3. Three bags contain 3 red, 7 black, 8 red, 2 black and 4 red and 6 black balls respectively. one 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.
Answers
Answer:
4/15
Step-by-step explanation:
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)}E
1
) = 3/10 , similarly p(A/\sf{E_2)}E
2
) = 8/10 , and p(A/\sf{E_3)}E
3
) = 4/10 . we required to find p(\sf{E_3/A)}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 .}}}
4/15.
Therefore, the probability that it is drawn from the third bag is 4/15