Question

the wheel on the tv program wheel of fortune is first divided equally into 24 compartments with one labeled"lose a turn" find the probability that the wheel stops a "lose turn" for each of the following a.single spinb.two successive spins c. three successive spins (complete solution)​

Answer 1

a)
`(1)/(24)` probability that the wheel stops a "lose turn" for a single spin.

b)
`(1)/(576)` probability that the wheel stops a "lose turn" for two successive spins.

c)
`(1)/(13824)` probability that the wheel stops a "lose turn" for three successive spins.

Explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

24 components on the wheel, one of which is "lose a turn"

This means that each time the wheel stops, there is a
`(1)/(24)` probability that it stops at lose a turn.

a.single spin

Only one spin, so:

`((1)/(24))^1 = (1)/(24)`

`(1)/(24)` probability that the wheel stops a "lose turn" for a single spin.

b.two successive spins

Two spins, so:

`((1)/(24))^2 = (1)/(576)`

`(1)/(576)` probability that the wheel stops a "lose turn" for two successive spins.

c. three successive spins

Three spins, so:

`((1)/(24))^3 = (1)/(13824)`

`(1)/(13824)` probability that the wheel stops a "lose turn" for three successive spins.