Let p be the probability of success that people is right handed
⇒p=

90

100

=

9

10

and q=1−p=1−

9

10

=

1

10

∴ X has a binomial distribution with n=10,p=

9

10

,q=

1

10

∴P(X=r)=nCr(q)n−rpr
Required probability = P(at most 6 of 10 people are right handed)
= P (X ≤ 6) = 1 – P (7 ≤ X ≤ 10)
=1−

10

∑

r=7

10Cr(

9

10

)r(

1

10

)10−r
=1−

10

∑

r=7

10Cr(0.9)r(0.1)10−r