IB Math Analysis & Approaches Question bank-Topic: SL 4.7 Concept of discrete random variables SL Paper 1

Question

A biased four-sided die, A, is rolled. Let X be the score obtained when die A is rolled. The
probability distribution for X is given in the following table.

(a) Find the value of p . [2]
(b) Hence, find the value of E (X ) . [2]
A second biased four-sided die, B, is rolled. Let Y be the score obtained when die B is rolled.
The probability distribution for Y is given in the following table.

(c)      (i)      State the range of possible values of r .

(ii)      Hence, find the range of possible values of q .      [3]

(d) Hence, find the range of possible values for E (Y ) .     [3]

Agnes and Barbara play a game using these dice. Agnes rolls die A once and Barbara rolls die B once. The probability that Agnes’ score is less than Barbara’s score is \(\frac{1}{2}\)

(e) Find the value of E (Y ) .       [6]

Answer/Explanation

Ans

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

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

The probability distribution of a discrete random variable X is given by \[{\rm{P}}(X = x) = \frac{{{x^2}}}{{14}}{\text{, }}x \in \left\{ {1{\text{, }}2{\text{, }}k} \right\}{\text{, where}} k > 0\] .

Write down \({\rm{P}}(X = 2)\) .

[1]
a.

Show that \(k = 3\) .

[4]
b.

Find \({\rm{E}}(X)\) .

[2]
c.
Answer/Explanation

Markscheme

\({\rm{P}}(X = 2) = \frac{4}{{14}}\) \(\left( { = \frac{2}{7}} \right)\)     A1     N1

[1 mark]

a.

\({\rm{P}}(X = 1) = \frac{1}{{14}}\)     (A1)

\({\rm{P}}(X = k) = \frac{{{k^2}}}{{14}}\)     (A1)

setting the sum of probabilities \( = 1\)     M1

e.g. \(\frac{1}{{14}} + \frac{4}{{14}} + \frac{{{k^2}}}{{14}} = 1\) , \(5 + {k^2} = 14\)

\({k^2} = 9\) (accept \(\frac{{{k^2}}}{{14}} = \frac{9}{{14}}\) )     A1

\(k = 3\)     AG     N0

[4 marks]

b.

correct substitution into \({\rm{E}}(X) = \sum {x{\rm{P}}(X = x)} \)     A1

e.g. \(1\left( {\frac{1}{{14}}} \right) + 2\left( {\frac{4}{{14}}} \right) + 3\left( {\frac{9}{{14}}} \right)\)

\({\rm{E}}(X) = \frac{{36}}{{14}}\) \(\left( { = \frac{{18}}{7}} \right)\)    A1     N1

[2 marks]

c.

Question

The random variable X has the following probability distribution.

Given that \({\rm{E}}(X) = 1.7\) , find q .

Answer/Explanation

Markscheme

correct substitution into \({\rm{E}}(X) = \sum {px} \) (seen anywhere)     A1

e.g. \(1s + 2 \times 0.3 + 3q = 1.7\) , \(s + 3q = 1.1\)

recognizing \(\sum {p = 1} \) (seen anywhere)     (M1)

correct substitution into \(\sum {p = 1} \)     A1

e.g. \(s + 0.3 + q = 1\)

attempt to solve simultaneous equations     (M1)

correct working     (A1)

e.g. \(0.3 + 2q = 0.7\) , \(2s = 1\)

\(q = 0.2\)     A1     N4

[6 marks]

Question

The random variable X has the following probability distribution, with \({\rm{P}}(X > 1) = 0.5\) .


 

Find the value of r .

[2]
a.

Given that \({\rm{E}}(X) = 1.4\) , find the value of p and of q .

[6]
b.
Answer/Explanation

Markscheme

attempt to substitute \({\rm{P}}(X > 1) = 0.5\)     (M1)

e.g. \(r + 0.2 = 0.5\)

\(r = 0.3\)     A1     N2

[2 marks]

a.

correct substitution into \({\rm{E}}(X)\) (seen anywhere)     (A1)

e.g. \(0 \times p + 1 \times q + 2 \times r + 3 \times 0.2\)

correct equation     A1

e.g. \(q + 2 \times 0.3 + 3 \times 0.2 = 1.4\) , \(q + 1.2 = 1.4\)

\(q = 0.2\)     A1      N1

evidence of choosing \(\sum {{p_i} = 1} \)     M1

e.g. \(p + 0.2 + 0.3 + 0.2 = 1\) , \(p + q = 0.5\)

correct working     (A1)

\(p + 0.7 = 1\) , \(1 – 0.2 – 0.3 – 0.2\) , \(p + 0.2 = 0.5\)

\(p = 0.3\)     A1 N2

Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award the final A mark for an inappropriate value of p.

[6 marks]

b.

Question

Let \(\boldsymbol{A} = \left( {\begin{array}{*{20}{c}}
  0&3 \\
  { – 2}&4
\end{array}} \right)\) and  \(\boldsymbol{B} = \left( {\begin{array}{*{20}{c}}
  { – 4}&0 \\
  5&1
\end{array}} \right)\).

Find AB .

[3]
a.

Given that \({\boldsymbol{X}} – 2{\boldsymbol{A}} = {\boldsymbol{B}}\), find X.

[3]
b.
Answer/Explanation

Markscheme

evidence of multiplying     (M1)

e.g. one correct element, \((0 \times – 4) + (3 \times 5)\)

\({\boldsymbol{AB}} = \left( {\begin{array}{*{20}{c}}
{15}&3\\
{28}&4
\end{array}} \right)\)     A2     N3

Note: Award A1 for three correct elements.

[3 marks]

a.

finding \(2{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}}
0&6\\
{ – 4}&8
\end{array}} \right)\)     (A1)

adding 2\({\boldsymbol{A}}\) to both sides (may be seen first)     (M1)

e.g. \({\boldsymbol{X}} = {\boldsymbol{B}} +2{\boldsymbol{A}}\)

 \({\boldsymbol{X}} = \left( {\begin{array}{*{20}{c}}
{ – 4}&6\\
1&9
\end{array}} \right)\)     A1     N2

[3 marks]

b.

Question

The following table shows the probability distribution of a discrete random variable X .

Find the value of k .

[3]
a.

Find \({\text{E}}(X)\) .

[3]
b.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

e.g. \(\sum\limits_{}^{} {p = 1{\text{, }}0.3 + k + 2k + 0.1 = 1} \)

correct working     (A1)

e.g. \(0.4 + 3k{\text{, }}3k = 0.6\)

\(k = 0.2\)    A1     N2

[3 marks]

a.

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

e.g. \(0(0.3) + 2(k) + 5(2k) + 9(0.1){\text{, }}12k + 0.9\)

correct working 

e.g. \(0(0.3) + 2(0.2) + 5(0.4) + 9(0.1){\text{, }}0.4 + 2.0 + 0.9\)     (A1)

\({\text{E}}(X)\)= 3.3     A1     N2

[3 marks]

b.

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

A discrete random variable \(X\) has the following probability distribution.

Find \(p\).

[3]
a.

Find \({\text{E}}(X)\).

[3]
b.
Answer/Explanation

Markscheme

summing probabilities to 1     (M1)

eg,\(\;\;\;\sum { = 1,{\text{ }}3 + 4 + 2 + x = 10} \)

correct working     (A1)

\(\frac{3}{{10}} + \frac{4}{{10}} + \frac{2}{{10}} + p = 1,{\text{ }}p = 1 – \frac{9}{{10}}\)

\(p = \frac{1}{{10}}\)     A1     N3

[3 marks]

a.

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

eg\(\;\;\;0\left( {\frac{3}{{10}}} \right) +  \ldots  + 3(p)\)

correct working     (A1)

eg\(\;\;\;\frac{4}{{10}} + \frac{4}{{10}} + \frac{3}{{10}}\)

\({\text{E}}(X) = \frac{{11}}{{10}}\;\;\;(1.1)\)     A1     N2

[3 marks]

Total [6 marks]

b.

Question

The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).

M17/5/MATME/SP1/ENG/TZ1/10

Show that \(\cos \theta  = \frac{3}{4}\).

[6]
a.

Given that \(\tan \theta  > 0\), find \(\tan \theta \).

[3]
b.

Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[6]
c.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(\sum {p = 1} \)

correct equation     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2\cos 2\theta  = 1\)

correct equation in \(\cos \theta \)     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2(2{\cos ^2}\theta  – 1) = 1,{\text{ }}4{\cos ^2}\theta  + \cos \theta  – 3 = 0\)

evidence of valid approach to solve quadratic     (M1)

eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}\)

correct working, clearly leading to required answer     A1

eg\(\,\,\,\,\,\)\((4\cos \theta  – 3)(\cos \theta  + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}\)

correct reason for rejecting \(\cos \theta  \ne  – 1\)     R1

eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta  > 0\)

Note:     Award R0 for \(\cos \theta  \ne  – 1\) without a reason.

\(\cos \theta  = \frac{3}{4}\)    AG  N0

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     

(A1)

eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)

\(\tan \theta  = \frac{{\sqrt 7 }}{3}\)     A1     N2

[3 marks]

b.

attempt to substitute either limits or the function into formula involving \({f^2}\)     (M1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)

correct substitution of both limits and function     (A1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\tan x\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} – \tan \theta \)

Note:     Award M0 if they substitute into original or differentiated function.

\(\tan \frac{\pi }{4} = 1\)    (A1)

eg\(\,\,\,\,\,\)\(1 – \tan \theta \)

\(V = \pi  – \frac{{\pi \sqrt 7 }}{3}\)     A1     N3

[6 marks]

c.

Leave a Reply