Question

A game is played in which you spin a 10-segment spinner as shown above. All segments are the same size. Find the probabilities below.

(a) Find the probability that you spin 4:

P(spin 4) = (round to one decimal place)

(b) Find the probability that you spin either 9 or 10:

P(spin 9 or 10) = (round to one decimal place)

(c) X is a binomial random variable. Suppose we define spinning 9 or 10 as "success", and we decide to spin the spinner 5 times.
(i) X = number of successes in trials.
(ii) What outcomes would be considered "failure"?

1 or 2 1, 2, 3, 4, 5, 6, 7, or 8     8, 9, or 10 1, 2, or 3 7, 8, 9, or 10

(iii) What is p, the probability of success in a single spin? p = (round to one decimal place)
(iv) What is the probability of failure in a single spin? (round to one decimal place)

(d) Using the binomial distribution formula, find the probability of seeing 2 successes in 5 spins.

P(X = 2) = (round to two decimal places)
Note: 2 successes in 5 trials would mean that on two our of the five spins, you would get either a 9 or 10 on the spinner. On the other three spins you would get any of the other results (1 to 8).

(e) Find the probability of seeing 0 successes in 5 spins.

P(X = 0) = (round to two decimal places)
Note: Zero successes in five trials would mean that on each of the five spins, you would not see either a 9 or 10.

Answer 1

a) Since there are 10 segments and each no occurs with equal probability since the segments are of the same size, then Probability of getting a 4= 1/10 = 0.1

b) P(Spin 9 or 10) = P(spin 9) + P(spin 10) = 0.1+0.1= 0.2

c) I) X= number of success ie getting 9 or 10 in the 5 trials.

ii) Failure: Getting 1,2,3,4,5,6,7 or 8

iii) Probability of success in a single spin= 0.2 (found in b above)

iv) Probability of failure= 1- Probability of success= 0.8

d) Pdf of a Binomial(5,0.2) distribution is:

P(X=x) = 5Cx*(0.2)^x*(0.8)^5-x

P(X=2) =5C2*(0.2)^2*(0.8)^3 = 0.2048

e) P(X=0) 5C0*(0.8)^5 = 0.32768