This question was previously asked in

Territorial Army Paper I : Official Practice Test Paper - 4

Option 4 : \(\frac{5}{8}\)

Given:

A bag contains 5 red balls and 3 blue balls.

One ball is drawn at random.

Concept Used:

The probability of the occurrence of an event A out of a total possible outcomes N, is given by: P(A) = \(\rm \frac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.

The number of ways in which r distinct objects can be selected from a group of n distinct objects, is: ^{n}C_{r} = \(\rm \frac {n!}{r!(n-r)!}\).

n! = 1 × 2 × 3 × ... × n.

Calculation:

The number of ways in which any 1 ball can be drawn from all the 8 balls = N = ^{8}C_{1} = 8.

The number of ways in which 1 red ball will be drawn = n(A) = ^{5}C_{1} = 5.

The required probability = \(\rm \frac{n(A)}{N}=\frac{5}{8}\).