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Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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20 Jun 2008, 00:58

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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?

OG C# 231 Xavier, Yvonne, Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2, and 5/8, repectively, what is the probability that xavier and yvonne, but not zelda, will solve the problem? Please provide solution with explanation. THanks a. 11/8 b. 7/8 c. 9/64 d. 5/64 e. 3/64

P(Xavier will solve)=1/4 P(Yvonne will solve)=1/2 P(Zelda will NOT solve) = 1- 5/8 = 3/8.

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

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

I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?
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Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

I'm not great with probability so double check whatever I say, but we're not looking for the order in which X and Y get it right and Z get's it wrong. We're just looking for the probability of X and Y getting it right while Z get's it wrong. It doesn't matter in what order they get it right or wrong, all that matters is how they perform. Think of it like flipping a coin a few times and recording the probability of getting heads. It doesn't matter if you get two heads in a row then one tails, or one tails then two heads in a row. The probability is the same either way. This is in contrast to say, the chances of gettign a certain color gum ball in a certain order.

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

I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?

In this type of probability question, we're asked for a specific outcome. To solve this problem, we'll have to deal with each piece individually, then multiply the outcomes together.

We're asked for 3 things:

Xavier solves the problem Yvonne solves the problem Zelda does NOT solve the problem.

Xavier's probability to solve = 1/4 Yvonne's probability to solve = 1/2 Zelda's probability to NOT solve = 1 - 5/8 = 3/8

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

We are first given the individual probabilities that Xavier, Yvonne, and Zelda WILL solve the problem. We list these below:

P(Xavier will solve) = ¼

P(Yvonne will solve) = ½

P(Zelda will solve) = 5/8

However, we see the question asks for the probability that Xavier and Yvonne, but not Zelda, will solve the problem. Thus we must determine the probability that Zelda WILL NOT solve the problem. "Solving" and "not solving" are complementary events. When two events are complementary, knowing the probability that one event will occur allows us to calculate the probability that the complement will occur. That is, P(A) + P(Not A) = 1. In the case of Zelda, the probability that she WILL NOT solve the problem, is the complement of the probability that she WILL solve the problem.

P(Zelda will solve) + P(Zelda will not solve) = 1

5/8 + P(Zelda will not solve) = 1

P(Zelda will not solve) = 1 – 5/8 = 3/8

Now we can determine the probability that Xavier and Yvonne, but not Zelda, will solve the problem. Since we need to determine three probabilities that all must take place, we must multiply the probabilities together. Thus we have:

P(Xavier will solve) x P(Yvonne will solve) x P(Zelda will not solve)

¼ x ½ x 3/8

1/8 x 3/8 = 3/64

The answer is E
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Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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22 Jul 2017, 08:33

EMPOWERgmatRichC wrote:

Hi All,

In this type of probability question, we're asked for a specific outcome. To solve this problem, we'll have to deal with each piece individually, then multiply the outcomes together.

We're asked for 3 things:

Xavier solves the problem Yvonne solves the problem Zelda does NOT solve the problem.

Xavier's probability to solve = 1/4 Yvonne's probability to solve = 1/2 Zelda's probability to NOT solve = 1 - 5/8 = 3/8

The question does not state anything about what the "order" of the outcomes must be - just the probability of each of the individual outcomes (and the specific 'overall' outcome that we're looking to calculate). Thus, it's not a permutation question (meaning that it does NOT matter who attempts to solve the problem first, second or third) - and we don't have to do anything besides multiply the individual probabilities together to get the correct answer.

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

P(Z solves the problem) = 1 - P(Z doesn't solve the problem) So, 5/8 = 1 - P(Z doesn't solve the problem) So, P(Z doesn't solve the problem) = 3/8

The question asks us to find P(Xavier and Yvonne solve problem, but Zelda does not solve problem) So, we want: P(X solves problem AND Y solves problem AND Z does not solve) = P(X solves problem) x P(Y solves problem) x P(Z does not solve) = 1/4 x 1/2 x3/8 = 3/64

Re: Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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05 Nov 2017, 01:09

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if x can solve with probability of 25 % and Y can at 50 % then combined together how come it is lesser than both????????? ie here 1/8.. i cannot feel this physically . could somebody explain this??

Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]

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08 Nov 2017, 10:42

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Cheryn wrote:

if x can solve with probability of 25 % and Y can at 50 % then combined together how come it is lesser than both????????? ie here 1/8.. i cannot feel this physically . could somebody explain this??

Cheryn, maybe what follows will help, because intuition is important: Think about how restrictively success is defined. (Success = desired outcome = win = passing the test.)

The probability of "success" here is restrictive because BOTH have to pass. If only X OR Y had to pass, success would be easier.

To win, "BOTH this AND that must happen." That is a narrower definition of success than "to win, EITHER this OR that must happen."

Look at the difference: if the question were "What is the probability of X or Y passing the test?" The answer: \(\frac{1}{2} + \frac{1}{4}=\frac{3}{4}\) One or the other must pass? Easier, less restrictive than "both must pass."

Different scenario, but it works exactly the same way.

A coin toss. What is the probability that one coin, flipped twice, will land on tails both times? Success = tails on the first flip AND on the second flip

P (tails) on the first flip is \(\frac{1}{2}\) P (tails) on second flip = \(\frac{1}{2}\)

Events are independent. Multiply:\(\frac{1}{2}* \frac{1}{2}=\frac{1}{4}\)

The probability of having both flips come up tails, \(\frac{1}{4}\), is less than the probability of having just one flip come up tails (\(\frac{1}{2}\)).

Again, that is because success is defined more restrictively. It is harder to get two tails on two flips than it is to get one tail on two flips; you have to "beat the odds" twice, not once. Lower probability.

Most probabilities are fractions between 0 and 1. When those fractions are multiplied, they get smaller. That fits.

It can be a little counterintuitive if you focus on AND. "And" might seem as if it should produce a better chance of success than "or." Maybe focus instead on: the definition of success, and how success is achieved.

Almost always, (BOTH must win) will be harder (lower probability) than (ONE OR THE OTHER must win).