Mathematics: 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

8.5 and 12.5. when you try to find a number that lies in between two numbers. add the two uo then divide by 2...

Xavier, yvonne, and zelda each try independently

samueldfhung

15.01.2022

The probability that Xavier and Yvonne can solve a problem but Zelda cannot is $\frac{3}{64}$

Step-by-step explanation:

We are given:

Probability of success of Xavier, $P(S)_{Xavier}=\frac{1}{4}$

Probability of failure of Xavier, $P(F)_{Xavier}=1-\frac{1}{4}=\frac{3}{4}$

Probability of success of Yvonne, $P(S)_{Yvonne}=\frac{1}{2}$

Probability of failure of Yvonne, $P(F)_{Yvonne}=1-\frac{1}{2}=\frac{1}{2}$

Probability of success of Zelda, $P(S)_{Zelda}=\frac{5}{8}$

Probability of failure of Zelda, $P(F)_{Zelda}=1-\frac{5}{8}=\frac{3}{8}$

We need to calculate:

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

$P(S)_{Xavier}\times P(S)_{Yvonne}\times P(F)_{Zelda}=\frac{1}{4}\times \frac{1}{2}\times \frac{3}{8}=\frac{3}{64}$

Hence, the probability that Xavier and Yvonne can solve a problem but Zelda cannot is $\frac{3}{64}$

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