Question

A  game of chance consists of spinning and arrow which is equally likely to come to the rest pointing to one of the numbers 1,2,3,4,.....,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 favorable outcomes = 1

Therefore, 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 favorable outcomes = 6

Therefore, P(arrow 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 favorable outcomes = 5

Therefore, P(arrow pointing at a prime number) = P(E₃) = \(\frac{5}{12}\)

(iv) out of the given numbers, the numbers that are multiple of 5 are and 10 only.

Let E₄ be the event of the arrow pointing at a multiple of 5

Then, number of favorable outcomes = 2

Therefore, P(arrow pointing at a number that is a multiple of 5) = P(E₄) = \(\frac{2}{12}\) = \(\frac{1}{6}\)