Probability 1 Question 21

29. Six boys and six girls sit in a row at random. Find the probability that

(i) the six girls sit together.

(ii) the boys and girls sit alternatively.

(1978, 3M)

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Answer:

Correct Answer: 29. (i) $\frac{1}{132}$ (ii) $\frac{1}{462}$

Solution:

  1. (i) The total number of arrangements of six boys and six girls $=12$ !

$\therefore$ Required probability $=\frac{6 ! \times 7 !}{(12) !}=\frac{1}{132}$

[since, we consider six girls at one person]

(ii) Required probability $=\frac{2 \times 6 ! \times 6 !}{(12) !}=\frac{1}{462}$



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