14 randomly chosen students have each won a free ticket to play a game of chance....

14 randomly chosen students have each won a free ticket to play a game of chance. In this game a wheel is spun that has been equally divided among 20 values, and when the wheel stops spinning a pointer will have selected one of the 20 values randomly (think Wheel-of-Fortune style). So therefore we will assume that each of the 20 outcomes is equally likely. Before spinning, the player chooses which value they think it will stop on. If it stops on the value chosen, they win $300. If it stops on the position to the left or right of the value chosen, they still win $100 for getting close. Otherwise, they win nothing. What is the probability that exactly 3 of the students will win some money? Leave your answer as a decimal.

Solutions

Expert Solution

Number of students randomly chosen to play a game of chance = 14

Number of values equally divided on the wheel = 20

Total number of outcomes =20

Player will win some money, if it stops on the value chosen or if it stops on the position to the left or right or the value chosen.

Number of outcomes that favor the Player will win some money= 3

p: Probability that a player will some money

= Number of outcomes that favor the Player will win some money/ Total number of outcomes = 3/20=0.15

X : Number of the students will win some money

X follows Binomial distribution with n= 14 and p =0.15

Probability mass function of X is given by :

Probability that 'r' students will win some money = P(X=r)

probability that exactly 3 of the students will win some money = P(X=3)

probability that exactly 3 of the students will win some money = 0.205581174872691

orchestra answered 1 year ago

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