If ten coins are tossed, then what is the probability of getting atleast 8 heads?

Explanation

In this case, we have to find out the probability of getting atleast 8 heads. Let X is the random variable for getting a head.
Here, n=10, r`ge8`
i.e., r=8,9,10,p=`1/2,q=1/2`
We know that, `P(X=r)=""^(n)C_(r)p^(r )q^(n-r)`
P(X=r)=P(r=8)+P(r=9)+P(r=10)
`=""^(10)C_(8)(1/2)^(8)(1/2)^(10-8)+""^(10)C_(9)(1/2)^(9)(1/2)^(10-9)+""^(10)C_(10)(1/2)^(10)(1/2)^(10-10)`
`=(10!)/(8!2!)(1/2)^(10)+(10!)/(9!1!)(1/2)^(10)+(10!)/(0!10!)(1/2)^(10)`
`=(1/2)^(10)[(10xx9)/2+10+1]`
`=(1/2)^(10)cdot56=1/(2^(7)cdot2^(3))cdot56=7/128`