The probability of rain showers in Barcelona on any given day is 0.4. What is the probability that it will rain on exact

[B BTGModeratorVI]

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

[Brent@GMATPrepNow]

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

[Scott@TargetTestPrep]

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

Solution:

We are given that the probability of rain in Barcelona is 0.4, and thus the probability of no rain is 1 - 0.4 = 0.6. We need to determine the probability of rain 1 day out of 3 consecutive days. If it rains on the first day, we can’t have rain the next two days. Thus, the initial probability is:

0.4 x 0.6 x 0.6 = 0.144

However, there are 3 different ways for it to rain on exactly 1 out of 3 days:

R - N - N

N - R - N

N - N - R

Each of these 3 ways has the same probability of occurring. Thus, the total probability is 3 x 0.144 = 0.432.

Answer: C