac{6}{8} \times \frac{5}{7}$$ ; $${\rm{P}}(2B) = \frac{1}{4} \times \frac{1}{4}$$ , $$\frac{2}{8} \times \frac{1}{7}$$

$${\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)$$     A1     N2

[3 marks]

c.

## Question

Celeste wishes to hire a taxicab from a company which has a large number of taxicabs.

The taxicabs are randomly assigned by the company.

The probability that a taxicab is yellow is 0.4.

The probability that a taxicab is a Fiat is 0.3.

The probability that a taxicab is yellow or a Fiat is 0.6.

Find the probability that the taxicab hired by Celeste is not a yellow Fiat.

Answer/Explanation

## Markscheme

recognize need for intersection of Y and F     (R1)

eg     $${\text{P}}(Y \cap F){\text{, }}0.3 \times 0.4$$

valid approach to find $${\text{P}}(Y \cap F)$$     (M1)

eg     $${\text{P}}(Y) + {\text{P}}(F) – {\text{P}}(Y \cup F)$$, Venn diagram

correct working (may be seen in Venn diagram)     (A1)

eg     $$0.4 + 0.3 – 0.6$$

$${\text{P}}(Y \cap F) = 0.1$$     A1

recognize need for complement of $$Y \cap F$$     (M1)

eg     $$1 – {\text{P}}(Y \cap F){\text{, }}1 – 0.1$$

$${\text{P}}\left( {(Y \cap F)’} \right) = 0.9$$     A1     N3

[6 marks]

[N/A]

## Question

Adam travels to school by car ($$C$$) or by bicycle ($$B$$). On any particular day he is equally likely to travel by car or by bicycle.

The probability of being late ($$L$$) for school is $$\frac{1}{6}$$ if he travels by car.

The probability of being late for school is $$\frac{1}{3}$$ if he travels by bicycle.

This information is represented by the following tree diagram.

Find the value of $$p$$.

[2]
a.

Find the probability that Adam will travel by car and be late for school.

[2]
b.

Find the probability that Adam will be late for school.

[4]
c.

Given that Adam is late for school, find the probability that he travelled by car.

[3]
d.

Adam will go to school three times next week.

Find the probability that Adam will be late exactly once.

[4]
e.
Answer/Explanation

## Markscheme

correct working     (A1)

eg$$\;\;\;1 – \frac{1}{6}$$

$$p = \frac{5}{6}$$     A1     N2

[2 marks]

a.

multiplying along correct branches     (A1)

eg$$\;\;\;\frac{1}{2} \times \frac{1}{6}$$

$${\text{P}}(C \cap L) = \frac{1}{{12}}$$     A1     N2

[2 marks]

b.

multiplying along the other branch     (M1)

eg$$\;\;\;\frac{1}{2} \times \frac{1}{3}$$

adding probabilities of their $$2$$ mutually exclusive paths     (M1)

eg$$\;\;\;\frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{3}$$

correct working     (A1)

eg$$\;\;\;\frac{1}{{12}} + \frac{1}{6}$$

$${\text{P}}(L) = \frac{3}{{12}}\;\;\;\left( { = \frac{1}{4}} \right)$$     A1     N3

[4 marks]

c.

recognizing conditional probability (seen anywhere)     (M1)

eg$$\;\;\;{\text{P}}(C|L)$$

correct substitution of their values into formula     (A1)

eg$$\frac{{\frac{1}{{12}}}}{{\frac{3}{{12}}}}$$

$${\text{P}}(C|L) = \frac{1}{3}$$     A1     N2

[3 marks]

d.

valid approach     (M1)

eg  $$X \sim B$$ $$\left({3,{\text{ }}\frac{1}{4}} \right),{\text{ }}\left({\frac{1}{4}} \right){\left({\frac{3}{4}} \right)^2},{\text{ }}\left( {\begin{array}{*{20}{c}}3 \\1\end{array}} \right)$$, three ways it could happen

correct substitution     (A1)

eg  $$\;\;\;\left({\begin{array}{*{20}{c}}3\\1\end{array}}\right){\left({\frac{1}{4}}\right)^1}{\left({\frac{3}{4}} \right)^2},{\text{ }}\frac{1}{4}\times\frac{3}{4}\times\frac{3}{4} + \frac{3}{4} \times\frac{1}{4}\times\frac{3}{4} + \frac{3}{4}\times\frac{3}{4}\times\frac{1}{4}$$

correct working     (A1)

eg$$\;\;\;3\left( {\frac{1}{4}} \right)\left( {\frac{9}{{16}}} \right),{\text{ }}\frac{9}{{64}} + \frac{9}{{64}} + \frac{9}{{64}}$$

$$\frac{{27}}{{64}}$$     A1     N2

[4 marks]

Total [15 marks]

e.

## Question

Ann and Bob play a game where they each have an eight-sided die. Ann’s die has three green faces and five red faces; Bob’s die has four green faces and four red faces. They take turns rolling their own die and note what colour faces up. The first player to roll green wins. Ann rolls first. Part of a tree diagram of the game is shown below.

Find the probability that Ann wins on her first roll.

[2]
a.

(i)     The probability that Ann wins on her third roll is $$\frac{5}{8} \times \frac{4}{8} \times p \times q\ \times \frac{3}{8}$$.

Write down the value of $$p$$ and of $$q$$.

(ii)     The probability that Ann wins on her tenth roll is $$\frac{3}{8}{r^k}$$ where $$r \in \mathbb{Q},{\text{ }}k \in \mathbb{Z}$$.

Find the value of $$r$$ and of $$k$$.

[6]
b.

Find the probability that Ann wins the game.

[7]
c.
Answer/Explanation

## Markscheme

recognizing Ann rolls green     (M1)

eg$$\;\;\;{\text{P(G)}}$$

$$\frac{3}{8}$$     A1     N2

[2 marks]

a.

(i)     $$p = \frac{4}{8},{\text{ }}q = \frac{5}{8}$$ or $$q = \frac{4}{8},{\text{ }}p = \frac{5}{8}$$     A1A1     N2

(ii)     recognizes Ann and Bob lose $$9$$ times     (M1)

eg$$\;\;\;\(\overbrace {{A_L}{B_\(\overbrace {{A_L}{B_ \ldots \(\overbrace {{A_L}{B_{\text{ 9 times, }}\underbrace {\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \ldots \times \left( {\frac{5}{8} \times \frac{4}{8}} \right)}_{{\text{9 times}}}$$

$$k = 9\;\;\;$$(seen anywhere)     A1     N2

correct working     (A1)

eg$$\;\;\;{\left( {\frac{5}{8} \times \frac{4}{8}} \right)^9} \times \frac{3}{8},{\text{ }}\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \ldots \times \left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \frac{3}{8}$$

$$r = \frac{{20}}{{64}}\;\;\;\left( { = \frac{5}{{16}}} \right)$$     A1     N2

[6 marks]

b.

recognize the probability is an infinite sum     (M1)

eg$$\;\;\;$$Ann wins on her $${{\text{1}}^{{\text{st}}}}$$ roll or $${{\text{2}}^{{\text{nd}}}}$$ roll or $${{\text{3}}^{{\text{rd}}}}$$ roll…, $${S_\infty }$$

recognizing GP     (M1)

$${u_1} = \frac{3}{8}\;\;\;$$(seen anywhere)     A1

$$r = \frac{{20}}{{64}}\;\;\;$$(seen anywhere)     A1

correct substitution into infinite sum of GP     A1

eg$$\;\;\;\frac{{\frac{3}{8}}}{{1 – \frac{5}{{16}}}},{\text{ }}\frac{3}{8}\left( {\frac{1}{{1 – \left( {\frac{5}{8} \times \frac{4}{8}} \right)}}} \right),{\text{ }}\frac{1}{{1 – \frac{5}{{16}}}}$$

correct working     (A1)

eg$$\;\;\;\frac{{\frac{3}{8}}}{{\frac{{11}}{{16}}}},{\text{ }}\frac{3}{8} \times \frac{{16}}{{11}}$$

$${\text{P (Ann wins)}} = \frac{{48}}{{88}}\;\;\;\left( { = \frac{6}{{11}}} \right)$$     A1     N1

[7 marks]

Total [15 marks]

c.

## Question

In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.

(i)     Find the number of boys who play both sports.

(ii)    Write down the number of boys who play only rugby.

[3]
a.

One boy is selected at random.

(i)     Find the probability that he plays only one sport.

(ii)    Given that the boy selected plays only one sport, find the probability that he plays rugby.

[4]
b.

Let A be the event that a boy plays football and B be the event that a boy plays rugby.

Explain why A and B are not mutually exclusive.

[2]
c.

Show that A and B are not independent.

[3]
d.
Answer/Explanation

## Markscheme

(i) evidence of substituting into $$n(A \cup B) = n(A) + n(B) – n(A \cap B)$$     (M1)

e.g. $$75 + 55 – 100$$ , Venn diagram

30     A1     N2

(ii) 45     A1     N1

[3 marks]

a.

(i) METHOD 1

evidence of using complement, Venn diagram     (M1)

e.g. $$1 – p$$ , $$100 – 30$$

$$\frac{{70}}{{100}}$$ $$\left( { = \frac{7}{{10}}} \right)$$     A1     N2

METHOD 2

attempt to find P(only one sport) , Venn diagram     (M1)

e.g. $$\frac{{25}}{{100}} + \frac{{45}}{{100}}$$

$$\frac{{70}}{{100}}$$ $$\left( { = \frac{7}{{10}}} \right)$$     A1     N2

(ii) $$\frac{{45}}{{70}}$$ $$\left( { = \frac{9}{{14}}} \right)$$     A2     N2

[4 marks]

b.

valid reason in words or symbols     (R1)

e.g. $${\rm{P}}(A \cap B) = 0$$ if mutually exclusive, $${\rm{P}}(A \cap B) \ne 0$$ if not mutually exclusive

correct statement in words or symbols     A1     N2

e.g. $${\rm{P}}(A \cap B) = 0.3$$ , $${\rm{P}}(A \cup B) \ne {\rm{P}}(A) + {\rm{P}}(B)$$ , $${\rm{P}}(A) + {\rm{P}}(B) > 1$$ , some students play both sports, sets intersect

[2 marks]

c.

valid reason for independence     (R1)

e.g. $${\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)$$ , $${\rm{P}}(B|A) = {\rm{P}}(B)$$

correct substitution     A1A1     N3

e.g. $$\frac{{30}}{{100}} \ne \frac{{75}}{{100}} \times \frac{{55}}{{100}}$$ , $$\frac{{30}}{{55}} \ne \frac{{75}}{{100}}$$

[3 marks]

d.

## Question

The letters of the word PROBABILITY are written on 11 cards as shown below.

Two cards are drawn at random without replacement.

Let A be the event the first card drawn is the letter A.

Let B be the event the second card drawn is the letter B.

Find $${\rm{P}}(A)$$ .

[1]
a.

Find $${\rm{P}}(B|A)$$ .

[2]
b.

Find $${\rm{P}}(A \cap B)$$ .

[3]
c.
Answer/Explanation

## Markscheme

$${\rm{P}}(A) = \frac{1}{{11}}$$     A1    N1

[1 mark]

a.

$${\rm{P}}(B|A) = \frac{2}{{10}}$$     A2     N2

[2 marks]

b.

recognising that $${\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B|A)$$     (M1)

correct values     (A1)

e.g. $${\rm{P}}(A \cap B) = \frac{1}{{11}} \times \frac{2}{{10}}$$

$${\rm{P}}(A \cap B) = \frac{2}{{110}}$$     A1     N3

[3 marks]

c.

## Question

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

(i)     Copy and complete the following tree diagram.

(ii)    Find the probability that two white balls are chosen.

[5]
a(i) and (ii).

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$ .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Find the probability that the two balls are white.

[5]
b.

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\frac{2}{7}$$ .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Given that both balls are white, find the probability that they were chosen from bag A.

[4]
c.
Answer/Explanation

## Markscheme

(i)

$$\frac{4}{6},\frac{3}{6}{\rm{and}}\frac{3}{6}\left( {\frac{2}{3},\frac{1}{2}{\rm{and}}\frac{1}{2}} \right)$$    A1A1A1     N3

(ii) multiplying along the correct branches (may be seen on diagram)     (A1)

e.g. $$\frac{3}{7} \times \frac{2}{6}$$

$$\frac{6}{{42}}\left( { = \frac{1}{7}} \right)$$     A1     N2

[5 marks]

a(i) and (ii).

$${\rm{P(bag A) = }}\frac{2}{6}\left( { = \frac{1}{3}} \right)$$ , $${\rm{P(bag B) = }}\frac{4}{6}\left( { = \frac{2}{3}} \right)$$ (seen anywhere)     (A1)(A1)

appropriate approach    (M1)

e.g. $${\rm{P(}}WW \cap A) + {\rm{P}}(WW \cap B)$$

correct calculation     A1

e.g. $$\frac{1}{3} \times \frac{1}{7} + \frac{2}{3} \times \frac{2}{7}$$ , $$\frac{2}{{42}} + \frac{8}{{42}}$$

$${\rm{P}}(2W) = \frac{{60}}{{252}}\left( { = \frac{5}{{21}}} \right)$$    A1     N3

[5 marks]

b.

recognizing conditional probability     (M1)

e.g. $$\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}$$ , $${\rm{P}}(A|WW) = \frac{{{\rm{P}}(WW \cap A)}}{{{\rm{P}}(WW)}}$$

correct numerator     (A1)

e.g. $${\rm{P}}(A \cap WW) = \frac{6}{{42}} \times \frac{2}{6},\frac{1}{{21}}$$

correct denominator     (A1)

e.g. $$\frac{6}{{252}},\frac{5}{{21}}$$

probability $$\frac{{84}}{{420}}\left( { = \frac{1}{5}} \right)$$    A1     N3

[4 marks]

c.

## Question

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, without replacement.

Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is $$1$$ or $$2$$, a marble is drawn from jar A. Otherwise, a marble is drawn from jar B.

Find the probability that

(i)     none of the marbles are green;

(ii)     exactly one marble is green.

[5]
a.

Find the expected number of green marbles drawn from the jar.

[3]
b.

(i)     Write down the probability that the marble is drawn from jar B.

(ii)     Given that the marble was drawn from jar B, write down the probability that it is red.

[2]
c.

Given that the marble is red, find the probability that it was drawn from jar A.

[6]
d.
Answer/Explanation

## Markscheme

(i)     attempt to find $${\rm{P(red)}} \times {\rm{P(red)}}$$     (M1)

eg   $$\frac{3}{8} \times \frac{2}{7}$$ , $$\frac{3}{8} \times \frac{3}{8}$$ , $$\frac{3}{8} \times \frac{2}{8}$$

$${\text{P(none green)}} = \frac{6}{{56}}$$ $$\left( { = \frac{3}{{28}}} \right)$$     A1     N2

(ii)     attempt to find $${\rm{P(red)}} \times {\rm{P(green)}}$$     (M1)

eg   $$\frac{5}{8} \times \frac{3}{7}$$ , $$\frac{3}{8} \times \frac{5}{8}$$ , $$\frac{{15}}{{56}}$$

recognizing two ways to get one red, one green     (M1)

eg   $$2{\rm{P}}(R) \times {\rm{P}}(G)$$ , $$\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}$$ , $$\frac{3}{8} \times \frac{5}{8} \times 2$$

$${\text{P(exactly one green)}} = \frac{{30}}{{56}}$$ $$\left( { = \frac{{15}}{{28}}} \right)$$     A1     N2

[5 marks]

a.

$${\text{P(both green)}} = \frac{{20}}{{56}}$$ (seen anywhere)     (A1)

correct substitution into formula for $${\rm{E}}(X)$$     A1

eg   $$0 \times \frac{6}{{56}} + 1 \times \frac{{30}}{{56}} + 2 \times \frac{{20}}{{56}}$$ , $$\frac{{30}}{{64}} + \frac{{50}}{{64}}$$

expected number of green marbles is $$\frac{{70}}{{56}}$$ $$\left( { = \frac{5}{4}} \right)$$     A1     N2

[3 marks]

b.

(i)     $${\text{P(jar B)}} = \frac{4}{6}$$ $$\left( { = \frac{2}{3}} \right)$$     A1     N1

(ii)     $${\text{P(red| jar B)}} = \frac{6}{8}$$ $$\left( { = \frac{3}{4}} \right)$$     A1     N1

[2 marks]

c.

recognizing conditional probability     (M1)

eg   $${\rm{P}}(A|R)$$ , $$\frac{{{\text{P(jar A and red)}}}}{{{\rm{P(red)}}}}$$ , tree diagram

attempt to multiply along either branch (may be seen on diagram)     (M1)

eg   $${\text{P(jar A and red)}} = \frac{1}{3} \times \frac{3}{8}$$ $$\left( { = \frac{1}{8}} \right)$$

attempt to multiply along other branch     (M1)

eg   $${\text{P(jar B and red)}} = \frac{2}{3} \times \frac{6}{8}$$ $$\left( { = \frac{1}{2}} \right)$$

adding the probabilities of two mutually exclusive paths     (A1)

eg   $${\rm{P(red)}} = \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}$$

correct substitution

eg   $${\text{P(jar A|red)}} = \frac{{\frac{1}{3} \times \frac{3}{8}}}{{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}}}$$ , $$\frac{{\frac{1}{8}}}{{\frac{5}{8}}}$$     A1

$${\text{P(jar A|red)}} = \frac{1}{5}$$     A1     N3

[6 marks]

d.

## Question

Two standard six-sided dice are tossed. A diagram representing the sample space is shown below.

Let $$X$$ be the sum of the scores on the two dice.

(i)     Find $${\rm{P}}(X = 6)$$ .

(ii)    Find $${\rm{P}}(X > 6)$$ .

(iii)   Find $${\rm{P}}(X = 7|X > 6)$$ .

[6]
a(i), (ii) and (iii).

Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which the game is fair.

[8]
b.
Answer/Explanation

## Markscheme

(i) number of ways of getting $$X = 6$$ is 5     A1

$${\rm{P}}(X = 6) = \frac{5}{{36}}$$     A1     N2

(ii) number of ways of getting $$X > 6$$ is 21     A1

$${\rm{P}}(X > 6) = \frac{{21}}{{36}}\left( { = \frac{7}{{12}}} \right)$$     A1     N2

(iii) $${\rm{P}}(X = 7|X > 6) = \frac{6}{{21}}\left( { = \frac{2}{7}} \right)$$     A2     N2

[6 marks]

a(i), (ii) and (iii).

attempt to find $${\rm{P}}(X < 6)$$     M1

e.g. $$1 – \frac{5}{{36}} – \frac{{21}}{{36}}$$

$${\rm{P}}(X < 6) = \frac{{10}}{{36}}$$     A1

fair game if $${\rm{E}}(W) = 0$$ (may be seen anywhere)     R1

attempt to substitute into $${\rm{E}}(X)$$ formula     M1

e.g. $$3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) – k\left( {\frac{{10}}{{36}}} \right)$$

correct substitution into $${\rm{E}}(W) = 0$$     A1

e.g. $$3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) – k\left( {\frac{{10}}{{36}}} \right) = 0$$

work towards solving     M1

e.g. $$15 + 21 – 10k = 0$$

$$36 = 10k$$     A1

$$k = \frac{{36}}{{10}}( = 3.6)$$     A1     N4

[8 marks]

b.