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.
$P({E_1}) = \frac{{13}}{{52}} = \frac{1}{4},P({E_2}) = \frac{{39}}{{52}} = \frac{3}{4}$
$P\left( {\frac{A}{{{E_1}}}} \right) = \frac{{12{C_2}}}{{51{C_2}}} = \frac{{12 \times 11}}{{51 \times 50}}$
$P\left( {\frac{A}{{{E_2}}}} \right) = \frac{{13{C_2}}}{{51{C_2}}} = \frac{{13 \times 12}}{{51 \times 50}}$
$P\left( {\frac{{{E_1}}}{A}} \right) = \frac{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right)}}{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\frac{A}{{{E_2}}}} \right)}}$ $=\frac{\frac{13}{52}\times \frac{12\times 11}{51\times 50}}{\frac{13}{52}\times \frac{12\times 11}{51\times 50}+\frac{3}{4}\times \frac{13\times 12}{51\times 50}}$
$ = \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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