Find the mean number of heads in three tosses of a fair coin.

Answer

N/a

Explanation

Probability of getting head or tail in a toss of one coin is
`P(H)=P(T)=1/2`
Sample space in three throw of an unbiased coin is
`S={"HH,HHT,HTH,THH,THT,TTH,HTT,TTT"}`
The possible values of random variable `X` are 0,1,2 or 3.
`:.P(X=0)=P` (getting all tails)
`=P{"TTT"}=1/2xx1/2xx1/2=1/8`
`P(X=1)=P` (getting one head and two tails)
`=P("TTH")+P(THT)+P("HTT")`
`=1/2xx1/2xx1/2+1/2xx1/2xx1/2+1/2xx1/2xx1/2`
`=1/8+1/8+1/8=3/8`
`P(X=2)=P` (getting two heads and one tail)
`P(HHT)+P(HTH)+P(THH)`
`=1/2xx1/2xx1/2+1/2xx1/2xx1/2+1/2xx1/2xx1/2`
`=1/8+1/8+1/8=3/8`
`P(X=3)=P` (gettting three heads)
`=P("HHH")=1/2xx1/2xx1/2=1/8`
Therefore the required probability distribution is:

Therefore mean
`=sumXP(X)=0xx1/8+1xx3/8+2xx3/8+3xx1/8`
`=3/8+6/8+3/8=12/8=3/2=1.5`