Question

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64

Answer 1

Answer: The probability that Xavier and Yvonne can solve a problem but Zelda cannot is
`(3)/(64)`

Explanation:

We are given:

Probability of success of Xavier,
`P(S)_(Xavier)=(1)/(4)`

Probability of failure of Xavier,
`P(F)_(Xavier)=1-(1)/(4)=(3)/(4)`

Probability of success of Yvonne,
`P(S)_(Yvonne)=(1)/(2)`

Probability of failure of Yvonne,
`P(F)_(Yvonne)=1-(1)/(2)=(1)/(2)`

Probability of success of Zelda,
`P(S)_(Zelda)=(5)/(8)`

Probability of failure of Zelda,
`P(F)_(Zelda)=1-(5)/(8)=(3)/(8)`

We need to calculate:

The probability that Xavier and Yvonne can solve the problem but Zelda cannot, we use:

`P(S)_(Xavier)* P(S)_(Yvonne)* P(F)_(Zelda)=(1)/(4)* (1)/(2)* (3)/(8)=(3)/(64)`

Hence, the probability that Xavier and Yvonne can solve a problem but Zelda cannot is
`(3)/(64)`