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Mathematics Question on Probability

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

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

(i) Probability of getting a red ball in the first draw=818=49\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=818=49=\frac{8}{18}=\frac{4}{9}
Therefore, probability of getting both the balls red=49×49=1681=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}

(ii) Probability of getting first ball black=1018=59=\frac{10}{18}=\frac{5}{9}
The ball is replaced after the first draw.
Probability of getting second ball as red=818=49=\frac{8}{18}=\frac{4}{9}
Therefore, probability of getting first ball as black and second ball as red=59×49=2081=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}

(iii) Probability of getting first ball as red =818=49=\frac{8}{18}=\frac{4}{9}
The ball is replaced after the first draw.
Probability of getting second ball as black=1018=59=\frac{10}{18}=\frac{5}{9}
Therefore, probability of getting first ball as black and second ball as red=59×49=2081=\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=2081+2081=4081=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}