What is the probability of the Head appearing exactly twice if a coin is flipped thrice? 
$1$. What is the probability of the Head appearing exactly twice if a coin is flipped thrice?
Solution: There are total $8$ combinations of Head and Tail if a coin is flipped thrice, where $2$ heads appear only thrice. So, the probability is, $\frac{3}{8}$.

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$2.$ What is the probability of the Head appearing in all three occasions if a coin is flipped thrice?
Solution: $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $

How can I solve #$1$ in the $2$nd method?
 A: You could use Combinations for example for the first question it would be:
$^3C_2 \times \frac12\times \frac12\times \frac12$
Which is:
$3\times \frac12\times \frac12\times \frac12 = \frac38$ 
The second question would be:
$^3C_3\times \frac12\times \frac12\times \frac12$
Which is:
$1\times \frac12\times \frac12\times \frac12 = \frac18$
A: You can use the formula in almost all such type of problems $^nC_k.p^k.q^{n-k}$
where $n=$number of times the coin flipped
$p=$ the probability that the coin landed head which is always $p=\frac12$
$q=$ the probability that the coin landed tail which is always $q=\frac12$
$k=$number of outcomes (Here we need two heads as outcome)
In this case $p=\frac12,q=\frac12,n=3,k=2$
1) $$^3C_2.\left(\frac{1}{2}\right)^2.\left(\frac{1}{2}\right)^{3-2}$$
$$^3C_2.\left(\frac{1}{2}\right)^2.\left(\frac{1}{2}\right)=\frac38$$
2) 
Here $P(E)=\frac{n(E)}{n(S)}$
where $n(S)=2^3=8$, we took $2^3$ because we need all the three heads
$n(E)=1$, since there is only way to get all the heads in $3$ flips that is $H,H,H$
S0, the answer is $\frac18$
Or in another way
1) The possibilities with two heads are 
$HHT$
$HTH$
$THH$
All the other possibilities are
$HHT$
$HTH$
$THH$
$HHH$
$TTT$
$TTH$
$THT$
$HTH$
So, there are $8$ total possibilities but we want only $2$ heads when coin is flipped thrice.
So, the probability is $\frac38$
