Question

. 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 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
​

Answer 1

$\huge\boxed{\fcolorbox{white}{pink}{Answer}}$

Total number of possible outcomes = 8

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of favourable events (i.e. 8) = 1

∴ P (pointing at 8) = ⅛ = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = ½ = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

Answer 2

Answer:

The total number of points = 8

So the total number of possible outcomes = 8

(i) Let A be the event that the number is 8

So the total number of possible outcomes for the event A is = 1

So the required probability = 1/8

(ii) Let B be the event that the number is ODD

So the total number of possible outcomes for the event A is = 3

So the required probability = 3/8

(iii) Let G be the event that the number is greater than 2

So the total number of possible outcomes for the event G is = 6

So the required probability = 6/8 = 3/4

(iv) Let H be the event that the number is less than 9

So the total number of possible outcomes for the event H is = 8

So the required probability = 8/8 = 1