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Q.

Box-I contains 3 cards bearing numbers 1, 2, 3; Box II contains 5 cards bearing numbers 1, 2, 3, 4, 5 and Box III contains 7 cards bearing numbers 1, 2, 3, 4, 5, 6, 7. One card is drawn at random from each of the boxes. If $x_{i}$ be the number on the card drawn from the ith box, i = 1, 2, 3, then the probability that $x_{1} + x_{2} + x_{3}$ is odd equal to

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a

$\frac{53}{105}$

b

$\frac{43}{105}$

c

$\frac{33}{105}$

d

$\frac{23}{105}$

answer is B.

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Detailed Solution

I	II	III
1,2,3	1,2,3,4,5	1,2,3,4,5,6,7

one card drawn from each box

$x_{1} =$ card drawn first box

$x_{2} =$ card drawn second box

$x_{3} =$ card drawn third box

$x_{1} + x_{2} + x_{3}$ in odd 2 even 1 odd (or) 3 odd

I	II	III	No of ways
e	e	o	$\frac{1}{3} \times \frac{2}{5} \times \frac{4}{7}$	$\frac{8}{105}$
e	o	e	$\frac{1}{3} \times \frac{3}{5} \times \frac{3}{7}$	$\frac{9}{105}$
o	e	e	$\frac{2}{3} \times \frac{2}{5} \times \frac{3}{7}$	$\frac{12}{105}$
o	o	o	$\frac{2}{3} \times \frac{3}{5} \times \frac{4}{7}$	$\frac{24}{105}$

$∴$ Total $= \frac{8}{105} + \frac{9}{105} + \frac{12}{105} + \frac{24}{105}$

$= \frac{53}{105}$

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