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A hiking club’s trip will be cancelled if, on the first day of the tri
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17 Apr 2018, 04:22
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A hiking club’s trip will be cancelled if, on the first day of the tri
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17 Apr 2018, 04:49
Bunuel wrote: A hiking club’s trip will be cancelled if, on the first day of the trip, the temperature is less than 20 degrees Fahrenheit or it is snowing. If there is a 15% chance that temperatures will be below 20 degrees and a 60% chance that it will not snow, what is the likelihood that the trip will be cancelled?
A. 40% B. 49% C. 55% D. 66% E. 75% Probability of temp at or above 20 degrees = 100%15% = 85% The likelihood that the trip will NOT be cancelled = prob of temp at or above 20 degrees * prob that it will not snow = 85% * 60% = 51% The likelihood that the trip will be cancelled = 100%  The likelihood that the trip will NOT be cancelled = 100% 51% = 49% Hence option B = 49% is the answer.
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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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19 Apr 2018, 05:52
A hiking club’s trip will be cancelled if, on the first day of the trip, the temperature is less than 20 degrees Fahrenheit or it is snowing. If there is a 15% chance that temperatures will be below 20 degrees and a 60% chance that it will not snow, what is the likelihood that the trip will be cancelled?
A. 40% B. 49% C. 55% D. 66% E. 75%
P(Trip cancelled )= P(Temp less than 20) +(OR) P(Snowing)
(P(Temp less than 20) = 15/100
P(Snowing) = 40/100
P(Trip cancelled )= 55/100
Please correct me if i am wrong !



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A hiking club’s trip will be cancelled if, on the first day of the tri
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Updated on: 27 Apr 2018, 11:16
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)
Originally posted by IdiomSavant on 19 Apr 2018, 09:11.
Last edited by IdiomSavant on 27 Apr 2018, 11:16, edited 1 time in total.



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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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19 Apr 2018, 15:51
Bunuel wrote: A hiking club’s trip will be cancelled if, on the first day of the trip, the temperature is less than 20 degrees Fahrenheit or it is snowing. If there is a 15% chance that temperatures will be below 20 degrees and a 60% chance that it will not snow, what is the likelihood that the trip will be cancelled?
A. 40% B. 49% C. 55% D. 66% E. 75% 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
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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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21 Apr 2018, 03:05
Hey Bunuel / JeffTargetTestPrepNeed 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
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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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21 Apr 2018, 04:35
P(a) + P(b)  P(aUb) = 49%



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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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21 Apr 2018, 23:45
pikolo2510 wrote: Hey Bunuel / JeffTargetTestPrepNeed 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 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%
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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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21 Jul 2018, 05:52
renjana wrote: A hiking club’s trip will be cancelled if, on the first day of the trip, the temperature is less than 20 degrees Fahrenheit or it is snowing. If there is a 15% chance that temperatures will be below 20 degrees and a 60% chance that it will not snow, what is the likelihood that the trip will be cancelled?
A. 40% B. 49% C. 55% D. 66% E. 75%
P(Trip cancelled )= P(Temp less than 20) +(OR) P(Snowing)
(P(Temp less than 20) = 15/100
P(Snowing) = 40/100
P(Trip cancelled )= 55/100
Please correct me if i am wrong ! 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)
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Re: A hiking club’s trip will be cancelled if, on the first day of the tri
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09 Aug 2018, 00:38
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




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