First slide

Introduction to probability

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

Moderate

Question

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

A

$\frac{29}{105}$
B

$\frac{53}{105}$
C

$\frac{57}{105}$
D

$\frac{1}{2}$

Solution

For, x₁ + x₂ + x₃ = odd
Case 1 : One odd, two even
(OEE) or (EOE) or (EEO)
Total number of ways = 2 x 2 x 3 + 1 x 3 x 3 + 1 x 2 x 4 = 29.
Case II: All three odd
Number of ways = 2 x 3 x 4 = 24
$∴$ Favorable ways = 53
$∴$ Required probability $= \frac{53}{3 \times 5 \times 7} = \frac{53}{105}$

Question

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

A

$\frac{9}{105}$
B

$\frac{10}{105}$
C

$\frac{11}{105}$
D

$\frac{7}{105}$

Solution

Here, $2 x_{2} = x_{1} + x_{3}$
$\Rightarrow x_{1} + x_{3} = even$
So, either x₁ and x₃ are both odd or both even.
Hence, number of favorable ways $=^{2} C_{1} \cdot^{4} C_{1} +^{1} C_{1} \cdot^{3} C_{1} = 11$ .
Therefore, required probability is $\frac{11}{105}$ .

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