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!
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