Question

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3,..., 12 as shown in the figure. What is the probability that it will point to
(i) 6?
(ii) an even number?
(iii) a prime number?
(iv) a number which is a multiple of 5?

Answer 1

The possible outcomes are 1,2, 3,4, 5................12.
Number of all possible outcomes = 12

(i) Let E₁ be the event that the pointer rests on 6.
Then, number of favourable outcomes = 1​
∴ P (arrow pointing at 6) = P( E₁) = $\frac{1}{12}$

(ii) Out of the given numbers, the even numbers are

2, 4, 6, 8,10 and 12

Let E₂ be the event of getting an even number.
Then, number of favourable outcomes = 6​
∴ P (arrow pointing at an even number) = P( E₂) = $\frac{6}{12}=\frac{1}{2}$

(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7 and 11.
Let E₃ be the event of the arrow pointing at a prime number.
Then, number of favourable outcomes =​ 5
∴ P (arrow pointing at a prime number) = P( E₃) = $\frac{5}{12}$

(iv) Out of the given numbers, the numbers that are multiples of 5 are 5 and 10 only.
Let E₄ be the event of the arrow pointing at a multiple of 5.

Then,​ number of favourable outcomes = 2
∴ P(arrow pointing at a number that is a multiple of 5) = P( E₄) =$\frac{2}{12}=\frac{1}{6}$