Abox contains 6 black balls and 4 white balls. if two balls are selected at random without replacement, what is the probability that both balls are white ball?

The probability that both balls are white = 11/15

Explanation:

pr = n(possible outcome)/n(total outcome)

Where pr = probability, n = number

If two balls are selected are random without replacement

pr(both balls are white) = pr(1)×pr(2) Equation

pr(1) = probability of picking the first white ball, pr(2) = probability of picking the second with ball

pr(1) = n(possible outcome)/n(total outcome)

n(possible outcome) = 4

n(total outcome) = 10

pr(1) = 4/10 = 2/5

After the first pick without replacement, the total number of ball will decrease.

pr(2) =  n(possible outcome)/n(total outcome)

n(possible outcome) = 3

n(total outcome) = 9

pr(2) = 3/9 = 1/3

Therefore,

pr(both balls are white) =(2/5) + (1/3)

Adding both fraction,

pr(both balls are white)  = (6+5)/15

pr(both balls are white) = 11/15.

Thus the probability that both balls are white = 11/15

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