Question

a bag contains 5 red and 7 white balls.four are drawn out one by one and not replaced. what is the probability that they are alternative of different colours?​

Answer 1

Answer:

Let's take x for red ball and y for white balls .

First Red = (5/12)

Second white = (7/11)

Third Red = (4/10) { As you have already withdrawn one red ball}

Fourth White = ( 6/9)

Hence overall probability = (5/12)×(7/11)×(4/10)×(6/9) = (7/99)

x y x y : You withdraw white first and then red , white and finally red. The probability:

White first = (7/12)

Second Red = ( 5/11)

Third White = (6/10)

Fourth Red = (4/9)

Hence overall probability= (7/12)×(5/11)×(6/10)×(4/9) = (7/99)

The total probability = (7/99)+(7/99) = (14/99)