Question:
One bag contains 3 red and 5 black balls another bag contains 6 red and 4 black balls.
i. find probability of getting two red balls from first bag.
ii. find probability of getting two black balls from second bag
Answer:
(i) Let A be the event of getting two red balls from the first bag
P(A) = P(getting first red ball) x P(getting second red ball | first ball was red)
The probability of getting the first red ball from the bag is 3/8, since there are 3 red balls out of a total of 8 balls in the bag. After taking out one red ball, there are 7 balls left, out of which 2 are red. Therefore, the probability of getting a second red ball given that the first ball was red is 2/7.
So, P(A) = (3/8) x (2/7) = 3/28
(ii) Let B be the event of getting two black balls from the second bag
P(B) = P(getting first black ball) x P(getting second black ball | first ball was black)
The probability of getting the first black ball from the bag is 4/10, since there are 4 black balls out of a total of 10 balls in the bag. After taking out one black ball, there are 9 balls left, out of which 3 are black. Therefore, the probability of getting a second black ball given that the first ball was black is 3/9 or 1/3.
So, P(B) = (4/10) x (1/3) = 2/15
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