There is a 50% chance Jen will visit Chile this year, while there is a

Our desired probability would be
(Probability of visiting Chile) * (Probability of not visiting Madagascar) + (Probability of not visiting Chile) * (Probability of visiting Madagascar)
= \(\frac{1}{2}*\frac{3}{4} + \frac{1}{2}*\frac{1}{4}\) \(=\frac{4}{8}\) \(=\frac{1}{2}\)
= 50 %

Answer:-B

This question was written to help people understand the use of sets in probability. Think of the venn diagram with two overlapping sets.

n(A or B) = n(A) + n(B) - n(A and B) ........(I)
(We subtract n(A and B) because it is double counted. The overlapping part was counted twice, once in each circle so it is subtracted out.)

n(A or B but not Both) = n(A) + n(B) - n(A and B) - n(A and B) ............(II)
(We subtract another n(A and B) so that 'Both' is subtracted out. The overlapping part was counted once in (I); now it is totally removed)

P (visit C or M but not Both) = 1/2 + 1/4 - (1/2*1/4) - (1/2*1/4) = 1/2

Answer (B)
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\(\frac{1}{2}+\frac{1}{4}-\frac{1}{2}*\frac{1}{4}*2\)=\(\frac{1}{2}\)..
B
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The thing to note in the question is she would not be visiting both.

Thus p(Chile)*(1-p(Madagascar))+p(Madagascar)*(1-p(Chile))
which is
(1/2*3/4) + (1/4*1/2),
3/8*1/8=1/2

1 - (the probability of choosing neither) - (probability of choosing both)

Probability of choosing neither = 1/2 * 3/4 = 3/8
Probability of choosing both = 1/2 * 1/4 = 1/8

1 - 3/8 - 1/8 = 4/8 or 1/2

B

VERITAS PREP OFFICIAL SOLUTION:

Putting aside her peculiar travel schedule for a moment, (visiting Western South America and South East Africa without stopping by the Falklands?) what is the probability that Jen travels to one of these locales but not the other? Dutifully applying our memorized probability formula: the probability of visiting Chile is 0.5, and the probability of visiting Madagascar is 0.25, we get 0.5 + 0.25 – (0.5*0.25) or 0.75 – 0.125 = 0.625. Or answer choice C. (As Admiral Ackbar tried to warn us in 1983: It’s a Trap!)

But why doesn’t the formula yield the correct choice? The answer lies in the question stem. This particular question asks us the probability of visiting Chile or Madagascar, but not both. The formula gives us the probability of visiting either, implicitly allowing the choice of visiting both. The probability of visiting either will indeed be 62.5% (or 5/8), but this is the probability of visiting Chile, Madagascar, or both. Verifying the converse, the probability of visiting neither is P(¬Chile) and P(¬Madagascar), or 0.5 * 0.75 = 0.375 (or 3/8), confirming that our 0.625 is merely the probability of not staying home this year.

The obvious question now is: Why doesn’t the formula work? Didn’t the formula already account for the possibility of both? How do I solve this question correctly? (Admittedly, these are three questions and not one). The key to answering all of them is the same, though. Let’s go through the logic of the formula P(C) + P(M) – P(C&M):

The first argument allows for all possibilities of visiting Chile, regardless of what happens with Madagascar.

The second argument allows for all possibilities of visiting Madagascar, regardless of what happens with Chile (or the Falklands)

The third argument is the possibility of both occurring.

The formula works because P(C) accounts for the both choice, and P(M) accounts for the both choice as well, indicating that this option has been double counted. In order to count it only once, we need to remove one instance of it. This is why the formula works and is popular; it addresses the inherent problem in probability, double counting certain situations. The same logic applies to Venn diagrams and other similar question types where counting the same argument twice (or thrice) can occur.

In practice, this question could then be solved using the default formula of P(C) + P(M) – P(C&M) and then subtracting P(C&M) again… or simply P(C) + P(M) – 2*P(C&M). Simply put, the first two arguments in the formula account for the “both” possibility twice, so we must remove it twice to answer the question at hand.
A somewhat similar yet more straight forward alternative is to go bottom up instead of top down. The probability of going only to Chile is P(C) * P(¬M) = 0.5 * 0.75 = 0.375. The probability of going only to Madagascar is P(M) * P(¬C) = 0.25 * 0.5 = 0.125. Adding those two probabilities together yields the correct answer of 0.5 or 50%. The revised formula also yields this result (0.5 + 0.25 – 2 (0.125) = 0.5), so we can feel confident in our answer choice and not any of the other tempting answer choices.

The answer is (B).
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I have a question about the P(A or B) formula.

So does this mean that by using the P(A) + P(B) - P(A and B) we are not subtracting the overlap to get the "pure" "A or B and not A and B at the same time" but rather we are simply correcting a mathematical error that otherwise would have occurred if we didn't subtract the overlap of "A and B" to get "A or B"?

Another observation is that in quotidian language when we say "A or B" we mean "or A, or B, but not both". eg: "I will have wine or beer" so here I want to say that I will have beer or wine, but not both. So I am implicitly excluding the possibility of having both.

However in the context of probability, I would have to be more explicit if wanted to say "A or B, but not both"

So by simply saying "A or B" in the context of math, we are still leaving in the possibility of having both A and B in contrast of saying the same thing in the quotidian language. To completely eliminate the possibility of "A and B" happening, we would have to first make the correct mathematical calculation by P(A) + P(B) - P(A and B) to get the "A or B and maybe both" expression and then subtract the overlap ones again to actually remove it for good and get the "isolated" "A or B but not both".

Am I correct here? And it seems that after so many years I finally understood the meaning of the P(A) + P(B) - P(A and B) formula...

Another question: is the P(A) + P(B) - P(A and B) and "At least one" so 1- P(of what we want to happen not happening) the exact same thing?

Thank you!

There are 4 total possibilities: (M,C) ; (M) ;(C) ;(No visit to any of two). Now Jen visiting M or C but not both is possible in only 2 of the 4 scenario. So the chances are 50%.

Hi,

Remember that the OVERLAP is part of both P(A) and P(B)....
so for removing the repetition, we subtract P(A and B) once...
this gives us P(A) or P(B)... ofcourse this will include the overlap..

when you have to completely remove the overlap..
remove it from P(A) and P(B),
so P(A) - P(A and B) + P(B) - P(A and B) = P(A) + P(B) - 2P(A and B).....
so our answer becomes P(A) + P(B) - 2P(A and B) = \(0.5+0.25 - 2(0.5*0.25) = 0.75 - 2(0.125) = 0.75 - 0.25 = 0.5\)... or 50%

Hope it helps you
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Chetan,

Thank you so much for the clarification! I was confused with the meaning of the formula..

Cheers!

This explanation really made sense for me!

There are two different scenarios that satisfy the requirement:

(1) She visits Chile but NOT Madagascar. The probability of this event is 1/2 x 3/4 = 3/8.

(2) She doesn’t visit Chile but she DOES visit Madagascar. The probability is 1/2 x 1/4 = 1/8.

We add the probabilities, obtaining 3/8 + 1/8 = 4/8 = ½, or 50%.

Answer: B
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A tip: Use Venn diagrams in this type of questions. Your life will become much easier.

Hi Hero8888, would you please draw the Venn diagram and suggest the logic? I want to ensure that I understood the point. Thank you

Posted from my mobile device

Here we go

.

I have a question, basis the same logic, let's say, There is a 50% chance Jen will win the Gold Medal, while there is a 25% chance that Barry will win the Gold Medal. What is the probability that either Jen will win the Gold Medal or Barry will win the Gold Medal, but NOT both?


Now, what would be the solution to this problem? 0.5 or 0.625? And why?

VeritasKarishma could you please help? I'm quite confused here.

Assuming that Jen winning the Gold and Barry winning the Gold are independent events, the probability of either Jen winning or Barry winning but not both winning is 0.5 (exactly like the logic above)

If we were asked the probability of Jen or Barry winning the Gold (Both could also win), then probability = .625


But if Jen and Barry are running the same race and hence their winning the Gold are not independent events (only one of them can win the Gold)
P(Both) = 0
P(Jen or Barry) = 0.5 + 0.25 = 0.75
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Assuming that Jen winning the Gold and Barry winning the Gold are independent events, the probability of either Jen winning or Barry winning but not both winning is 0.5 (exactly like the logic above)

If we were asked the probability of Jen or Barry winning the Gold (Both could also win), then probability = .625


But if Jen and Barry are running the same race and hence their winning the Gold are not independent events (only one of them can win the Gold)
P(Both) = 0
P(Jen or Barry) = 0.5 + 0.25 = 0.75[/quote]

Couldn't have found a better explanation. Thank you very much, VeritasKarishma. Much appreciated.

What's the difference between this question and the following question: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?

VeritasKarishma Bunuel ScottTargetTestPrep