5 Marks Questions

Question 15 Marks

Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

Box

Marble colour

Red

White

Black

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?

Answer

Let R represents the drawing of red ball and the four boxes are represented by A, B, C and D.
So $P(R|A) = \frac{1}{{10}}$, $P(R|B) = \frac{6}{{10}}$,
$P(R|C) = \frac{8}{{10}},P(R|D) = \frac{0}{{10}} = 0$
Since there are 4 bags.
Therefore, P (A) = $\frac{1}{4},$ P (B) = $\frac{1}{4},$ P (C) = $\frac{1}{4},$ P (D) = $\frac{1}{4}$
$P(A|R) = \frac{{P(A) \cdot P(R|A)}}{{P(A) \cdot P(R|A) + P(B) \cdot P(R|B) + P(C) \cdot P(R|C) + P(D) \cdot (R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{1}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}}$
$ = \frac{{\frac{1}{{10}}}}{{\frac{1}{{10}} + \frac{6}{{10}} + \frac{8}{{15}}}} = \frac{1}{{15}}$
$P(B/R) = \frac{{P(B) \cdot P(R|B)}}{{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C) + P(D)P(R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{6}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}} = \frac{6}{{15}} = \frac{2}{5}$
$P(C|R) = \frac{{P(C) \cdot P(R|C)}}{{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C) + P(D)P(R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{8}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}} = \frac{8}{{15}}$

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Question 25 Marks

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i. e if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer

$E_1$ : the person has the disease
$E_2$ : the person is healthy
A : test is positive
$P(E_1) = 0.1 = \frac{1}{{10}}, P(E_2) = 1 - \frac{1}{{10}} = \frac{9}{{10}}$
P($\frac {A}{E_1}$) = $\frac{{99}}{{100}}$, $P\left( {\frac{A}{{{E_2}}}} \right) $ = 0.005
$ = \frac{5}{{1000}}$
$P\left( {\frac{{{E_1}}}{A}} \right) = \frac{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right)}}{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\frac{A}{{{E_2}}}} \right)}}$
$ = \frac{{\frac{{99}}{{100}} \times \frac{1}{{10}}}}{{\frac{{99}}{{100}} \times \frac{1}{{10}} + \frac{5}{{1000}} + \frac{9}{{10}}}}$
$ = \frac{{22}}{{23}}$

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Question 35 Marks

A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer

Let A be the event that ball drawn is red and let $E_1$ and $E_2$ be the events that the ball drawn is from the first bag and second bag respectively.
$P (E_1) = \frac{1}{2}, P (E_2) = \frac{1}{2}$,
P$\left( {A|{E_1}} \right)$= P (drawing a red ball from bag I) = $\frac{4}{{4 + 4}}$=$\frac{4}{8} = \frac{1}{2}$
P$\left( {A|{E_2}} \right)$ = P (drawing a red ball from bag II) = $\frac{2}{{4 + 4}}$=$\frac{2}{8} = \frac{1}{4}$
Therefore, by Bayes’ theorem,
P$\left( {{E_1}|A} \right)$ = P (red ball drawn from bag I) = $\frac{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)}}$
= $\frac{{\frac{1}{2} \times \frac{1}{2}}}{{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}}} = \frac{{\frac{1}{4}}}{{\frac{1}{4} + \frac{1}{8}}} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}$

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Question 45 Marks

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer

$E_1$ : lost card is diamond
$E_2$ : lost card is not diamond
let A: two cards drawn from the remaining pack are diamonds.
$P({E_1}) = \frac{{13}}{{52}} = \frac{1}{4},P({E_2}) = \frac{{39}}{{52}} = \frac{3}{4}$
$P\left( {\frac{A}{{{E_1}}}} \right) = \frac{{12{C_2}}}{{51{C_2}}} = \frac{{12 \times 11}}{{51 \times 50}}$
$P\left( {\frac{A}{{{E_2}}}} \right) = \frac{{13{C_2}}}{{51{C_2}}} = \frac{{13 \times 12}}{{51 \times 50}}$
$P\left( {\frac{{{E_1}}}{A}} \right) = \frac{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right)}}{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\frac{A}{{{E_2}}}} \right)}}$$=\frac{\frac{13}{52}×\frac{12×11}{51×50}}{\frac{13}{52}×\frac{12×11}{51×50}+\frac{3}{4}×\frac{13×12}{51×50}}$
$ = \frac{{11}}{{50}}$

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Question 55 Marks

A manufacturer has three machine operators $A, B$ and $C$. The first operator A produces $1\%$ defective items, where as the other two operators $B$ and $C$ produce $5\%$ and $7\%$ defective items respectively. A is on the job for $50\%$ of the time, $B$ is on the job for $30\%$ of the time and $C$ on the job for $20\%$ of the time. $A$ defective item is produced, what is the probability that it was produced by $A?$

Answer

Let $E_1 =$ the item is manufactured by the operator $A, E_2 =$ the item is manufactured by the operator $B, E_3 =$ the item is manufactured by the operator $C$ and $A =$ the item is defective
Now $P (E_1) = \frac{{50}}{{100}}$, $P(E_2) = \frac{{30}}{{100}}$, $P(E_3) = \frac{{20}}{{100}}$
Now $P\left( {A|{E_1}} \right)$ = P(item drawn is manufactured by operator A) = $\frac{1}{{100}}$
Similarly, $P\left( {A|{E_2}} \right)$ = $\frac{5}{{100}}$ and $P\left( {A|{E_3}} \right)$ = $\frac{7}{{100}}$
Now Required probability = Probability that the item is manufactured by operator. A given that the item drawn is defective
$P(\frac {{E_1}} {A})$ = $\frac{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A|{E_3}} \right)}}$= $\frac{{\frac{{50}}{{100}} \times \frac{1}{{100}}}}{{\frac{{50}}{{100}} \times \frac{1}{{100}} + \frac{{30}}{{100}} \times \frac{5}{{100}} \times \frac{7}{{100}}}}$
= $\frac{{50}}{{50 + 150 + 140}} = \frac{5}{{34}}$

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