If the probability of rain on any given day in City X is 50 percent

Question

If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

(A) 8/125
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

Bunuel · Accepted Answer

The probability of rain each day is 1/2 and the probability of no rain is also 1/2. C^3_5=10 represent ways to choose on which 3 days out of 5 there will be a rain, so P=C^3_5*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16} . Answer: C. Or think about it this way: we want the probability of the following event: RRRNN, where R represent rain day and N represents no-rain day. Now, each R and each N have individual probability of 1/2, so (\frac{1}{2})^5 . But the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, ... basically it will be equla to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R's and 2 identical N's. That # of arrangements is \frac{5!}{3!2!} , (notice that it's the same as C^3_5 ). So, finally P=\frac{5!}{3!2!}*(\frac{1}{2})^5=\frac{5}{16} . Answer: C. For more on this topic check Combinations and Probability chapters of Math Book: math-combinatorics-87345.html math-probability-87244.html Also check similar questions to practice: if-the-probability-of-rain-on-any-given-day-is-50-what-is-99577.html on-saturday-morning-malachi-will-begin-a-camping-vacation-100297.html what-is-the-probability-that-a-field-gun-will-hit-thrice-on-127334.html there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html combination-ps-55071.html the-probability-that-a-family-with-6-children-has-exactly-88945.html? Hope it helps.

srivas · Answer

If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

(A) 8/125
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4

Soln:
Let Y represent raining on a day and N represent Not raining on a day
So for it to rain 3 out of 5 days we have YYYNN.
Since the order in which it rains in 5 days matters hence we find the total number of possibilities
But since Y has a happened 3 times and N has happened 2 times we cannot distinguish between them and hence divide by the repetitions . Thus we need to divide by (3! * 2!)
=> 5!/(2! * 3!)
=> 10 ways

Now since on each day it can rain or it cannot rain. Thus total outcomes is 
=> 2 * 2 * 2 * 2 * 2
=> 32

So probability that it rains on exactly 3 days is
= 10/32
= 5/16

Ans is  C

ashkrs · Answer

welcome to the forum

total possibilties 2^5 = 32
rain combinations 5c3 = 10
10/32 = 5/16

powerka · Answer

Binomial distribution formula: C(n,k) * p^k * (1-p)^(n-k)

Given that the probability of Rain happening is p (=1/2) and not happening is 1-p (=1-1/2=1/2), 
=> Probability of Rain happening k times (=3) in n repeated tests (=5) 
= C(5,3) * (1/2)^3 * (1-1/2)^(5-3) 
= C(5,3) * (1/2)^5
= 10/32 
= 5/16

KarishmaB · Answer

Think of it this way:
R - Rainy day
N - Non rainy day
P(R) = 1/2
P(N) = 1/2

3 of the 5 days are rainy can happen in many ways:
RRRNN
RNRRN
NNRRR
etc...

What is the probability of RRRNN? It is  (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = (1/2)^5 
This probability can be satisfied in the number of ways in which you can arrange RRRNN (some arrangements are shown above)
You have to arrange 5 things of which 2 are same and another 3 are same. This can be done in 5!/2!*3! ways
Probability of exactly 3 rainy days is  5!/(2!*3!) * (1/2)^5 = 5/16

flokki · Answer

thank you for your answer!

I just started with permutations and probability.

I still dont understand why we use the combination formula and not the permutation one.

Is it because we do not differentiate between one rainy day and the other? are they all the same for us?

Are most Gmat questions at this level?

Bunuel · Answer

Yes, we do not differentiate between RRR and NN: for example RRRNN means that it was raining on the first three days and we have no reason to differentiate between them. Also notice that most GMAT combination/probability questions are fairly straightforward and can be solved in several ways. This problem is also dealing with a simple concept explained in the links I sited above, so I'd recommend to brush your fundamentals on combinations before you move to the problems.

ScottTargetTestPrep · Answer

P(RRRNN) = (1/2)^5 =1/32

Since RRRNN can be arranged in 5!/(3! x 2!) = 5 x 2 = 10 ways, the overall probability is 10/32 = 5/16.

Answer: C

CrackverbalGMAT · Answer

Solution:

Probability of rain each day =1/2

=> Probability of no rain on a day =1-1/2 = 1/2

Number of ways to select 3 of 5 days = 5C3 =5! / 3! 2! =10

=>Probability(Rains on exactly 3 days in a 5-day period) = 10 x (1/2)^3 x(1/2)^2

= 5/16  (option c) 
 
 Devmitra Sen
GMAT SME

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If the probability of rain on any given day in City X is 50 percent

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