Search any question & find its solution
Question:
Answered & Verified by Expert
A boy spcaks truth in 3 out of 5 times. If he throws a die and tells that the number appeared on it is five, then the probability that it is actually five, is
Options:
Solution:
1998 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{13}$
Let $E_1$ : Event of getting a five
$E_2$ : Event of not getting a five and $A$ : Event that the boy reports that it is a five. So, required probabilltv by Baye's theorem
$\begin{aligned} P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)} \\ & =\frac{\frac{1}{6} \times \frac{3}{5}}{\left(\frac{1}{6} \times \frac{3}{5}\right)+\left(\frac{5}{6} \times \frac{2}{5}\right)}=\frac{3}{3+10}=\frac{3}{13}\end{aligned}$
Hence, option (d) is correct.
$E_2$ : Event of not getting a five and $A$ : Event that the boy reports that it is a five. So, required probabilltv by Baye's theorem
$\begin{aligned} P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)} \\ & =\frac{\frac{1}{6} \times \frac{3}{5}}{\left(\frac{1}{6} \times \frac{3}{5}\right)+\left(\frac{5}{6} \times \frac{2}{5}\right)}=\frac{3}{3+10}=\frac{3}{13}\end{aligned}$
Hence, option (d) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.