melissawlim wrote:

A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

(A) 5/8

(B) 3/4

(C) 7/8

(D) 57/64

(E) 15/16

It would be easier to calculate the probability of opposite event and subtract it from 1.

Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways (*see more about this case below);

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

\(P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8}\)

Answer: C.

* Counting probability of getting 1 tail when flipped 6 times: 1 tail can occur in 6 different ways:

THHHHH

HTHHHH

HHTHHH

HHHTHH

HHHHTH

HHHHHT

Generally probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\)

In our case \(k=1\) and \(n=6\), so we get:

\(P = C^6_1*\frac{1}{2}*\frac{1}{2^5}=6*\frac{1}{2^6}=\frac{6}{64}\)

So \(\frac{1}{64}\) should be multiplied by \(C^6_1\), which is \(6\).

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