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-

Probability fo solving specific problem independently by A and B are $\frac{1}{2} a n d \frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that
Exactly one of them solves the problem

Q. A problem in statistics is given to solve to three students independently. Their chances of solving problem are

\frac{1}{3}, \frac{1}{4}, \frac{1}{5}

respectively. Find the probability that the problem is solved.

Q. Probability of solving specific problem independently by

A a n d B a r e \frac{1}{2} a n d \frac{1}{3}

respectively. If both try to solve problem independently, find the probability.

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Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 14,12 and 58, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

The correct option is C 364P(Xavier will solve)=14P(Yvonne will solve)=12P(Zelda will NOT solve) =1−58=38.Now, we need to multiply all this Ps to find an answer:p=(14)∗(12)∗(38)=364.

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

The correct option is C $\frac{3}{64}$
P(Xavier will solve) $= \frac{1}{4}$
P(Yvonne will solve) $= \frac{1}{2}$
P(Zelda will NOT solve) $= 1 - \frac{5}{8} = \frac{3}{8} .$

Now, we need to multiply all this Ps to find an answer:
$p = (\frac{1}{4}) * (\frac{1}{2}) * (\frac{3}{8}) = \frac{3}{64} .$