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The probability of rain showers in Barcelona on any given day is 0.4.

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by BTGModeratorVI » Fri Apr 10, 2020 8:11 am

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Answers

A

B

C

D

E

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Difficulty—

The probability of rain showers in Barcelona on any given day is 0.4. What is the probability that it will rain on exactly one out of three straight days in Barcelona?

A. 0.144
B. 0.072
C. 0.432
D. 0.72
E. 0.288

Answer: C
Source: Economist GMAT
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Re: The probability of rain showers in Barcelona on any given day is 0.4.

by deloitte247 » Sun Apr 12, 2020 3:44 am
The probability of raining on a given day is 0.4
The probability of not raining on a given day is 1 - 0.4 = 0.6

Considering 3 consecutive days
Number of possible ways for it to rain on exactly $$1\ out\ of\ 3\ days=\frac{3!}{\left(3\ -\ 1\ \right)!}=\ \frac{3!}{2!}=\frac{3\ \cdot\ 2\ \cdot\ 1}{2\ \cdot\ 1}=\ \frac{6}{2}=\ 3$$
3 different possible ways for it to rain on exactly 1 out of 3 days

Therefore, the probability of raining on exactly 1 out of 3 days = ( 0.4 * 0.6 * 0.6 ) * 3
= 0.144 * 3 = 0.432

Answer = C
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Re: The probability of rain showers in Barcelona on any given day is 0.4.

by Brent@GMATPrepNow » Sun Apr 12, 2020 9:47 am
BTGModeratorVI wrote: ↑
Fri Apr 10, 2020 8:11 am
 The probability of rain showers in Barcelona on any given day is 0.4. What is the probability that it will rain on exactly one out of three straight days in Barcelona?

A. 0.144
B. 0.072
C. 0.432
D. 0.72
E. 0.288

Answer: C
Source: Economist GMAT
P(Rain) = 0.4 = 2/5
So, P(no rain) = 0.6 = 3/5

Let R represent Rain, and let N represent no rain

So, P(Rain exactly once) = P(R-N-N OR N-R-N OR N-N-R)
= P(R-N-N) + P(N-R-N) + P(N-N-R)
= (2/5)(3/5)(3/5) + (3/5)(2/5)(3/5) + (3/5)(3/5)(2/5)
= 18/125 + 18/125 + 18/125
= 54/125
NO DECIMAL CONVERSION NEEDED
Notice that 62.5/125 = 1/2 = 0.5, so 54/125 = a little less than 0.5

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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