Question

a bag contains two coins ,one biased and the other unbiased.when tossed the biased coin has a 60% chances of showing heads.One of the coins is selected at random and on tossing it shows tails . what is the probability it was an unbiased coin?

Answer 1

Answer:

5/9

Explanation:

Answer 2

Given:

A bag contains two coins ,one biased and the other unbiased.when tossed the biased coin has a 60% chances of showing heads.

To find:

The probability it was an unbiased coin.

Solution:

Let E1 be the event of choosing a biased coin

Let E2 be the event of choosing an unbiased coin

⇒ P (E1) = P (E2) = 1/2

The probability of biased coin has the chance of showing heads is 60%

The probability of biased coin has the chance of showing tails is 40%

(100% - 60% = 40%)

Let A be the event of showing tails

P (A/E1) = P (biased coin showing tails) = 40/100 = 2/5

P (A/E2) = P (unbiased coin showing tails) = 1/2

Using Baye's theorem, we get,

$P\left ( \dfrac{E_2}{A} \right )=\dfrac{P(E_2) \times P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}$

$P\left ( \dfrac{E_2}{A} \right )=\dfrac{ \frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times\frac{2}{5} + \frac{1}{2} \times \frac{1}{2} }$

$P\left ( \dfrac{E_2}{A} \right )=\dfrac{ \frac{1}{4}}{\frac{1}{5} + \frac{1}{4} }$

$P\left ( \dfrac{E_2}{A} \right )=\dfrac{ \frac{1}{4}}{\frac{9}{20} }$

$P\left ( \dfrac{E_2}{A} \right )= \dfrac{1}{4} \times \dfrac{20}{9}$

$P\left ( \dfrac{E_2}{A} \right )=\dfrac{5}{9}$

Therefore, the probability it was an unbiased coin is 5/9 = 0.55