Problem:
A bag contains 3 black and 4 red balls. Two balls are drawn at random, one at a time, without replacement. Find the probability that the first ball is black if the second ball is known to be red.

Solution:
To find the probability that the first ball is black if the second ball is known to be red, we need to use conditional probability.

Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. Conditional probability is calculated by dividing the probability of the joint event by the probability of the conditioning event.

Formula:
P(A|B) = P(A and B) / P(B)

Where,
P(A|B) = Probability of event A given that event B has occurred
P(A and B) = Probability of joint event of A and B occurring
P(B) = Probability of event B occurring

Solution:
Let A be the event of drawing a black ball first and B be the event of drawing a red ball second.

P(B) = Probability of drawing a red ball second = 4/7

To find P(A and B), we need to consider two cases:

Case 1: The first ball drawn is black and the second ball drawn is red.
P(A and B) = Probability of drawing a black ball first and a red ball second
= (3/7) * (4/6) [Since we are drawing without replacement]
= 2/7

Case 2: The first ball drawn is red and the second ball drawn is black.
P(A' and B) = Probability of drawing a red ball first and a black ball second
= (4/7) * (3/6) [Since we are drawing without replacement]
= 2/7

P(A and B) + P(A' and B) = Probability of drawing a red ball second
= 4/7

P(A and B) = (4/7) - (2/7)
= 2/7

Now, we can use the formula for conditional probability to find the probability that the first ball is black given that the second ball is red.

P(A|B) = P(A and B) / P(B)
= (2/7) / (4/7)
= 1/2

Therefore, the probability that the first ball is black given that the second ball is red is 1/2 or 0.5.

Let probability of first ball be black ,P(E)= 3/7 and then probability of first ball be red,P(E')= 4/7. now let 'A' be the probability of second ball be red.firstly if first ball is black then, P(A/E)=4/6 and if first ball is red then, P(A/E') = 3/6. then we need to find out P(E/A). therefore by Bayes' theorem,. P(E/A) = {P(A/E)× P(E)}/P(A/E)×P(E)+P(A/E')×P(E')= (4/6×3/7)/(4/6×3/7)+(3/6×4/7). = 12/12+12= 12/24 = 1/2.hence required P(E/A) = 1/2 .