Question

Contestants on a game show spin a wheel with 24 equally-sized segments. Most of those segments show different prize amounts but 2 of them are labeled 'bankrupt'. Suppose that a contestant is going to spin the wheel twice in a row and the outcome of one spin doesn't affect the outcome of future sprints - what is the probability that neither of the spins land on 'bankrupt'?

Answer 1

Explanation:

It is said, that the outcome of one spin doesn't affect the outcome of future spins. That means the spins are independent.Since they are independent we use the formulaP(A and B)=P(A)⋅P(B) P(A) is the probability the first spin doesn't land on "bankrupt" is 22/24.P(B) is the probability the second spin doesn't land on "bankrupt" is 22/24.So P(2 spins do not land on bankrupt) =2422⋅2422≈0.84. This is the answer.

Answer 2

Given:

A wheel has 24 equally-sized segments, Upon these segments 2 of them are labeled 'bankrupt'.

To find:

The probability that neither of the spins land on 'bankrupt'

Step-by-step solution:

P(not bankrupt) = 1 - P(bankrupt)

P(not bankrupt) = 1 - (2/24)

P(not bankrupt) = 22/24

Thus we can see that the probability of neither of spin land on the bankrupt is 22/24