IB DP Maths Topic 5.7 Concept of discrete random variables and their probability distributions SL Paper 2

 

https://play.google.com/store/apps/details?id=com.iitianDownload IITian Academy  App for accessing Online Mock Tests and More..

Question

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


Find the value of k.

[4]
a.

Find the expected value of X.

[3]
b.
Answer/Explanation

Markscheme

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

correct substitution     A1

e.g. \(10{k^2} + 3k + 0.6 = 1\) , \(10{k^2} + 3k – 0.4 = 0\)

\(k = 0.1\)     A2     N2

[4 marks]

a.

evidence of using \({\rm{E}}(X) = \sum {{p_i}{x_i}} \)     (M1)

correct substitution     (A1)

e.g. \( – 1 \times 0.2 + 2 \times 0.4 + 3 \times 0.3\)

\({\rm{E}}(X) = 1.5\)     A1     N2

[3 marks]

b.

Question

A test has five questions. To pass the test, at least three of the questions must be answered correctly.

The probability that Mark answers a question correctly is \(\frac{1}{5}\) . Let X be the number of questions that Mark answers correctly.

Bill also takes the test. Let Y be the number of questions that Bill answers correctly.

The following table is the probability distribution for Y .


(i)     Find E(X ) .

(ii)    Find the probability that Mark passes the test.

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

(i)     Show that \(4a + 2b = 0.24\) .

(ii)    Given that \({\rm{E}}(Y) = 1\) , find a and b .

[8]
b(i) and (ii).

Find which student is more likely to pass the test.

[3]
c.
Answer/Explanation

Markscheme

(i) valid approach     (M1)

e.g. \(np\) , \(5 \times \frac{1}{5}\)

\({\rm{E}}(X) = 1\)     A1     N2

(ii) evidence of appropriate approach involving binomial     (M1)

e.g. \(X \sim B\left( {5,\frac{1}{5}} \right)\)

recognizing that Mark needs to answer 3 or more questions correctly     (A1)

e.g. \({\rm{P}}(X \ge 3)\)

valid approach     M1

e.g. \(1 – {\rm{P}}(X \le 2)\) , \({\rm{P}}(X = 3) + {\rm{P}}(X = 4) + {\rm{P}}(X = 5)\)

\({\text{P(pass)}} = 0.0579\)     A1     N3

[6 marks]

a(i) and (ii).

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

e.g. \(0.67 + 0.05 + (a + 2b) + \ldots + 0.04 = 1\)

some simplification that clearly leads to required answer

e.g. \(0.76 + 4a + 2b = 1\)     A1

\(4a + 2b = 0.24\)     AG      N0

(ii) correct substitution into the formula for expected value     (A1)

e.g. \(0(0.67) + 1(0.05) + \ldots + 5(0.04)\)

some simplification     (A1)

e.g. \(0.05 + 2a + 4b + \ldots + 5(0.04) = 1\)

correct equation     A1

e.g. \(13a + 5b = 0.75\)

evidence of solving     (M1)

\(a = 0.05\) , \(b = 0.02\)     A1A1     N4

[8 marks]

b(i) and (ii).

attempt to find probability Bill passes     (M1)

e.g. \({\rm{P}}(Y \ge 3)\)

correct value 0.19     A1

Bill (is more likely to pass)     A1     N0

[3 marks]

c.

Question

Two fair 4-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1, 2, 3, 4. The score for each die is the number which lands face down.

List the pairs of scores that give a sum of 6.

[3]
a.

The probability distribution for the sum of the scores on the two dice is shown below.


Find the value of p , of q , and of r .

[3]
b.

Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5 on three or more throws.

Find the probability that Fred wins a prize.

[6]
c.
Answer/Explanation

Markscheme

three correct pairs     A1A1A1     N3

e.g. (2, 4), (3, 3), (4, 2) , R2G4, R3G3, R4G2

[3 marks]

a.

\(p = \frac{1}{{16}}\) , \(q = \frac{2}{{16}}\) , \(r = \frac{2}{{16}}\)     A1A1A1     N3

[3 marks]

b.

let X be the number of times the sum of the dice is 5

evidence of valid approach     (M1)

e.g. \(X \sim {\rm{B}}(n{\text{, }}p)\) , tree diagram, 5 sets of outcomes produce a win

one correct parameter     (A1)

e.g. \(n = 4\) , \(p = 0.25\) , \(q = 0.75\)

Fred wins prize is \({\rm{P}}(X \ge 3)\)     (A1)

appropriate approach to find probability     M1

e.g. complement, summing probabilities, using a CDF function

correct substitution     (A1)

e.g. \(1 – 0.949 \ldots \) , \(1 – \frac{{243}}{{256}}\) , \(0.046875 + 0.00390625\) , \(\frac{{12}}{{256}} + \frac{1}{{256}}\)

\({\text{probability of winning}} = 0.0508\) \(\left( {\frac{{13}}{{256}}} \right)\)     A1     N3

[6 marks]

c.

Question

A bag contains four gold balls and six silver balls.

Two balls are drawn at random from the bag, with replacement. Let \(X\) be the number of gold balls drawn from the bag.

Fourteen balls are drawn from the bag, with replacement.

(i)     Find \({\rm{P}}(X = 0)\) .

(ii)     Find \({\rm{P}}(X = 1)\) .

(iii)    Hence, find \({\rm{E}}(X)\) .

[8]
a.

Hence, find \({\rm{E}}(X)\) .

[3]
a.iii.

Find the probability that exactly five of the balls are gold.

[2]
b.

Find the probability that at most five of the balls are gold.

[2]
c.

Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correct to two decimal places.

[3]
d.
Answer/Explanation

Markscheme

METHOD 1

(i)     appropriate approach     (M1)

eg   \(\frac{6}{{10}} \times \frac{6}{{10}}\) , \(\frac{6}{{10}} \times \frac{5}{9}\) , \(\frac{6}{{10}} \times \frac{5}{{10}}\)

\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\)     A1     N2

(ii)     multiplying one pair of gold and silver probabilities     (M1)

eg   \(\frac{6}{{10}} \times \frac{4}{{10}}\) , \(\frac{6}{{10}} \times \frac{4}{9}\) , 0.24

adding the product of both pairs of gold and silver probabilities     (M1)

eg   \(\frac{6}{{10}} \times \frac{4}{{10}} \times 2\) , \(\frac{6}{{10}} \times \frac{4}{9} + \frac{4}{{10}} \times \frac{6}{9}\)

\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\)     A1     N3

(iii)

\({\rm{P}}(X = 2) = 0.16\) (seen anywhere)     (A1)

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

eg   \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1     N3

METHOD 2

(i)     evidence of recognizing binomial (may be seen in part (ii))     (M1)

eg   \(X \sim {\rm{B}}(2,0.6)\) , \(\left( \begin{array}{l}
2\\
0
\end{array} \right){(0.4)^2}{(0.6)^0}\)

correct probability for use in binomial     (A1)

eg   \(p = 0.4\) , \(X \sim {\rm{B}}(2,0.4)\) , \(^2{C_0}{(0.4)^0}{(0.6)^2}\)

\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\)     A1 N3

(ii)     correct set up     (A1)

eg     \(_2{C_1}{(0.4)^1}{(0.6)^1}\)

\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\)     A1     N2 

(iii)

attempt to substitute into \(np\)     (M1)

eg   \(2 \times 0.6\)

correct substitution into \(np\)     (A1)

eg   \(2 \times 0.4\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1 N3

[8 marks]

a.

METHOD 1

\({\rm{P}}(X = 2) = 0.16\) (seen anywhere)     (A1)

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

eg   \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1     N3

METHOD 2

attempt to substitute into \(np\)     (M1)

eg   \(2 \times 0.6\)

correct substitution into \(np\)     (A1)

eg   \(2 \times 0.4\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1 N3

[3 marks]

a.iii.

Let \(Y\) be the number of gold balls drawn from the bag.  

evidence of recognizing binomial (seen anywhere)     (M1)

eg   \(_{14}{C_5}{(0.4)^5}{(0.6)^9}\) , \({\rm{B}}(14,0.4)\)

\({\rm{P}}(Y = 5) = 0.207\)     A1     N2

[2 marks]

b.

recognize need to find \({\rm{P}}(Y \le 5)\)     (M1)

\({\rm{P}}(Y \le 5) = 0.486\)     A1     N2

[2 marks]

c.

Let \(Y\) be the number of gold balls drawn from the bag.

recognizing conditional probability     (M1)

eg   \({\rm{P}}(A|B)\) , \({\rm{P}}(Y = 5|Y \le 5)\) , \(\frac{{{\rm{P}}(Y = 5)}}{{{\rm{P}}(Y \le 5)}}\) , \(\frac{{0.207}}{{0.486}}\)

\({\rm{P}}(Y = 5|Y \le 5) = 0.42522518\)     (A1)

\({\rm{P}}(Y = 5|Y \le 5) = 0.43\) (to \(2\) dp)     A1     N2

[3 marks]

d.

Question

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

Find the value of \(k\).

[3]
a.

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

[2]
b.
Answer/Explanation

Markscheme

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

correct substitution     A1

eg\(\;\;\;0.15 + k + 0.1 + 2k = 1,{\text{ }}3k + 0.25 = 1\)

\(k = 0.25\)     A1     N2

[3 marks]

a.

correct substitution     (A1)

eg\(\;\;\;0 \times 0.15 + 1 \times 0.25 + 2 \times 0.1 + 3 \times 0.5\)

\({\text{E}}(X) = 1.95\)     A1     N2

[2 marks]

Total [5 marks]

b.

Question

A factory has two machines, A and B. The number of breakdowns of each machine is independent from day to day.

Let \(A\) be the number of breakdowns of Machine A on any given day. The probability distribution for \(A\) can be modelled by the following table.

M16/5/MATME/SP2/ENG/TZ1/08

Let \(B\) be the number of breakdowns of Machine B on any given day. The probability distribution for \(B\) can be modelled by the following table.

M16/5/MATME/SP2/ENG/TZ1/08.c+d

On Tuesday, the factory uses both Machine A and Machine B. The variables \(A\) and \(B\) are independent.

Find \(k\).

[2]
a.

(i)     A day is chosen at random. Write down the probability that Machine A has no breakdowns.

(ii)     Five days are chosen at random. Find the probability that Machine A has no breakdowns on exactly four of these days.

[3]
b.

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

[2]
c.

(i)     Find the probability that there are exactly two breakdowns on Tuesday.

(ii)     Given that there are exactly two breakdowns on Tuesday, find the probability that both breakdowns are of Machine A.

[8]
d.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(0.55 + 0.3 + 0.1 + k = 1\)

\(k = 0.05{\text{ (exact)}}\)    A1     N2

[2 marks]

a.

(i)     0.55     A1     N1

(ii)     recognizing binomial probability     (M1)

eg\(\,\,\,\,\,\)\(X:{\text{ }}B(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right),{\text{ }}{(0.55)^4}(1 – 0.55),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n – r}}\)

\(P(X = 4) = 0.205889\)

\(P(X = 4) = 0.206\)    A1     N2

[3 marks]

b.

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

eg\(\,\,\,\,\,\)\(0.2 + (2 \times 0.08) + (3 \times 0.02)\)

\({\text{E}}(B) = 0.42{\text{ (exact)}}\)    A1     N2

[2 marks]

c.

(i)     valid attempt to find one possible way of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)\(2A,{\text{ }}2B,{\text{ }}1A\) and \(1B\), tree diagram

one correct calculation for 1 way (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(0.1 \times 0.7,{\text{ }}0.55 \times 0.08,{\text{ }}0.3 \times 0.2\)

recognizing there are 3 ways of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)A twice or B twice or one breakdown each

correct working     (A1)

eg\(\,\,\,\,\,\)\((0.1 \times 0.7) + (0.55 \times 0.08) + (0.3 \times 0.2)\)

\({\text{P(2 breakdowns)}} = 0.174{\text{ (exact)}}\)     A1     N3

(ii)     recognizing conditional probability     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A|B),{\text{ P}}(2A|{\text{2breakdowns}})\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{{0.1 \times 0.7}}{{0.174}}\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402298\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402\)     A1     N2

[8 marks]

d.

Question

A jar contains 5 red discs, 10 blue discs and \(m\) green discs. A disc is selected at random and replaced. This process is performed four times.

Write down the probability that the first disc selected is red.

[1]
a.

Let \(X\) be the number of red discs selected. Find the smallest value of \(m\) for which \({\text{Var}}(X{\text{ }}) < 0.6\).

[5]
b.
Answer/Explanation

Markscheme

\({\text{P(red)}} = \frac{5}{{15 + m}}\)     A1     N1

[1 mark]

a.

recognizing binomial distribution     (M1)

eg\(\,\,\,\,\,\)\(X \sim B(n,{\text{ }}p)\)

correct value for the complement of their \(p\) (seen anywhere)     A1

eg\(\,\,\,\,\,\)\(1 – \frac{5}{{15 + m}},{\text{ }}\frac{{10 + m}}{{15 + m}}\)

correct substitution into \({\text{Var}}(X) = np(1 – p)\)     (A1)

eg\(\,\,\,\,\,\)\(4\left( {\frac{5}{{15 + m}}} \right)\left( {\frac{{10 + m}}{{15 + m}}} \right),{\text{ }}\frac{{20(10 + m)}}{{{{(15 + m)}^2}}} < 0.6\)

\(m > 12.2075\)     (A1)

\(m = 13\)     A1     N3

[5 marks]

b.

Question

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

N17/5/MATME/SP2/ENG/TZ0/04

Find the value of \(k\).

[4]
a.

Write down \({\text{P}}(X = 2)\).

[1]
b.

Find \({\text{P}}(X = 2|X > 0)\).

[3]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)total probability = 1

correct equation     (A1)

eg\(\,\,\,\,\,\)\(0.475 + 2{k^2} + \frac{k}{{10}} + 6{k^2} = 1,{\text{ }}8{k^2} + 0.1k – 0.525 = 0\)

\(k = 0.25\)     A2     N3

[4 marks]

a.

\({\text{P}}(X = 2) = 0.025\)     A1     N1

[1 mark]

b.

valid approach for finding \({\text{P}}(X > 0)\)     (M1)

eg\(\,\,\,\,\,\)\(1 – 0.475,{\text{ }}2({0.25^2}) + 0.025 + 6({0.25^2}),{\text{ }}1 – {\text{P}}(X = 0),{\text{ }}2{k^2} + \frac{k}{{10}} + 6{k^2}\)

correct substitution into formula for conditional probability     (A1)

eg\(\,\,\,\,\,\)\(\frac{{0.025}}{{1 – 0.475}},{\text{ }}\frac{{0.025}}{{0.525}}\)

0.0476190

\({\text{P}}(X = 2|X > 0) = \frac{1}{{21}}\) (exact), 0.0476     A1     N2

[3 marks]

c.

Question

A biased four-sided die is rolled. The following table gives the probability of each score.

Find the value of k.

[2]
a.

Calculate the expected value of the score.

[2]
b.

The die is rolled 80 times. On how many rolls would you expect to obtain a three?

[2]
c.
Answer/Explanation

Markscheme

evidence of summing to 1      (M1)

eg   0.28 + k + 1.5 + 0.3 = 1,  0.73 + k = 1

k = 0.27     A1 N2

[2 marks]

a.

correct substitution into formula for E (X)      (A1)
eg  1 × 0.28 + 2 × k + 3 × 0.15 + 4 × 0.3

E (X) = 2.47  (exact)      A1 N2

[2 marks]

b.

valid approach      (M1)

eg  np, 80 × 0.15

12     A1 N2

[2 marks]

c.

Question

The mass M of apples in grams is normally distributed with mean μ. The following table shows probabilities for values of M.

The apples are packed in bags of ten.

Any apples with a mass less than 95 g are classified as small.

Write down the value of k.

[2]
a.i.

Show that μ = 106.

[2]
a.ii.

Find P(M < 95) .

[5]
b.

Find the probability that a bag of apples selected at random contains at most one small apple.

[3]
c.

Find the expected number of bags in this crate that contain at most one small apple.

[3]
d.i.

Find the probability that at least 48 bags in this crate contain at most one small apple.

[2]
d.ii.
Answer/Explanation

Markscheme

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

eg   k + 0.98 + 0.01 = 1

k = 0.01     A1 N2

[2 marks]

a.i.

recognizing that 93 and 119 are symmetrical about μ       (M1)

eg   μ is midpoint of 93 and 119

correct working to find μ       A1

\(\frac{{119 + 93}}{2}\)

μ = 106     AG N0

[2 marks]

a.ii.

finding standardized value for 93 or 119      (A1)
eg   z = −2.32634, z = 2.32634

correct substitution using their z value      (A1)
eg  \(\frac{{93 – 106}}{\sigma } =  – 2.32634,\,\,\frac{{119 – 106}}{{2.32634}} = \sigma \)

σ = 5.58815     (A1)

0.024508

P(X < 95) = 0.0245      A2 N3

[5 marks]

b.

evidence of recognizing binomial    (M1) 

eg 10, ananaCpqn−=××and 0.024B(5,,)pnp= 

valid approach    (M1) 

eg P(1),P(0)P(1)XXX≤=+= 

0.976285 

0.976     A1 N2 

[3 marks]

c.

recognizing new binomial probability      (M1)
eg     B(50, 0.976)

correct substitution      (A1)
eg     E(X) = 50 (0.976285)

48.81425

48.8    A1 N2

[3 marks]

d.i.

valid approach      (M1)

eg   P(X ≥ 48), 1 − P(X ≤ 47)

0.884688

0.885       A1 N2

[2 marks]

d.ii.
Scroll to Top