Question

A fair spinner with 5 sides numbered 1, 2, 3, 4, 5 is spun repeatedly. The score on each spin is the number on the side on which the spinner lands.

(a) Find the probability that a score of 3 is obtained for the first time on the 8th spin.
(b) Find the probability that fewer than 6 spins are required to obtain a score of 3 for the first time.

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

A fair spinner has five sides and each side is numbered from 1,2,3,4,5

Probability of getting 3 = 1/5

Probability of not getting 3 = 4/5

a) probability that a score of 3 is obtained for the first time on the 8th spin:

We need 3 on the 8th spin means 3 should not come in the first 7 spins.

and 3 comes on the 8th spin.

so the probability will be (4/5)× (4/5)× (4/5)×(4/5) ×(4/5)× (4/5)× (4/5)×(1/5)

                                     =$\frac{4^7}{5^8}$

b) probability that fewer than 6 spins are required to obtain a score of 3 for the first time:

We have to find the probability of getting 3 in less than 6 spins means it should come in first or second or third or fourth or fifth spin.

So we have to find each probability and add them,

probability of getting 3 in first attempt, P(1) = 1/5

probability of getting 3 in second attempt, P(2) = (4/5)x(1/5)

probability of getting 3 in third attempt, P(3) = (4/5)²x(1/5)

probability of getting 3 in fourth attempt, P(4) = (4/5)³x(1/5)

probability of getting 3 in fifth attempt, P(2) = (4/5)⁴x(1/5)

overall probability = P(1) +  P(2) +  P(3) +  P(4) +  P(5)

                              =(1/5) +(4/5)x(1/5) + (4/5)²x(1/5) + (4/5)³x(1/5) + (4/5)⁴x(1/5)

                              =(1/5){1+ (4/5) + (4/5)² + (4/5)³ + (4/5)⁴}

                              =(1/5)[{1 - (4/5)⁵}/{1 - (4/5)}]

                              = (5⁵-4⁵)/(5⁵)

                              =2101/3125