$(a)\;\large\frac{2}{3}\qquad(b)\;\large\frac{2}{9}\qquad(c)\;\large\frac{1}{9}\qquad(d)\;\large\frac{4}{9}$

- 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 $E_1$: event that the coin is 2 headed, $E_2$ be the event that its biased with heads 75% of the time and $E_3$ be the fair coin.

All these three events are mutually exclusive and exchaustive, and are equally likely.

$\therefore, P(E_1) = P(E_2) = P(E_3) = \large\frac{1}{3}$

Let one of the coin chosen at random be tossed up.

Let $E$ be the event of chosen coin showing head.

$P ($coin shows head given that its 2 headed coin) $= P (E|E_1) = 1$.

$P ($coin shows head given that its 75% biased for heads) $= P (E|E_2) = \large\frac{3}{4}$

$P ($coin shows head given that its a fair coin)$ = P (E|E_3) = \large\frac{1}{2}$

We need to find the probability that it is two headed coin, given that it shows head.

We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)+ P(E_3)P(A|E_3)}$

Therefore $ P(E_1|E) = \large\frac{\frac{1}{3} }{\frac{1}{3} + \frac{1}{3}.\frac{3}{4} + \frac{1}{3}.\frac{1}{2}}$

$P(E_1|E) = \large\frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \large\frac{1}{\frac{4+3+2}{4}} = \frac{4}{9}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...