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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → P(X) = 1/21, 5/14, 10/21, 5/42 for X = 0, 1, 2, 3
Let X be the number of white balls drawn. Drawing three balls one by one without replacement gives the same probabilities as drawing three balls together, so each value of X is counted with combinations.
\(\text{Total ways} = {}^{9}C_{3} = 84\)
\( \displaystyle P(X=0) = \frac{{}^{4}C_{3}}{84} = \frac{4}{84} = \frac{1}{21} \)
\( \displaystyle P(X=1) = \frac{{}^{5}C_{1}\times{}^{4}C_{2}}{84} = \frac{5\times 6}{84} = \frac{5}{14} \)
\( \displaystyle P(X=2) = \frac{{}^{5}C_{2}\times{}^{4}C_{1}}{84} = \frac{10\times 4}{84} = \frac{10}{21} \)
\( \displaystyle P(X=3) = \frac{{}^{5}C_{3}}{84} = \frac{10}{84} = \frac{5}{42} \)
Check: \(4 + 30 + 40 + 10 = 84\), so the probabilities add up to 1.