A bag contains 12 white balls and 8 black balls,... | WAEC SSCE - General Mathematics - 1994 (Essay)

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

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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