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13.     Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

                (iii) one of them is black and other is red.

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Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw 

                                                                            =\frac{10}{18}=\frac{5}{9}

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

the probability that the first  ball is black and the second is red

                                                                                     =\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}    ...........................1

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

The probability of getting a black ball in the second  draw 

                                                                            =\frac{10}{18}=\frac{5}{9}

the probability that the first  ball is red and the second is black

                                                                                     =\frac{4}{9}\times \frac{5}{9}=\frac{20}{81}    ...........................2

Thus, 

The probability that one of them is black and the other is red = the probability that the first  ball is black and the second is red  + the probability that the first  ball is red and the second is black

                                                                                       =\frac{20}{81}+\frac{20}{81}=\frac{40}{81}

Posted by

seema garhwal

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