A hiking club’s trip will be cancelled if, on the first day of the tri

P(A) Cold = 15%
P(B) Snow = 40%

P(AuB) = 55% Snow and cold are not mutually exclusive. It can be too cold and snow simultaneously. Remove this overlap
P(AnB) = During the 40% chance of snow, 15% of the time it will be under 20 degrees. Remove 15% of 40% or 6%. (.15 * .4 = .06)
P(AuB) = P(A) + P(B) - (AnB) = 49% chance of cancellation

Although I don't like this problem, it's unrealistic.
If it's already going to snow, if that 40% is going to happen then the 15% chance of cold has already drastically improved its chances.
That 15% cold probability is taking into account whatever the entire forecast is that day lets say the 15% chance is based on reports that it could be anywhere from 15 degrees to 40 degrees. If the 40% snow is going to happen then its automatically under 32 degrees and the chance of cold has shifted from 15% to 48% chance.

You would need to limit the problem by saying its automatically cold enough to snow and that rain isn't an option.
Same thing the other way around if its already the 15% cold hasn't the snow changes improved?
Was the likelihood of the 40% weather snow forecast contingent on it being cold enough (100% chance of snow or rain, X% chance of too cold) or climate weather (x% chance of snow or rain, 100% chance of too cold)

We can use the equation:

P(trip being canceled) = 1 - P(trip not being canceled)

P(trip not being canceled) = 0.85 x 0.6 = 0.51

So, P(trip being canceled) = 1 - 0.51 = 0.49, or 49%.

Answer: B

Hey Bunuel / JeffTargetTestPrep

Need your help with the below approach
P(temp < 20) = P(A)
P(snow) = P(B)

P(A) = 15%
P(B) = 60%
P(A n B) = 15% x 60%
= 9%

The probability of trip being cancelled = P(A) + P(B) - P(A n B)
= 15% + 60% - 9%
= 66%

Can you let me know what is wrong in my approach

P(a) + P(b) - P(aUb) = 49%

Nothing wrong with the approach, just that 60% is the P(No Snow).
So in your calculations P(B) has to be 40%

P(A n B ) = 6%

P (Trip cancelled) = 55/100 - 6/100
P(Trip Cancelled) = 49%

The question is looking for the likelihood in which the trip will be canceled.
A: Temp less than 20 = 15%
B: Snowing = 40%

not A: Temp not less than 20 = 85%
not B: Not Snowing = 60%

The occurrence of A & B are parallel and Happening of either of them will cancel the trip.
The trip is going to be canceled under the following conditions:
A*B + A*(not B) + (not A)*B :
15% * 40% + 15% * 60% + 85% * 40% = 49%

Shortcut method:
Find the probability in which trip will happen and subtract from the total set of events, i.e., 1 - (when trip won't be canceled) : 1 - (not A)*(not B)

I think this question again reminds the fact to read the question very thoroughly..realized when i did not read through the lines and ended up messing it up.

This is how i solved it (after realizing my mistake )

this could be solved using the formula P(A OR B) = P(A) + P(B) - P(A and B)

Now, lets assume P(A) = T<20 degrees
P(B) = snowfall (Question stem gives info of NOT snowing)

So P(A) = 15/100 = 3/20
P(B) = 40/100 = 2/5

P(A OR B) = 3/20 + 2/5 - (3/20*2/5)= 49/100 which is 49%
So Answer = B

Why is P(A & B) = P(A)*P(B)? This simplification can only be used if A and B are independent? But how do we know that is the case here?

­Agreed. No way of knowing if they are independent. In real life, they are not independent, so I think this question is flawed as the answer could be A or B or C given the min and max of (<20 and Snow) being 0% and 15%, respectively.

Bunuel JeffTargetTestPrep gmatophobia

Why is it not 55%?
P ( Temperatures below 20 degrees ) = 15%
P ( Snow ) = 40%
P ( Temperatures below 20 degrees or Snow ) = 15% + 40% = 55%

 

­
iishim

We have also to consider a scenario when both events occur. That's a common overlap and we need to subtract that once 

Probability of both = 0.15 * 0.40 = 0.06 = 6%

Subtract this from 55%

55 - 6 = 49%