An Urn contains 7 red and 3 white balls. Two balls are drawn in succession without replacement. If the second ball is white, what is the probability that the first ball was white?
A box contain 3 red balls and 6 white balls. One ball was pulled and was seen its collar, and after that was pulled another ball.Find the probability that the first ball was white and the second was red ?
To find the probability that the first ball drawn was white, given that the second ball drawn is white, we can use Bayes' theorem.
Let's denote the events:
A: The first ball is white
B: The second ball is white
We are given that B has occurred, and we want to find the probability of A given B, denoted as P(A|B).
By Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) represents the probability of the second ball being white given that the first ball is white. Since one white ball has already been drawn, there will be 2 white balls left in the urn, out of a total of 9 remaining balls.
So, P(B|A) = 2/9
P(A) represents the probability of the first ball being white. We know that there are a total of 10 balls in the urn, 3 of which are white. Therefore, P(A) = 3/10.
P(B) represents the probability of the second ball being white. There are two ways that the second ball can be white: either the first ball drawn is white and the second ball drawn is also white, or the first ball drawn is red and the second ball drawn is white.
The probability of the first ball being white and the second ball being white can be calculated as:
(P(A) * P(B|A)) + (P(A')(P(B|A')))
where A' represents the event that the first ball is red.
P(A') represents the probability of the first ball being red, which can be calculated as:
1 - P(A)
Therefore, P(B) can be calculated as:
(P(A) * P(B|A)) + (P(A') * P(B|A'))
Substituting the values we know:
P(B) = (3/10 * 2/9) + (7/10 * 3/9)
Now, we can calculate P(A|B) using Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
Substituting the values we know:
P(A|B) = (2/9 * 3/10) / [(3/10 * 2/9) + (7/10 * 3/9)]
By simplifying this expression, we can find the probability that the first ball was white, given that the second ball drawn is white.
To find the probability that the first ball was white given that the second ball is white, we can use Bayes' theorem.
Let's break down the problem step by step:
Step 1: Determine the probability of drawing a white ball on the first draw.
Initially, there are 10 balls in the urn (7 red and 3 white). So, the probability of drawing a white ball on the first draw is 3/10 (since there are 3 white balls out of 10 total balls).
Step 2: Determine the probability of drawing a white ball on the second draw.
If the first ball drawn was white, there are now 9 balls remaining in the urn, out of which 2 are white. So, the probability of drawing a white ball on the second draw, given that the first ball was white, is 2/9.
Step 3: Apply Bayes' theorem to find the probability that the first ball was white, given that the second ball is white.
Using Bayes' theorem:
P(White First | White Second) = (P(White First) * P(White Second|White First)) / P(White Second)
P(White First | White Second) = (3/10 * 2/9) / (2/10)
Simplifying the expression, we have:
P(White First | White Second) = (6/90) / (2/10)
P(White First | White Second) = (6/90) * (10/2)
P(White First | White Second) = 6/9
P(White First | White Second) = 2/3
Therefore, the probability that the first ball was white, given that the second ball is white, is 2/3.