If three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls, then the probability distribution of the number of white balls drawn is Option 1: | X | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(X) | 30/91 | 45/91 | 15/91 | 1/91 | Option 2: | X | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(X) | 1/21 | 5/14 | 10/21 | 5/42 | Option 3: | X | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(X) | 1/8 | 3/8 | 3/8 | 1/8 | Option 4: | X | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(X) | 12/19 | 16/35 | 16/35 | 3/95 |

The answer is '| X | 0 | 1 | 2 | 3 | |---|---|---|---|---| | P(X) | 1/21 | 5/14 | 10/21 | 5/42 |'. View full solution:

Algebra > Probability

Medium

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If three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls, then the probability distribution of the number of white balls drawn is

X	0	1	2	3
P(X)	30/91	45/91	15/91	1/91

X	0	1	2	3
P(X)	1/21	5/14	10/21	5/42

X	0	1	2	3
P(X)	1/8	3/8	3/8	1/8

X	0	1	2	3
P(X)	12/19	16/35	16/35	3/95

✅ Correct Option: 2

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