A bag contains 3 red and 7 black balls.Two balls are drawn at random without replacement .if the second ball is red what is the probability that the first ball is also red
P(red,red) = 3/10 * 2/9 = 1/15
Apologies for the mistake. You are correct.
To find the probability that the first ball is red given that the second ball is red, we need to find the probability of both events occurring.
There are 10 balls in total, with 3 red and 7 black. If the first ball is red, there are 2 red balls left out of 9 total balls in the bag.
Therefore, the probability that the first ball is red is:
P(first ball is red) = 3/10
If the second ball is red, there is only 1 red ball left out of the remaining 9 balls in the bag.
Therefore, the probability that the second ball is red is:
P(second ball is red) = 2/9
The probability that both events occur is the product of the individual probabilities:
P(first ball is red and second ball is red) = (3/10) * (2/9) = 6/90 = 1/15
Therefore, the probability that the first ball is red given that the second ball is red is:
P(first ball is red | second ball is red) = (1/15) / (2/9) = 3/10
To find the probability that the first ball is red given that the second ball is red, we need to find the probability of both events occurring.
There are a total of 10 balls in the bag, with 3 red and 7 black. If the first ball is red, there are 2 red balls left out of 9 total balls in the bag.
Therefore, the probability that the first ball is red is:
P(first ball is red) = 3/10
If the second ball is red, there is only 1 red ball left out of the remaining 9 balls in the bag.
Therefore, the probability that the second ball is red is:
P(second ball is red) = 1/9
The probability that both events occur is the product of the individual probabilities:
P(first ball is red and second ball is red) = (3/10) * (1/9)
Therefore, the probability that the first ball is red given that the second ball is red is:
P(first ball is red | second ball is red) = (3/10) * (1/9) / (1/9) = 3/10