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28 Jan 2021, 04:52
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A
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Difficulty:
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Question Stats:
42% (02:26) correct
58%
(02:36)
wrong
based on 88
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Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie's wedding?
A. 1/9
B. 2/9
C. 1/3
D. 3/4
E. 5/9
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 1/9
B. 2/9
C. 1/3
D. 3/4
E. 5/9
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
fall2021
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Joined: 08 Apr 2018
Last visit: 25 Feb 2022
Posts: 233
Given Kudos: 89
Location: India
Concentration: Marketing, Strategy
Schools: Ross '23 (WL) CBS '23 (D) McDonough'23 (M$) Marshall'23 (A) Darden '23 (D) Stern '23 (WL) Booth '23 (D) Kellogg 1Year '22 (WL) Yale '23 (D) LBS '23 (D) Sloan Fellows '22 (D) Foster '23 (D)
GMAT 1: 750 Q50 V40 (Online)
GMAT 2: 760 Q50 V42
GPA: 3.2
WE:Marketing (Telecommunications)
Schools: Ross '23 (WL) CBS '23 (D) McDonough'23 (M$) Marshall'23 (A) Darden '23 (D) Stern '23 (WL) Booth '23 (D) Kellogg 1Year '22 (WL) Yale '23 (D) LBS '23 (D) Sloan Fellows '22 (D) Foster '23 (D)
GMAT 2: 760 Q50 V42
Posts: 233
28 Jan 2021, 07:26
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Assuming there are 365 days in the year (non-leap year)
We know that it rains 5 times a year. Weatherman predicts with 90% accuracy on these 5 days. So the numerator of our answer is the favourable outcome = 0.9*5 = 4.5
Denominator is favourable outcome (4.5) + Weatherman prediction going wrong (360*0.1 = 36)
Total = 40.5
Probability of it actually raining = 4.5/40.5 = 1/9
Option A
Posted from my mobile device
We know that it rains 5 times a year. Weatherman predicts with 90% accuracy on these 5 days. So the numerator of our answer is the favourable outcome = 0.9*5 = 4.5
Denominator is favourable outcome (4.5) + Weatherman prediction going wrong (360*0.1 = 36)
Total = 40.5
Probability of it actually raining = 4.5/40.5 = 1/9
Option A
Posted from my mobile device
General Discussion
08 Feb 2021, 12:17
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Bunuel
Solution:
This problem can be solved by the application of Bayes’ theorem, but a more intuitive solution is provided here.
Let’s first look at the 4 different possible outcomes that can happen concerning whether it rains (R) on Maire’s wedding day or it doesn’t rain (nR), and whether the weather status was predicted (P) or was not predicted (nP). We have:
(R, P) or (R, nP) or (nR, P) or (nR, nP)
Let’s calculate the probability of each outcome:
P(R, P) = 5/365 x 0.9 = 4.5/365
P(R, nP) = 5/365 x 0.1 = 0.5/365
P(nR, P) = 360/365 x 0.1 = 36/365
P(nR, nP) = 360/365 x 0.9 = 324/365
We see that only 2 of the 4 possible outcomes deal with the event “the weatherman has predicted rain for tomorrow,” and they are the following outcomes: (R,P) and (nR,P), and the total probability that either of these 2 outcomes will happen is the sum of their individual probabilities, which is 4.5/365 + 36/365 = 40.5/365.
But, out of these two outcomes, only (R, P), with probability 4.5/365, is relevant to answering the question “what is the probability that it will rain on Marie’s wedding day?” Thus, the probability that it will rain on Marie’s wedding day, given that the weatherman has predicted rain, is 4.5/365 / 40.5/365 = 4.5/40.5 = 45 / 405 = 1/9.
Answer: A
