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 x_i be the number on the card drawn from the i^th box, i = 1, 2, 3.

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Linked Question 1

The probability that x₁ + x₂ + x₃ is odd, is

$\frac{29}{105}$

$\frac{1}{2}$

$\frac{57}{105}$

$\frac{53}{105}$

x¹ + x² + x³

I 0 + 0 + 0 = 0

0 0 e = e

II 0 e e = 0

e e e = e

Total = 2 · 3 · 4 = 24

II^st

0 e e = 2 · 2 · 3 = 12

e 0 e = 1 · 3 · 3 = 9

e e 0 = 1 · 2 · 4 = 8

Total = 29

Total = 29 + 24 = 53

$\frac{53}{3 \cdot 5 \cdot 7} = \frac{53}{105}$

Linked Question 2

The probability that x₁, x₂, x₃ are in an arithmetic progression, is

$\frac{11}{105}$

$\frac{7}{105}$

$\frac{10}{105}$

$\frac{9}{105}$

x₁, x₂, x₃

2x₂ = x₁ + x₃

2 = 2 × 1 = 1 + 1

4 = 2 × 2 = 1 + 3 / 2 + 2 / 3 + 1

6 = 2 × 3 = 1 + 5 / 2 + 4 / 3 + 3

8 = 2 × 4 = 1 + 7 / 2 + 6 / 3 + 5

10 = 2 × 5 = 3 + 7

Total cases = 11

$probability = \frac{11}{105} .$