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

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

#### Solution

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=54=1

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

Concept: Mean of a Random Variable
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