Problem Solving

Xavier, Yvonne, and Zelda each try independently to
solve a problem. If their individual probabilities for
success are 1/4, 1/2, 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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = 3/8 = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?
so using the probability OR formula
3/8 = 1/4 + 1/2 - P(Xavier and Yvonne)
P (X&Y) = 6/8 - 3/8 = 3/8
What am i doing wrong?

Thanks in advance for your help

rchetan_18 wrote:Xavier, Yvonne, and Zelda each try independently to
solve a problem. If their individual probabilities for
success are 1/4, 1/2, 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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?

Thanks in advance for your help

The portion in red is incorrect.
Xavier, Yvonne and Zelda each try INDEPENDENTLY to solve the problem.
The probability that Zelda doesn't solve the problem does NOT depend on whether Xavier or Yvonne solves the problem.
It is entirely possible that ALL 3 solve the problem.
Thus, it is not necessarily true that P(Z doesn't solve the problem) = P(X or Y solves the problem).

this seemed easy... can any one tell me which level is this ... below 600 ? :/

finished it in 49 seconds... all the probability is given...

it's just P {X & Y & (not Z)}
= 1/4 * 1/2 * (1-5/8)
=1/4 * 1/2 * 3/8

3/64

Hi,

When I solved this problem I multiplied the result with 3 to consider the 3 different sequences in which this task can be done. But I now realize that the different orders are not required. How is this problem different from other similar problems which require us to do so ? Is it correct to say that for ALL independent events we do not consider the order ?

Consider the example where we have 3 Blue balls and 2 Green balls. If I have to take 2 Blue and 1 Green balls at random then the probability is 3/5 x 2/4 + 2/5 x 3/4.
Is it because this scenario has dependent events ?


Thanks
Karthik

GMATGuruNY wrote:

rchetan_18 wrote:Xavier, Yvonne, and Zelda each try independently to
solve a problem. If their individual probabilities for
success are 1/4, 1/2, 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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?

Thanks in advance for your help

The portion in red is incorrect.
Xavier, Yvonne and Zelda each try INDEPENDENTLY to solve the problem.
The probability that Zelda doesn't solve the problem does NOT depend on whether Xavier or Yvonne solves the problem.
It is entirely possible that ALL 3 solve the problem.
Thus, it is not necessarily true that P(Z doesn't solve the problem) = P(X or Y solves the problem).