Xavier, Yvonne, and Zelda each try independently to solve a

Yes - since they are independent events, you can multiply the probabilities together.

Be sure to remember to take the opposite probability for that last one.

Please see: https://www.gmatpill.com/gmat-practice-t ... stion/2387

[youtube]https://www.youtube.com/watch?v=ubENfsH0tLU[/youtube]

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

Hi Rich,

Could you please tell me why dont we multiply this by 3! ?
(X)(Y)(not Z)
(not Z)(X)(Y)....so on till 6 times.

Please clarify.

Thanks

Hi Sash143,

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.

GMAT assassins aren't born, they're made,
Rich

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 x 3/8
= 3/64

Answer:

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 strictly or 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 stricter definition of success than "to win, EITHER this OR that must happen."

In the first case, two people must succeed. In the second case (OR), only one person OR the other person must succeed. It's harder to get two wins than it is to get one win.

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 lower (smaller) 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).

Hope that helps.

Greetings

can anyone explain whats wrong with the following solution:

multiplied 1/4 * 1/2 * 5/8 = 5/64

than i thought we need to subtract 5/8 from 5/64

isnt it locically correct ?

thank you !

Hi dave13!

Were you going for an approach of calculating 1 - the probability?
In that case you would have needed to calculate (Xavier and Yvonne succeed) - (Xavier and Yvonne and Zelda succeed).
Which is to say 1/4*1/2 - 1/4 * 1/2 * 5/8 = 1/8 - 5/64 = 3/64, answer (E).

What you wrote above was (Xavier and Yvonne and Zelda succeed) - (Zelda succeeds).
This doesn't make sense because it is negative, but even if it were reversed then it would have given you a different result:
(Zelda succeeds) - (Xavier and Yvonne and Zelda succeed) is the probability that (Zelda succeeds and (one or more of Xavier and Yvonne fail))

Pay attention to what is your "1" and what is your "probability" when you calculate "1 - probability"!

Hope that helps.

CrackVerbalGMAT Can you please explain why we do not multiply the product of the individual events with the arrangement? (since the order is not specified here)

If I'm interested in the probability that, when flipping a coin twice in a row, I get a different result each time... I don't care about the order. These events are independent. This is essentially the same situation as the problem above.

P(Heads) = 1/2
P(Tails) = 1/2

By the arguments given above by numerous people, since we don't care about the order of the result, the answer should be P(H)*P(T) = 1/4

And this is absolutely wrong. The correct answer is P(H)*P(T) + P(T)*P(H) = 1/2 ... precisely BECAUSE we don't care about the order.

The answer to this problem should be (1/4)*(1/2)*(3/8)*3! = 9/32

What am I missing?

Hello,

What is the event for which we are trying to find the probability here? X and Y should be able to solve the problem while Z should not be able to solve the problem. As such, you need to worry about how considering the order (or NOT considering, for that matter) will affect the number of ways in which the event can be accomplished.

Remember, in a topic like P&C and Probability, you count a certain way of doing an activity/task/event only if it is distinct and hence has a bearing on the total number of ways.

Imagine that you have X, Y, and Z sitting in front of you and you gave them this problem to solve. You also are aware of the probabilities of X, Y and Z will solve the problem (and not solve the problem). Now, if Z got the wrong answer first and X got the right answer last, will it be different from X getting the right answer first and Z getting the wrong answer last? It wouldn’t be, isn’t it?

The only outcome that you have to be worried about is that X and Y should be able to solve the problem while Z should not be able to solve the problem. And that can happen only in 1 way.

Therefore, the required probability = 1/4 * 1/2 * 3/8 = 3/64.

Hope that helps!
Arvind.­

If Zelda is not participating, why can't we just eliminate her and have 1/4*1/2?
What can we get if we just do 1/4*1/2?

Hi graceo,

The first sentence tells us that ALL 3 people are trying to solve the problem (independently of one another) - so it's NOT correct to assume that Zelda is not participating. We're asked for a specific Probability calculation and we need 3 specific things to happen: X solves it, Y solves it and Z does NOT solve it. The calculation has to account for all 3 'events' - and your calculation accounts for just the first 2 events.

GMAT assassins aren't born, they're made,
Rich

Hi GMATters,

Here is my video solution to this problem:

Bunuel can you answer this? No one that I have gotten an answer from has given me a clear answer. Thanks.

­

 

My approach is slightly different:

1. The basic definition of probability of anything is: Ratio of (desired outcomes) to (all possible outcomes).­
2. For discrete events, we can count the number of outcomes in the numerator and denominator manually.
3. This method is especially useful when we're dealing with relatively small numbers of outcomes.
4. The core principle to keep in mind while counting outcomes is that the ORDER ALWAYS MATTERS.
5. Accordingly, for the 2-coin scenario, all possile outcomes are: HH, HT, TH, TT.
   Notice that TH and HT are counted as two different outcomes irrespective of the fact that the outcomes are independent.
   So, denominator = 4.
6. The desired outcome needs to very precisely interpreted/defined from the question stem.
   Some examples:
   
   1. Probability of getting both heads: Only one outcome matches this desired outcome, so probability = 1/4
   2. Probability of getting exactly one tail: Two out of the four outcomes match this desired outcome, so probability = 2/4 = 1/2
   3. Probability of getting at least one head: Three out of the four outcomes match this desired outcome, so probability = 3/4

Hope this helps.­