$\begin{array}{1 1}8/11 \\ 2/11 \\ 3/11 \\ 6/11 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)

Let $E_1$ be the event that the girl gets 5 or 6 on the roll. P($E_1$) = $\large\frac{2}{6} = \frac{1}{3}$

Let $E_2$ be the event that the girl gets 1,2,3 or 4 on the roll. P($E_1$) = $\large\frac{4}{6} = \frac{2}{3}$

Let A be the event that she obtained exactly one head.

If she tossed a coin 3 times and exactly 1 head showed up, then the total number of favourable outcomes = {(HTT), (THT), (TTH),} = 3.

P (A|$E_1$) = $\large\frac{3}{8}$ (Note: in 3 coin tosses, total sample space = 8)

If she tossed the coin only once and exactly 1 showed up, the total number of favourable outcomes = 1.

P (A|$E_2$) = $\large\frac{1}{2}$

We need to find the probability that she threw 1, 2, 3 or 4 with the die, given that she got exactly one head.

We can use Baye's theorem, according to which $P(E_2|A) = \large\frac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$

P($E_2$|A) = $\Large \frac {\frac{1}{2}.\frac{2}{3}}{\frac{1}{2}.\frac{2}{3} + \frac{3}{8}.\frac{1}{3}}$ = $\Large \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}$ = $\large\frac{8}{11}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...