question_answer

                

A speaks the truth 8 times out of 10 times. He tossed a die. He reports that it was 5. What is the probability that it was actually 5?

OR

A committee of 4 students is selected at random from a group consisting 8 boys and 4 girls. If there is atleast one girl on the committee, then calculate the probability that there are exactly 2 girls on the committee.

Answer:

Let us define the following events as

\[{{E}_{1}}:5\] occur on the die

\[{{E}_{2}}:5\] does not occur on the die

E : A reports that 5 occur in the throwing of the die.

Then, \[P({{E}_{1}})=\] Probability of getting 5 on a die

\[=\frac{1}{6}\]

\[P({{E}_{2}})=\] Probability of not getting 5 on a die

\[=1-\frac{1}{6}=\frac{5}{6}\]

\[P\left( \frac{E}{{{E}_{1}}} \right)=\] Probability that 'A reports that '5' occurs on the die, when 5 has actually occurred on the die, i.e. A is speaking truth \[=\frac{8}{10}=\frac{4}{5}\]

\[P\left( \frac{E}{{{E}_{2}}} \right)=\] Probability that 'A? reports that '5' occurs on a die when 5 has not actually occurred on the die, i.e. A is not speaking truth \[=1-\frac{4}{5}=\frac{1}{5}\]

Now, by Baye?s theorem, we have

\[P\left( \frac{{{E}_{1}}}{E} \right)=\frac{P({{E}_{1}})\cdot P\left( \frac{E}{{{E}_{1}}} \right)}{P({{E}_{1}})\cdot P\left( \frac{E}{{{E}_{1}}} \right)+P({{E}_{2}})\cdot P\left( \frac{E}{{{E}_{2}}} \right)}\]

On putting all the values, we get

\[P\left( \frac{{{E}_{1}}}{E} \right)=\frac{\frac{1}{6}\times \frac{4}{5}}{\left( \frac{1}{6}\times \frac{4}{5} \right)+\left( \frac{5}{6}\times \frac{1}{5} \right)}=\frac{\frac{4}{30}}{\frac{4}{30}+\frac{5}{30}}\]

\[=\frac{4}{4+5}=\frac{4}{9}\]

Hence, the required probability is \[\frac{4}{9}.\]

OR

Let A denotes the event that at least one girl will be chosen and B denotes the event that exactly 2 girls will be chosen. Then, to find P (B / A)

Now, \[P\,(A)=1-P\,(\bar{A})=1-P\] (no girl is chosen)

\[=1-P\] (4 boys are chosen)

\[=1-\frac{^{8}{{C}_{4}}}{^{12}{{C}_{4}}}=1-\frac{70}{495}=1-\frac{14}{99}=\frac{85}{99}\]

and  \[P(A\cap B)=P\] (2 boys and 2 girls are chosen)

\[=\frac{^{8}{{C}_{2}}{{\times }^{4}}{{C}_{2}}}{^{12}{{C}_{4}}}=\frac{28\times 6}{495}=\frac{56}{165}\]

Hence, \[P\left( \frac{B}{A} \right)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{56}{165}}{\frac{85}{99}}=\frac{56}{165}\times \frac{99}{85}=\frac{168}{425}\]