On the morning of day 1, Suzy began her tracking tour. She plans to re

Question

On the morning of day 1, Suzy began her tracking tour. She plans to return home at the end of the first day on which it rains. If for the first three days of the tour, the probability of rain on each day is 0.25, what is the probability that Suzy will return home at the end of the day 3?

(A) 1/8
(B) 9/64
(C) 27/64
(D) 37/64
(E) 1/64

(A) 1/8
(B) 9/64
(C) 27/64
(D) 37/64
(E) 1/64

CounterSniper · Answer

p(rain) =  \frac{25}{100} 
p(suzy will return at the end of third day ) =  \frac{75}{100}  * \frac{75}{100}  *  \frac{25}{100}  =  \frac{9}{64} 
B

MHIKER · Answer

The probability of raining for not raining of first two days and training on the third day  =\frac{75*75*25}{100*100*100}=\frac{9}{64}

The answer is  B

MathRevolution · Answer

As Suzy is returning on the 3rd day, that means for the first two there was no rain.

P(rain) = 0.25 =  \frac{1}{4}

P(no rain) = 1 - 0.25 = 0.75 =  \frac{3}{4}

P(3rd day return):  \frac{3}{4}  *  \frac{3}{4}  *  \frac{1}{4}  =  \frac{9}{64}

Answer B

ScottTargetTestPrep · Answer

Solution:

The probability that it will rain on any given day is 0.25, or 1/4, and the probability that it will not rain on any given day is 1 - 1/4 = 3/4.

In order for Suzy to return home at the end of the 3rd day, we know that it did not rain on the first or second day, but it did rain on the third day. This event has a probability of 3/4 x 3/4 x 1/4 = 9/64.

Answer: B

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