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Re: A certain coin with heads on one side and tails on the other [#permalink]
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sandy wrote:
A certain coin with heads on one side and tails on the other has a \(\frac{1}{2}\) probability of landing on heads. If the coin is flipped three times, what is the probability of flipping 2 tails and 1 head, in any order?

(A) \(\frac{1}{8}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{3}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{2}{3}\)



Easy way:

Let's examine ONE case in which we get exactly 2 tails: HTT

P(HTT)=\(\frac{1}{2}\) *\(\frac{1}{2}\)*\(\frac{1}{2}\)=\(\frac{1}{8}\)

This, of course, is just ONE possible way to get exactly 2 TAILS.

Another possible outcome is THT

Here, P(THT)= (\(\frac{1}{2}\))(\(\frac{1}{2}\))(\(\frac{1}{2}\)) =\(\frac{1}{8}\)

As you might guess, each possible outcome will have the same probability (\(\frac{1}{8}\)). So, the question becomes "In how many different ways can we get exactly 2 TAILS ?"

In other words, in how many different ways can we arrange the letters HTT?\(\frac{3!}{2!}\)=3 WAY.

So P(exactly 2 Tails) = (\(\frac{1}{8}\))(3) = \(\frac{3}{8}\)
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Re: A certain coin with heads on one side and tails on the other [#permalink]
Given that coin is flipped three times and we need to find what is the probability of flipping 2 tails and 1 head, in any order?

Coin is tossed 3 times => Total number of cases = \(2^3\) = 8

Out of these 8 cases there are three cases where we can get 2 Tail and 1 Head. HTT, THT, TTH

=> Probability of flipping 2 tails and 1 head, in any order = \(\frac{3}{8}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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Re: A certain coin with heads on one side and tails on the other [#permalink]
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