This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

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Let $P(A)=\large \frac{3}{7}$ (ie) Probability of A coming in time.

Therefore $P(A^1)=1-\large \frac{3}{7}$$=\large \frac{4}{7}$(ie) Probability of A not coming in time.

$P(B)=\large \frac{5}{7}$ (ie) Probability of B coming in time.

$P(B^1)=1-\large \frac{5}{7}$$=\large \frac{2}{7}$(ie) Probability of B not coming in time.

Therefore Probability of only one of them coming to school in time is

$P(A).P(B^1)+P(B)+P(A^1)$

$=\large \frac{3}{7}.\frac{2}{7}+\frac{5}{7}.\frac{4}{7}$

$=\large \frac{6}{49}+\frac{20}{49}=\frac{26}{49}$

Hence the probability of only one of them coming to school in time is $\large \frac{26}{49}$

The advantage of coming to school in time is we can relax ourselves and concentrate more during the class.

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