# A bag A contains 2 white and 2 red balls and another bag B contains 4 white and 5 red balls. A ball is drawn and is found to be red

6 views

closed

A bag A contains 2 white and 2 red balls and another bag B contains 4 white and 5 red balls. A ball is drawn and is found to be red.The probability that it was drawn from bag B is:

(a) $\frac{5}{19}$

(b) $\frac{21}{52}$

(c) $\frac{10}{19}$

(d) $\frac{25}{52}$

+1 vote
by (17.1k points)
selected by

Answer: (c)  $\frac{10}{19}$

Let the events E1, E2, and A be defined as follows:

E1 = Choosing bag A

E2 = Choosing bag B

A = Choosing red ball.

Then, P(E1) = P(E2) = $\frac{1}{2}$

(∵ There are two bags that have an equally likely chance of being chosen

P(A/E1) = P(Drawing red ball from bag A) = $\frac{2}{4} =\frac{1}{2}$  (∵ 2 red out of 4 balls)

P(A/E2) = P(Drawing red ball from bag B) = $\frac{5}{9}$        (5 red out of 9 balls)

∴ P(Red balls are drawn from bag B)

$=P(E_2/A) = \frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}$

$=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{1}{2}+ \frac{1}{2}\times \frac{5}{9}}$ = $\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}$ = $=\frac{5/18}{19/36}$ = $\frac{10}{19}$