One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from the first bag to the second bag, and then a ball is drawn from the second bag. Find the probability that the ball is drawn is red.
OR
If P(A) = 0.6, P(B) = 0.5 and P(A|B) = 0.3, then find P (Α∪ B).
We have 2 possibilities –
1. red transferred and red drawn
2. black transferred and red drawn
Total balls in first bag = 8
Total balls in bag 2 = 10
Case – 1:
Initially, red ball is transferred from the first bag.
So,
the probability of transferring a red ball from first bag![]()
Now, the second bag has 7 red and 4 black balls,
So,
probability of drawing a red ball from second bag![]()
So, required probability from case 1![]()
![]()
Case 2:
Initially, black ball is transferred from first bag.
So, probability of transferring a black ball from first bag![]()
Now, second bag has 6 red and 5 black balls,
So,
probability of drawing a red ball from second bag![]()
So,
required probability from case 2![]()
![]()
Also, Case 1 and Case 2 are independent,
So,
required probability![]()
![]()
OR
P(A) = 0.6,
P(B) = 0.5,
P(A|B) = 0.3
![]()
![]()
⇒ P(AꓵB) = 0.3(P(B))
⇒ P(AꓵB) = (0.3) (0.5)
⇒ P(AꓵB) = 0.15
Now, P(AꓴB) = P(A) + P(B) – P(AꓵB)
⇒ P(AꓴB) = 0.6 + 0.5 – 0.15
= 0.95
Therefore, P(AꓴB) = 0.95
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
