IB Math Analysis & Approaches Question bank-Topic: SL 4.7 Applications SL Paper 1

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

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

The following diagram shows a board which is divided into three regions \(A\), \(B\) and \(C\).

A game consists of a contestant throwing one dart at the board. The probability of hitting each region is given in the following table.

Find the probability that the dart does not hit the board.

[3]
a.

The contestant scores points as shown in the following table.

Given that the game is fair, find the value of \(q\).

[4]
b.
Answer/Explanation

Markscheme

evidence of summing probabilities to \(1\)     (M1)

eg\(\;\;\;\frac{5}{{20}} + \frac{4}{{20}} + \frac{1}{{20}} + p = 1,\;\;\;\sum { = 1} \)

correct working     (A1)

eg\(\;\;\;p = 1 – \frac{{10}}{{20}}\)

\(p = \frac{{10}}{{20}}\;\;\;\left( { = \frac{1}{2}} \right)\)     A1     N2

[3 marks]

a.

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

eg\(\;\;\;\frac{4}{{20}}(q) + \frac{1}{{20}}(10) + \frac{{10}}{{20}}( – 3)\)

valid reasoning for fair game (seen anywhere, including equation)     (M1)

eg\(\;\;\;{\text{E}}(X) = 0\), points lost\( = \)points gained

correct working     (A1)

eg\(\;\;\;4q + 10 – 30 = 0,\;\;\;\frac{4}{{20}}q + \frac{{10}}{{20}} = \frac{{30}}{{20}}\)

\(q = 5\)     A1     N2

[4 marks]

Total [7 marks]

b.

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.

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