Question

: Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.​

Answer 1

QUESTION :-

Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.

SOLUTION :-

$\implies$let the E1 ,E2,E3 and A are event defined as follows .

$\implies$E1 = first bag is chosen

$\implies$E2= second bag is chosen

$\implies$E3 = third bag is chosen

$\implies$A = ball drawn is red

$\implies$since , there are three bags and one of the bags is chosen at random, so

$\implies$ p(E1) = p(E2)=p(E3) = 1/3.

$\implies$if E1 has already occurred, then first bag has been chosen which contains 3 red and 7 black balls. the probability of drawing 1 red ball from it is 3/10. So, p(A/$\sf{E_1)}$= 3/10 , similarly p(A/$\sf{E_2)}$ = 8/10 , and p(A/$\sf{E_3)}$ = 4/10 . we required to find p($\sf{E_3/A)}$ i.e. given that the ball drawn is red, what is the probability that the ball is drawn from the third bag by baye's rule.

= ⅓ × ¼ / ⅓×3/10 + ⅓×8/10 + ⅓× 4/10

=$\red{\boxed{\underline{4/15 .}}}$

Therefore, the probability that it is drawn from the third bag is 4/15.