Question

Five numbers x₁, x₂, x₃, x₄, x₅ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order such that x₁ < x₂ < x₃ < x₄ < x₅. What is the probability that x₂ = 7 and x₄ = 11?

Options

• `26/51`

• `3/104`

• `1/68`

• `1/34`

Answer 1

`bb(1/68)`

Explanation:

The number of ways to choose 5 numbers from 18 is given by:

`(18/5)`

= `(18!)/(5!(18 - 5)!)`

= `(18!)/(5!*13!)`

= `(18 xx 17 xx 16 xx 15 xx 14)/(5 xx 4 xx 3 xx 2 xx 1)`

= 8568

For x₂ = 7 and x₄ = 11, we must choose x₁ < 7x (6 options), 7 < x₃ < 11 (3 options), and x₅ > 11 (7 options).

Total favorable outcomes = 6 × 3 × 7

= 126

Probability = `"Number of favorable outcomes"/"Total number of possible outcomes"`

= `126/8568`

= `1/68`