There is a 50% chance Walcir will visit Chile this year

Probability of visiting both Chile and Madagascar
\(P=\frac{1}{2} X \frac{1}{4}\)
\(P=\frac{1}{8}\)

Probability of not visiting Chile or Madagascar
\(P=\frac{1}{2} X \frac{3}{4}\)
\(P=\frac{3}{8}\)

In all other cases, Walcir will visit one or the other, but not both.

Probability of visiting Chile or Madagascar, but not both.
\(P=1 - (\frac{1}{8} + \frac{3}{8})\)
\(P=1 - \frac{4}{8}\)
\(P=\frac{1}{2}\)

50% Answer B

Please explain...
A= 50
B = 25

We should find out A&B .....then should add
(A - A&b) + (B-A&B) .....

i hope this is the solution approach

50/50 chances are fun because the probability of getting either answer is the same.

There are four basic possibilities here: C & M, C & ¬M, ¬C & M, ¬C & ¬M. These events are independent, so the probability of each is a straight multiplication.

C & ¬M = 0.5 * 0.75 = 0.375
¬C & M = 0.5 * 0.25 = 0.125
Add these two up and get 0.500. Or 50%.

For completion's sake, the other two probabilities are:

C & M = 0.5 * 0.25 = 0.125
¬C & ¬M = 0.5 * 0.75 = 0.375

The fact that Chile is a 50/50 makes all the math pretty straight forward. The typical question here is at least one, which would involve either calculating three probabilities or taking the compliment of visiting neither.

Hope this helps!
-Ron

Hi Ron,
Can you please tell me whats wrong if we find probability by doing: C OR M-(C&M)??

Hey Swarman, great question! A lot of people just plug in the formula and figure it will yield the right answer, but in this case you'd likely go directly to the trap answer C (without passing go or collecting 200$)

Let's look at the formula on a simple case. 2 fair coins, and we want to know if we'll get at least one Heads. We'd use the formula to say that the chance of getting heads on the first flip is 0.5, heads on the second is 0.5, and heads on both is 0.5x0.5 or 0.25. This gives us 0.5 + 0.5 - 0.25 = 0.75, which is correct. However, the question being asked here is "at least one head", which means HH is being counted in both calculations of 0.5. That's why we need to remove it afterwards, because it's being double counted, and should only be counted once.

In this example, the question is asking for exactly one or the other, which means both should be discarded. In the above coinflip example, there would be four possible outcomes: HH, HT, TH and TT. The question of at least one heads allows for HH, HT or TH. The question of exactly one heads allows for HT or TH, but not HH or TT. The question stem can dramatically change the question and send even the most experienced test taker towards a trap.

If you wanted to solve this by a formula, the same logic would apply, except it would be C or M - 2x(C&M). Logically, HH is being counted twice, when it should be counted zero times, so you must remove both instances of it from the equation. You can use this formula if you want, but the underlying logic is probably more helpful when faced with a tricky question on test day.

Hope this helps!
-Ron

Thank you so much Ron I completely got what u r trying to say,but since you mentioned that understanding of the concept is more reqd can u pls tell me if what i have deduced is right ? ''This gives us 0.5 + 0.5 - 0.25 = 0.75''.. you found this as the probability of finding Atleast one head - which could be in HT/HH +TH/HH-HH
and hence in the same question it has to be C or M - 2x(C&M) and NOT C or M -(C&M) as the latter would be STILL considering the option of both happening once.. right?

Hi swarman, exactly! C or M - (C*M) would mean you're still counting the possibility of both occurring once, which is why you have to remove it twice.

I mentioned that the understanding is more important than the formula(s) because the understanding will dictate how you use the formula(s), so you seem to be spot on here, which is great.

Try playing around with the question a little and seeing what it unlocks and what the test makers could try and lure you in with (example chances of going to Chile are 50% and Madagascar are 100%, or 0% and 50%, etc. You'll understand the problem from 360 degrees if you can feel comfortable with all the potential possibilities. That's what I try to do, but I'm a giant GMAT nerd

Hope this helps!
-Ron

After reading all posts above, i got messed up with the basic concepts.
"Let's look at the formula on a simple case. 2 fair coins, and we want to know if we'll get at least one Heads. We'd use the formula to say that the chance of getting heads on the first flip is 0.5, heads on the second is 0.5, and heads on both is 0.5x0.5 or 0.25. This gives us 0.5 + 0.5 - 0.25 = 0.75, which is correct. "
If at least one head is needed in two flips, then propabilty should be " prob of head occuring in first coin (HT--->0.5) + prob of head occuring in second coin (HT--->0.5) + prob of head occuring in both the flips HH--->0.25 ( because we need that at least one head is occurring).
So , all the three probabilties shud be added. Y do v subtract the last one...and how its like " the question being asked here is "at least one head", which means HH is being counted in both calculations of 0.5. That's why we need to remove it afterwards, because it's being double counted, and should only be counted once." ??? ....in ques it is mentioned that "AT LEAST" one head shud occur...so whats wrong in adding the third probabilty in which both heads are occurring.

Hi Perhaps,

I apologize if the example confused you, I'm trying to denote the difference between the standard question: at least one occurs and this particular question: Exactly one occurs

AT LEAST ONE

If at least one occurs, you use the formula P(1) + P(2) - P(1&2). This is because the probability of 1&2 occuring is accounted for once in P(1) and another time in P(2). Hence, it is double counted. You then have to remove one instance of both occuring in order to get the correct answer.

In the simple example of two fair coin flips. The chances of getting Heads on at least one is P(1) + P(2) - P(1&2) or 0.5 + 0.5 - 0.25, which is 0.75. Written out: HH, HT and TH work. TT does not.

EXACTLY ONE

In this question, they are asking the equivalent of getting exactly one Head. Using the same basic formula, we could just take the probability of at least one and subtract the probability of both occuring, which leaves us with P(1) + P(2) - 2xP(1&2). As previously mentioned, the case where both occurs is accounted for twice when it should be accounted for zero times.

For the coin flips we again get P(1) + P(2) - 2xP(1&2) = 0.5 + 0.5 - 2x0.25 = 0.5. Written out: HT and TH work. HH and TT don't work.


On simple examples like this you're probably better off using logic and writing out examples (until ~10 possibilities this is best), but if the exam started asking about 5 coin flips you're better served to use the formulae than writing out 32 possibilities. I hope this helped clear up any misconceptions. The same principle can be applied regardless of the percentages used, it will only change when the number of coinflips or requirements change.

Hope this helps!
-Ron

Responding to a pm:



The best way is the one that comes to your mind as long as it leads you to the answer in the stipulated time!

As for using venn diagrams here, you don't really need to as long as you understand the logic of venn diagrams (sets). There are two events - visiting Chile (1/2) and visiting Madagascar (1/4).
The probability that both may take place = (1/2)*(1/4) = 1/8 (independent events)

P(Chile or Madagascar or both) = 1/2 + 1/4 - 1/8 (from sets, we know this. We remove the extra common region once)

Now what do you do if you want to remove "both" from this? You subtract another 1/8 to totally remove the common region.
You get P(Chile only or Madagascar only) = 1/2 + 1/4 -1/8 - 1/8 = 1/2

Or another way to think about is this: P(Visiting Chile) = 1/2. From this remove the probability of both.
P(Visiting Madagascar) = 1/4. From this remove the probability of both.

P(Visiting Chile only or Visiting Madagascar only) = 1/2 - 1/8 + 1/4 - 1/8 = 1/2

(C && -M) || (-C && M)
=(0.5 * 0.75) + (0.5 * 0.25)
=0.5

this one was fun!
probability of getting to Chile is 50%.
he either gets there or not.
chances of getting to Madagascar is 25%.

so
C+ M+ => not interested
C+ M- => 1/2 * 3/4 = 3/8
C- M+ => 1/2 * 1/4 = 1/8
C- M- => not interested

now
3/8+1/8 = 4/8 = 1/2 = 50%

Solution:

Recall the formula for P(A or B) is:

P(A or B) = P(A) + P(B) - P(A and B)

The formula for P(A or B, but not both) is:

P(A or B, but not both) = P(A or B) - P(A and B)

P(A or B, but not both) = P(A) + P(B) - P(A and B) - P(A and B)

P(A or B, but not both) = P(A) + P(B) - 2 * P(A and B)

Using the above formula and assuming visiting one country is independent from visiting another, we have:

P(Chile (C) or Madagascar (M), but not both) = P(C) + P(M) - 2 * P(C and M)

P(C or M, but not both) = 0.5 + 0.25 - 2 * (0.5 * 0.25)

P(C or M, but not both) = 0.75 - 0.25

P(C or M, but not both) = 0.5

Answer: B
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W=50%
W'=50%
M=25%
M'=75%

M or W but not both = W*M' + W'*M = (0.5)(0.75)+(0.5)(0.25) = 50%
1/2*3/4 + 1/2*1/4 = 3/8+1/8 = 4/8=1/2

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