If the probability that an unfair coin displays heads after it’s flipp

Question

If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

Bunuel · Accepted Answer

If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

A fair coin has the probability of heads equal to the probability of tails, so P(tail) = P(head) = 1/2. We are told that the coins is unfair so P(tail) ≠ P(head).

The probability that the coin displays heads after it’s flipped is three times the probability that it will display tails: P(h) = 3*P(t). Sin, the sum of these probabilities must be 1, then P(h) + P(t) = 1:

3*P(t) + P(t) = 1;

P(t) = 1/4 and P(h) = 3/4.

The question ask to find the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times. Exactly 3 heads can occur in several different ways:
HHHTT
HHTHT
HTHHT
THHHT
THHTH
THTHH
TTHHH
HHTTH
HTTHH
HTHTH
Basically, this is permutations of 5 letters HHHTT, out of which 3 H's and 2 T's are identical: 5!/(3!2!) = 10.

Now, each of the above 10 cases have the probability of (3/4)^3*(1/4)^2. Thus, the overall probability of P(HHHTT) is  10*(\frac{3}{4})^3*(\frac{1}{4})^2 =\frac{135}{512} .

Answer: B.

Hope it's clear.

pr3mk5 · Answer

Probability of getting a head is 3 times the probability of getting a tail.

If probability of Head is X then Tails is (1-X)

x = 3(1-x) -> 3/4 = 3(1/4)

Probability of head = 3/4 
probability of tail = 1/4

We are tossing 5 times for 3 heads:

5!/(3!*2!) * (1/4)^2 * (3/4)^3

= 10* (27/4^5) = (27*5)/4^4*2 = 135/512

OA-B.

prags1989 · Answer

I still don't understand this question. Can someone help?

Bunuel · Answer

22. Probability 
     Theory   
 Math: Probability 
 Probability Made Easy! 
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     Questions   
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 Advanced Topic: Probability and Combinations Questions With Solutions 
 DS Questions 
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For more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

swap6098 · Answer

Basically this coins does not have equal chance of getting Head and Tails. Chance of getting Head is three times than getting Tail.
You have to consider different chances of getting Head and Tail.

Helium · Answer

This problem can be solved by binomial probability formula.

Probability of doing a thing x times out of total n ways=  nCx * (p^x) ( 1- p )^ (n –x )

where n denotes the number of trials and p denotes the success probability.

Required probability =  (5C3) * (3/4)^3 * (1/4)^2

= 135/512

ScottTargetTestPrep · Answer

Since P(H) = 3P(T) and P(H) + P(T) = 1, then P(H) = ¾, and P(T) = ¼. Therefore, the probability of getting heads 3 times followed by tails 2 times is:

P(HHHTT) = (¾)^3 x (¼)^2 = 27/64 x 1/16 = 27/1024

However, since 3 heads and 2 tails can be arranged in 5!/(3!2!) = 10 ways, the probability of getting 3 heads and 2 tails is:

P(3 H’s and 2 T’s) = 10 x P(HHHTT) = 10 x 27/1024 = 5 x 27/512 = 135/512

Answer: B

CrackverbalGMAT · Answer

Since we are dealing with an unfair coin, we understand that getting a head and a tail will not be equally likely, which is what the first statement of the question says as well.

Let the probability of getting a head = P(h) and the probability of getting a tail = P(t). Given that P(h) = 3* P(t). 
The sum of the probabilities, P(h) + P(t) = 1. 
Solving the two equations above, we get P(h) = ¾ and P(t) = ¼.

Now, in 5 coin flips, we need heads exactly three times. This means that the other two outcomes should be tails. So, the required combination of 3 heads and 2 tails is {H, H, H, T, T}.

There are two things that we need to note about the combination shown above:

1) The probability of this combination is  \frac{3}{4}^3  *  \frac{1}{4}^2  =  \frac{27}{1024}

2) The heads and tails can be permuted in  \frac{5!}{{3!*2!}}  = 10 ways i.e. there are 10 different cases of obtaining 3 heads and 2 tails when the coin is flipped 5 times.

Therefore, the total probability = Probability of the combination * Number of permutations.
Substituting the respective values, required probability =  \frac{27 }{ {1024}}  * 10 =  \frac{135 }{ 512} .

The correct answer option is B .

Hope that helps!

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=G7a_MBqZjTY

AbhaGanu · Answer

ScottTargetTestPrep .. how did you get 3/4?

IanStewart · Answer

If two numbers are in a 3 to 1 ratio, the first is 3/4 of the total, the second 1/4 of the total. Here the total is 1, because we have two probabilities that cover every possible outcome, so one probability is 3/4, the other 1/4.

This now becomes a standard binomial probability question ("standard" does not mean "easy" though; binomial probability questions are essentially always going to be high-level questions on the GMAT). Using whatever method you use to solve binomial probability, you'll find the answer is

5C3 * (3/4)^3 * (1/4)^2

which looks like a mess. We don't need to do much calculation though; this equals:

10*(3^3) / 4^5

and since 4^5 is only divisible by 2s, the only thing that will cancel from this fraction is a single '2' from the '10'. So we're looking for an answer with a numerator of 5*27 = 135, and B must be right (or if you notice that A is way too small, and that almost no cancellation will happen, B is clearly the only answer with a large enough denominator -- there's no need to compute 4^4 or 4^5 here).

bumpbot · Answer

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