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 xi be the number on the card drawn from the i textth box, i=1,2,3 The probability that x1+x2+x3 is odd, is

Question

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 xi be the number on the card drawn from the i textth box, i=1,2,3 The probability that x1+x2+x3 is odd, is (A) (29/105) (B) (53/105) (C) (57/105) (D) (1/2)

Tardigrade Admin · Accepted Answer

Case I: One odd, 2 even Total number of ways =2 × 2 × 3+1 × 3 × 3+1 × 2 × 4=29. Case II: All 3 odd Number of ways =2 × 3 × 4=24 Favourable ways =53 Required probability =(53/3 × 5 × 7)=(53/105)

$\frac{29}{105}$

$\frac{53}{105}$

$\frac{57}{105}$

$\frac{1}{2}$

Case I : One odd, $2$ even
Total number of ways $= 2 \times 2 \times 3 + 1 \times 3 \times 3 + 1 \times 2 \times 4 = 29$ .
Case II: All $3$ odd
Number of ways $= 2 \times 3 \times 4 = 24$
Favourable ways $= 53$
Required probability $= \frac{53}{3 \times 5 \times 7} = \frac{53}{105}$

10529​

10553​

10557​

21​

Case I : One odd, 2 even Total number of ways =2×2×3+1×3×3+1×2×4=29. Case II: All 3 odd Number of ways =2×3×4=24 Favourable ways =53 Required probability =3×5×753​=10553​

$\frac{29}{105}$

$\frac{53}{105}$

$\frac{57}{105}$

$\frac{1}{2}$

Case I : One odd, $2$ even
Total number of ways $= 2 \times 2 \times 3 + 1 \times 3 \times 3 + 1 \times 2 \times 4 = 29$ .
Case II: All $3$ odd
Number of ways $= 2 \times 3 \times 4 = 24$
Favourable ways $= 53$
Required probability $= \frac{53}{3 \times 5 \times 7} = \frac{53}{105}$