A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each b...
Question 1 Report
A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :
(a) all four balls are black ;
(b) exactly one of the four balls is white.
Answer Details
Setup. Bag A: 12 white, 8 black (20 balls). Bag B: 10 white, 15 black (25 balls). Two balls are drawn from each bag without replacement.
Useful probabilities per bag
Bag A, two black: \(\dfrac{8}{20}\times\dfrac{7}{19} = \dfrac{14}{95}\)
Bag A, exactly one white: \(2\times\dfrac{12}{20}\times\dfrac{8}{19} = \dfrac{48}{95}\)
Bag B, two black: \(\dfrac{15}{25}\times\dfrac{14}{24} = \dfrac{7}{20}\)
Bag B, exactly one white: \(2\times\dfrac{10}{25}\times\dfrac{15}{24} = \dfrac{1}{2}\)
(a) All four balls black
Both bags must give two blacks:
\[P = \frac{14}{95}\times\frac{7}{20} = \frac{98}{1900} = \frac{49}{950} \approx 0.052\]
(b) Exactly one of the four balls is white
The single white must come from one bag while the other bag yields two blacks.
White from A, two black from B: \(\dfrac{48}{95}\times\dfrac{7}{20} = \dfrac{84}{475}\)
Two black from A, white from B: \(\dfrac{14}{95}\times\dfrac{1}{2} = \dfrac{7}{95} = \dfrac{35}{475}\)
Add the mutually exclusive cases:
\[P = \frac{84}{475} + \frac{35}{475} = \frac{119}{475} \approx 0.251\]
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