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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)​

Sagot :

Answer:

a) [tex]\frac{1}{24}[/tex] probability that the wheel stops a "lose turn" for a single spin.

b) [tex]\frac{1}{576}[/tex] probability that the wheel stops a "lose turn" for two successive spins.

c) [tex]\frac{1}{13824}[/tex] probability that the wheel stops a "lose turn" for three successive spins.

Step-by-step 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 [tex]\frac{1}{24}[/tex] probability that it stops at lose a turn.

a.single spin

Only one spin, so:

[tex](\frac{1}{24})^1 = \frac{1}{24}[/tex]

[tex]\frac{1}{24}[/tex] probability that the wheel stops a "lose turn" for a single spin.

b.two successive spins

Two spins, so:

[tex](\frac{1}{24})^2 = \frac{1}{576}[/tex]

[tex]\frac{1}{576}[/tex] probability that the wheel stops a "lose turn" for two successive spins.

c. three successive spins

Three spins, so:

[tex](\frac{1}{24})^3 = \frac{1}{13824}[/tex]

[tex]\frac{1}{13824}[/tex] probability that the wheel stops a "lose turn" for three successive spins.

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