Question

a wheel contains eight slices, numbered sequentially 1 through 8. the probability of landing on each slice is equal. if the wheel is spun three times, what is the probability that the first spin is 4 or less, the second spin is 6 or less, and the third spin is 8 or less?

Answer 1

4 or less: 4/8 so 1/2
6 or less: 6/8 so 3/4
8 or less: 8/8 so 1

Answer 2

Answers:

1.  $\frac{1}{2}$

2.  $\frac{3}{4}$

3. 1

Probability:

The probability equation of an event E , P(E) = n(A) / n(S)

P(A) is the probability of an event “A” n(A) is the number of favourable outcomes. n(S) is the total number of events in the sample space.

Here, (Given)

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = P(7) = P(8) = 1 / 8

1. Probability that first spin is 4 or less = n(A) / n(S)

Here n(S) is the total number of slices = 8

n(A) is the number of favourable outcomes (i.e., it can be 1, 2, 3 or 4 ) = 4

P = 4 / 8 = 1 / 2

2.  Probability that second spin is 6 or less = n(A) / n(S)

Here n(S) is the total number of slices = 8

n(A) is the number of favourable outcomes (i.e., it can be 1, 2, 3, 4, 5 or 6 ) = 6

P = 6 / 8 = 3 / 4

3. Probability that third spin is 8 or less = n(A) / n(S)

Here n(S) is the total number of slices = 8

n(A) is the number of favourable outcomes (i.e., it can be 1, 2, 3, 4, 5, 6, 7 or 8 ) = 8

P = 8 / 8 = 1 / 1 = 1

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