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

The bag contains 5 white balls and 4 red balls, for a total of 9 balls. Three balls are drawn one by one without replacement. X represents the number of white balls drawn.

The total number of ways to draw 3 balls from 9 balls:

$C(9,3) = \dfrac{9!}{3! \times 6!}$

$= \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}$

$= \dfrac{504}{6}$

$= 84$

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For X = 0 (no white balls, all 3 red):

Number of ways to choose 0 white from 5: $C(5,0) = 1$

Number of ways to choose 3 red from 4: $C(4,3) = 4$

Favorable outcomes $= 1 \times 4 = 4$

$P(X = 0) = \dfrac{4}{84} = \dfrac{1}{21}$

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For X = 1 (exactly 1 white ball):

Number of ways to choose 1 white from 5: $C(5,1) = 5$

Number of ways to choose 2 red from 4: $C(4,2) = 6$

Favorable outcomes $= 5 \times 6 = 30$

$P(X = 1) = \dfrac{30}{84} = \dfrac{5}{14}$

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For X = 2 (exactly 2 white balls):

Number of ways to choose 2 white from 5: $C(5,2) = 10$

Number of ways to choose 1 red from 4: $C(4,1) = 4$

Favorable outcomes $= 10 \times 4 = 40$

$P(X = 2) = \dfrac{40}{84} = \dfrac{10}{21}$

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For X = 3 (all 3 white balls):

Number of ways to choose 3 white from 5: $C(5,3) = 10$

Number of ways to choose 0 red from 4: $C(4,0) = 1$

Favorable outcomes $= 10 \times 1 = 10$

$P(X = 3) = \dfrac{10}{84} = \dfrac{5}{42}$

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The probability distribution is:

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

Therefore, the answer is Option 2.