Question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.Find the probability that:
(i)both balls are red.
(ii)first ball is black and second is red.
(iii)one of them is black and other is red.

Answer 1

Total number of balls=18
Number of red balls=8
Number of black balls=10

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(i) Probability of getting a red ball in the first draw=\(\frac{8}{18}=\frac{4}{9}\)
The ball is replaced after the first draw.
\(\therefore\) Probability of getting a red ball in the second draw\(=\frac{8}{18}=\frac{4}{9}\)
Therefore, probability of getting both the balls red\(=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}\)

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(ii) Probability of getting first ball black\(=\frac{10}{18}=\frac{5}{9}\)
The ball is replaced after the first draw.
Probability of getting second ball as red\(=\frac{8}{18}=\frac{4}{9}\)
Therefore, probability of getting first ball as black and second ball as red\(=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}\)

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(iii) Probability of getting first ball as red \(=\frac{8}{18}=\frac{4}{9}\)
The ball is replaced after the first draw.
Probability of getting second ball as black\(=\frac{10}{18}=\frac{5}{9}\)
Therefore, probability of getting first ball as black and second ball as red\(=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}\)
Therefore, probability that one of them is black and other is red= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black\(=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}\)