JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Probability

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Box 1 contains three cards bearing numbers 1,2,3; box 2 contains five cards bearing numbers 1,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,7. A card is drawn from each of the boxes. Let

x_{i}

be the number on the card drawn from the

i^{th}

box,

i = 1, 2, 3

.

The probability that

x_{1} + x_{2} + x_{3}

is odd, is

(A) $\frac{29}{105}$
(B) $\frac{53}{105}$ ✓ Correct
(C) $\frac{57}{105}$
(D) $\frac{1}{2}$

Answer: (B)

Solution

For the sum to be odd, we need either one odd number or three odd numbers.

Box 1: 2 odd (1,3), 1 even (2)
Box 2: 3 odd (1,3,5), 2 even (2,4)
Box 3: 4 odd (1,3,5,7), 3 even (2,4,6)

Total outcomes:

3 \times 5 \times 7 = 105

.

Favorable outcomes:

If $x_{1} = 1$ (odd): $x_{2} + x_{3}$ must be even = (odd,odd) or (even,even) = $3 \times 4 + 2 \times 3 = 18$
If $x_{1} = 2$ (even): $x_{2} + x_{3}$ must be odd = (odd,even) or (even,odd) = $3 \times 3 + 2 \times 4 = 17$
If $x_{1} = 3$ (odd): $x_{2} + x_{3}$ must be even = $3 \times 4 + 2 \times 3 = 18$

Total favorable =

18 + 17 + 18 = 53

.

Probability =

\frac{53}{105}

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Probability

Solution

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