Question

An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained.

Answer 1

 

Let X denote the number of heads in the four tosses of the coin.

Then X is a random variable that can take the values 0, 1, 2, 3 or 4.

P(X=0)=Probability of getting no head (TTTT)

`=1/16`

P(X=1)=Probability of getting one head (HTTT, THTT, TTHT, TTTH)             

`=4×1/16=1/4`

P(X=2)=Probability of getting two heads (HHTT, HTHT, HTTH, THHT, THTH, TTHH)             

`=6×1/16=3/8`

P(X=3)=Probability of getting three heads (HHHT, HHTH, HTHH, THHH)             

`=4×1/16=1/4`

P(X=4)=Probability of getting four heads (HHHH)            

`=1/16`

The probability distribution of X is

X	0	1	2	3	4
P(X)	1/16	1/4	3/8	1/4	1/16

 

 

xi	pi	pixi	pixi2
0	1/16	0	0
1	1/4	1/4	1/4
2	3/8	3/4	3/2
3	1/4	3/4	9/4
4	1/16	1/4	1
		∑pixi=2	∑pixi2=5

 

Mean, E(X) = ∑pixi=2

Var(X)= ∑pixi²−(∑pixi)²=5−4=1

Therefore, the mean and variance of the number of heads obtained are 2 and 1, respectively.