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

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

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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.

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

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

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!

Answer: Option B

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ScottTargetTestPrep.. how did you get 3/4?

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).

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