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Bunuel
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?
P(Rain) = 20% = 20/100 = 1/5
P(No Rain) = 1 - P(Rain) = 1 - 1/5 = 4/5

Required Probability = 7C2 * P(Rain)^2 * P(No Rain)^5
= 7C2 * (1/5)^2 * (4/5)^5
= 21 * 1/5^2 * ( 4^5/5^5 )
= 21 * ( 4^5 ) / 5^(2+5)
= 21 * 4^5 / 5^7
= 21 * 2^10 / 5^7
= 21 * ( 2^10 * 2^7 ) / 5^7 *2^7
= 21 * 2^17 / 10^7

Hence, D.

Can someone help me to understand why we are multiplying by 2^7 in the highlighted step? I am fine up until that point. Thank you!
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Can someone help me to understand why we are multiplying by 2^7 in the highlighted step? I am fine up until that point. Thank you!
ehill12
As we can see that all the options have 10^7 as the denominator, hence multiplying 2^7 to both numerator and denominator means no change, and helps us get the answer.
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ehill12
Can someone help me to understand why we are multiplying by 2^7 in the highlighted step? I am fine up until that point. Thank you!

As we can see that all the options have 10^7 as the denominator, hence multiplying 2^7 to both numerator and denominator means no change, and helps us get the answer.

Ah, I see. I thought I was incorrectly calculating something. The same denominator in all answer choices was the first thing that I noticed in this problem. Thank you again!
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Possibility of Rain (R): 1/5
Possibility of No Rain (N): 4/5

No. of ways we can arrange RRNNNNN = 7!/5!2! = 21.

So we have 21*(1/5)^2*(4/5)^5 = (21* 4^5)/ 5^7 = (21*2^10)/5^7

Since everything in denominator is in powers of 10

We can rewrite this as (21*2^10*2^7)/(5^7*2^7) = (21*2^17)/10^17.

Option D
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Why cant we solve it in format - No of desired / total outcomes ?
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Why cant we solve it in format - No of desired / total outcomes ?

The probability of each outcome is not the same. The probability that it will rain is different from the probability that it will not rain.

Take another example: Say what is the probability that you will get an even number on the throw of a die?

There are 6 possible outcomes of a die throw.

Favorable cases = 3 (you get 2/4/6)
Total cases = 6 (you get 1/2/3/4/5/6)
The probability of getting any outcome is the same.
Hence probability = 3/6 = 1/2

Here, the probability of a rainy day is 1/5 and probability of a non-rainy day is 4/5.

Favorable cases are RRNNNNN, RNRNNNN, NRNRNNN ... etc
All cases are NNNNNNN, RNNNNNN, NRNNNNN, ... etc

Probability of each outcome is not the same since the probabilities of R and N are not the same.

Probability of getting RRNNNNN = (2/10)*(2/10)*(8/10)*(8/10)*(8/10)*(8/10)*(8/10) = 2^{17} / 10^7

Probability of getting RNRNNNN = 2^{17} / 10^7

There will be 7!/2!*5! = 21 such arrangements

Hence total probability = 21 * 2^{17} / 10^7

Answer (D)
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Using the binomial distribution:

P(desired event) = (number of desired occurrences selected from a lot) * P(Success) * P(Failure)
desired event: Probably that it'll rain for exactly 2 days out of 7 days

P(rain) = 0.2, P(not rain) = 0.8
number of desired occurrences selected from a lot = 7C2 = 21
P(It will rain for exactly 2 days) = 21 * (0.2)^2 * (0.8)^5

Solving this will give that option (D) is the correct answer.
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Oh i thought the probability of it raining 7 days in a a row was 20% lol
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