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If there is a 20% chance of rain every day for the next 7

Expert replies
by BTGmoderatorLU » Mon Feb 11, 2019 3:27 pm

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Answers

A

B

C

D

E

Stats

Difficulty

Source: Veritas Prep

If there is a 20% chance of rain every day for the next 7 days, what is the probability that it will rain exactly 2 days out of the next 7 days?

$$A. \ \frac{21}{10^7}$$
$$B.\ \frac{2^7}{10^7}$$
$$C.\ \frac{21^7}{10^7}$$
$$D.\ \frac{21\cdot2^7}{10^7}$$
$$E.\ \frac{42\cdot2^7}{10^7}$$
The OA is D
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Source: — Problem Solving |

by GMATGuruNY » Mon Feb 11, 2019 4:16 pm
Options D and E should appear as shown below:
BTGmoderatorLU wrote:Source: Veritas Prep

If there is a 20% chance of rain every day for the next 7 days, what is the probability that it will rain exactly 2 days out of the next 7 days?

$$A. \ \frac{21}{10^7}$$
$$B.\ \frac{2^7}{10^7}$$
$$C.\ \frac{21^7}{10^7}$$
$$D.\ \frac{21\cdot2^{17}}{10^7}$$
$$E.\ \frac{42\cdot2^{17}}{10^7}$$
P(exactly n times) = P(one way) * all possible ways.

Let R = rain and N = no rain.
Since P(R) = 20% = 2/10, P(N) = 80% = 8/10 = 2³/10.

P(one way):
One way to get exactly 2 days of rain is to have rain on the first 2 days but not on the last 5 days.
P(RRNNNNN) = 2/10 * 2/10 * 2³/10 * 2³/10* 2³/10 * 2³/10 * 2³/10 = 2¹�/10�.

All possible ways:
Any arrangement of the letters RRNNNNN will yield exactly 2 days of rain.
Thus, to account for all the ways to get exactly 2 days of rain, the result above must be multiplied by the number of ways to arrange RRNNNNN.
Number of ways to arrange RRNNNNN = 7!/(2!5!) = 21.

Multiplying the results above, we get:
P(exactly 2 days of R) = (21 * 2¹�)/10�.

The correct answer is D.

For an even trickier rain problem, check here:

https://www.beatthegmat.com/rain-check-t79099.html
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by Scott@TargetTestPrep » Wed Feb 13, 2019 6:34 pm
BTGmoderatorLU wrote:Source: Veritas Prep

If there is a 20% chance of rain every day for the next 7 days, what is the probability that it will rain exactly 2 days out of the next 7 days?

$$A. \ \frac{21}{10^7}$$
$$B.\ \frac{2^7}{10^7}$$
$$C.\ \frac{21^7}{10^7}$$
$$D.\ \frac{21\cdot2^7}{10^7}$$
$$E.\ \frac{42\cdot2^7}{10^7}$$
The OA is D
We can assume the scenario is RRNNNNN (i.e., it rains the first 2 days and doesn't rain for the next 5). Therefore, the probability of this scenario is 2/10 x 2/10 x 8/10 x 8/10 x 8/10 x 8/10 x 8/10 = (2/10)^2 x (8/10)^5 = (2^2 x 8^5)/10^7 = (2^2 x 2^15)/10^7 = 2^17/10^7.

However, this is not the only scenario in which it will rain exactly 2 days out of the 7 days. For example, it can rain on the third and fifth days and not on the other 5 days (i.e., NNRNRNN). In fact, there are 7!/(2! x 5!) ways it can rain exactly 2 days out of the 7 days since that is the number of ways we can arrange the 2 R's and 5N's.

Since 7!/(2! x 5!) = (7 x 6)/2 = 21, then the probability that it will rain exactly 2 days out of the 7 days is:

2^17/10^7 x 21 = (21 x 2^17)/10^7

Answer: D

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