If John flips a coin three times, what is the probability that he will

Hi Bunuel,

I understand how to calculate P(H=0) is (\frac{1}{2})^3.
What does (\frac{1}{2}) + (\frac{1}{2}) + (\frac{1}{2}) mean ?

Could you please help me explain in detail how to calculate P(H=2) ?

Thanks in advance,
tanzzt

The probability that the coin will turn up heads at least twice is the sum of the probability that it turn up heads twice and the probability that it turn up heads thrice: P(H=2) + P(H=3):

\(P(H=2) =\frac{3!}{2!}*(\frac{1}{2})^2*\frac{1}{2}\): Heads, Heads, Tails, can occur in several ways: HHT, HTH, THH, hence 3!/2! (the number of permutations of three letters HHT, where two H's are identical). Each of them has the equal probabilities: \((\frac{1}{2})^2*\frac{1}{2}=(\frac{1}{2})^3\)

\(P(H=3) = (\frac{1}{2})^3\).

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Hope it helps.

Hi Bunuel,

First of all, thanks for your reply.
Sorry that I'm still confused with the probability that it turns up heads exactly twice.
Does \frac{3!}{2!} mean flipping a coin three times & turning up exactly two heads?
So, if we have another question flipping a coin five times & turning up exactly three heads, it should be \frac{5!}{3!}. is that right?

Thanks.

Two heads and one tail can occur in three ways: HHT, HTH, THH. So basically the number of ways this particular scenario can happen is the number of permutations of three letters HHT, where two H's are identical.

If it were 3 heads and 2 tails, then the number of ways this scenario can happen would be 5!/(3!2!) = 10:

HHHTT;
HHTHT;
HTHHT;
THHHT;
THHTH;
THTHH;
TTHHH;
HTTHH;
HHTTH;
HTHTH.

Again, this is the number of permutations of 5 letters HHHTT, out of which 3 H's and 2 T's are identical.

THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

Hope it's clear.

P.S. For more follow the links in my previous post.

It was very clear Bunuel,
Many thanks for your explanation. You helped me a lot.