JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive.

Choose the correct answer:

Explanation

1. Total Number of Ways

The total number of ways to choose 3 numbers from a set of 20 is given by the combination formula $\binom{n}{r}$ :

n(S) = \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1}

n(S) = 20 \times 19 \times 3 = 1140

2. Number of Consecutive Triplets

A triplet of numbers $(x, y, z)$ is consecutive if they are of the form $(n, n+1, n+2)$ .

The possible consecutive triplets from the set $\{1, 2, 3, \dots, 20\}$ are:

(1, 2, 3), (2, 3, 4), (3, 4, 5), \dots, (18, 19, 20)

The total number of such triplets is:

n(E') = 20 - 3 + 1 = 18

3. Number of Non-Consecutive Selections

The number of ways to choose three numbers such that they are not all consecutive is:

n(E) = n(S) - n(E')

n(E) = 1140 - 18 = 1122

4. Probability Calculation

The probability $P(E)$ is the ratio of favorable outcomes to total outcomes:

P(E) = \frac{n(E)}{n(S)} = \frac{1122}{1140}

Divide both the numerator and the denominator by 6:

P(E) = \frac{187}{190}

Final Result

The probability that the three chosen numbers are not consecutive is:

\frac{187}{190}

Choose the correct answer:

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