Question

For a specific location in a particularly rainy city, the time a new thunderstorm begins to produce rain (first drop time) is uniformly distributed throughout the day and independent of this first drop time for the surrounding days. Given that it will rain at some point both of the next two days, what is the probability that the first drop of rain will be felt between 8:25 AM and 3:30 PM on both days? a) O 0.2951 b) 0.9137 c) 0.0871 d) 0.2938 e) 0.0863 f) None of the above.

Answer 1

Answer:

c) 0.0871

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X between c and d is given by the following formula.

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

On a single day:

24 hours, so $a = 0, b = 24$

8:25 P.M. is 8:25 = 8.4167h

3:30 P. M. is the equivalent to 12 + 3:30 = 15:30 = 15.5h

Probability of rain between these times:

$P(8.4167 \leq X \leq 15.5) = \frac{15.5 - 8.4167}{24 - 0} = 0.2951$

On both days:

Two days, each with a 0.2951 probability

0.2951*0.2951 = 0.0871

The correct answer is:

c) 0.0871