A sample space contains three outcomes, A, B, and C. Which of the following could be a legitimate assignment of probabilities to the outcomes?
A) $P ( \mathrm {~A} ) = 0.2 \quad P ( \mathrm {~B} ) = 0.4 \quad P ( \mathrm { C } ) = 0.6$
B) $P ( \mathrm {~A} ) = 0.2 \quad P ( \mathrm {~B} ) = 0.2 \quad P ( \mathrm { C } ) = 0.6$
C) $P ( A ) = 2 \quad P ( B ) = 3 \quad P ( C ) = 1$
D) $P ( \mathrm {~A} ) = 0.3 \quad P ( \mathrm {~B} ) = 0.3 \quad P ( \mathrm { C } ) = 0.3$

Question

Quizplus · Accepted Answer

The sum of the probabilities of all the outcomes must equal 1. Therefore, choice A and D cannot be legitimate since the sum is not equal to 1. Additionally, in choice C, the probabilities are not in the range of 0 to 1. Therefore, the only legitimate choice is B where the probabilities sum up to 1 and are within the range of 0 to 1.

A Sample Space Contains Three Outcomes, A, B, and C P( A)=0.2P( B)=0.4P(C)=0.6P ( \mathrm {~A} ) = 0.2 \quad P ( \mathrm {~B} ) = 0.4 \quad P ( \mathrm { C } ) = 0.6P( A)=0.2P( B)=0.4P(C)=0.6

A Sample Space Contains Three Outcomes, A, B, and C $P ( \mathrm {~A} ) = 0.2 \quad P ( \mathrm {~B} ) = 0.4 \quad P ( \mathrm { C } ) = 0.6$