A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 (see figure 16.1) and there are equally likely outcomes. What is the probability that it will point at

(1) 7?

(2) a number greater than 9?

(3) an odd number?

(4) an even number?

(5) a number less than 5?

Total numbers are from 1 to 12 i.e, 12 numbers.

Now, let the event that the arrow will point at 7 be A.

Event that arrow will point at a number greater than 5 be B.

Event that arrow will point at an odd number be C.

Event that arrow will point at an even number be D.

Event that arrow will point at a number less than 5 be E.

Number of favourable outcomes of event A = 1

Favourable outcomes of event B i.e., a number greater than 5 are 6, 7, 8 ,9 10, 11, 12.

Number of favourable outcomes of event B = 7

Favourable outcomes of event C i.e., an odd number are 1,3,5,7,9,11.

Number of favourable outcomes of event C = 6

Favourable outcomes of event D i.e., an even number are 2,4,5,8,10,12.

Number of favourable outcomes of event D = 6

Favourable outcomes of event E i.e., a number less than 5 are 1,2 ,3,4.

Number of favourable outcomes of event E = 4

Total number of favourable outcomes = 12

(i) Probability that the arrow will point at 7 = Probability of occurrence of event A = P(A)

P(A) = = 0.0833

(ii) Probability that arrow will point at a number greater than 5 = Probability of occurrence of event B = P(B)

P(B) = = 0.58333

(iii) Probability that arrow will point at an odd number = Probability of occurrence of event C = P(C)

P(C) = = 0.5

(iv) Probability that arrow will point at an even number = Probability of occurrence of event D = P(D)

P(D) = = 0.5

(v) Probability that arrow will point at a number less than 5 = Probability of occurrence of event E = P(E)

P(E) = = 0.3333