A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

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Solution

Q.
A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

1 $\frac{9}{50}$

2 $\frac{1}{5}$

3 $\frac{11}{50}$

4 $\frac{1}{4}$

Ans.
(2)

$Probability of getting any number on a die = \frac{1}{6}$

$Probability of getting all white balls when 1 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{1}}}{{}^{10}{C_{1}}} = \frac{1}{6} \times \frac{6}{10} = \frac{1}{10}$

$Probability of getting all white balls when 2 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{2}}}{{}^{10}{C_{2}}} = \frac{1}{6} \times \frac{6 \times 5}{10 \times 9} = \frac{1}{18}$

$Probability of getting all white balls when 3 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{3}}}{{}^{10}{C_{3}}} = \frac{1}{6} \times \frac{6 \times 5 \times 4}{10 \times 9 \times 8} = \frac{1}{36}$

$Probability of getting all white balls when 4 appears on die = \frac{1}{84}$

$Probability of getting all white balls when 5 appears on die = \frac{1}{252}$

$Probability of getting all white balls when 6 appears on die = \frac{1}{1260}$

$Required Probability = \frac{1}{10} + \frac{1}{18} + \frac{1}{36} + \frac{1}{84} + \frac{1}{252} + \frac{1}{1260} = \frac{1}{5}$

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