14.Probability
medium

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that One of them is black and other is red.

A

$\frac{40}{81}$

B

$\frac{40}{81}$

C

$\frac{40}{81}$

D

$\frac{40}{81}$

Solution

Total number of balls $=18$

Number of red balls $=8$

Number of black balls $=10$

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 $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$

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

$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$

Standard 11
Mathematics

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