Let X be the number of coins showing heads Then X∼B(100,p) We have P(X=51)=P(X=50) Since, P(X=r)=nCTprqn−T we get ⇒100C51p51q49=100C50p50q50 ⇒‌
p
q
=‌
100!
50!â‹…50!
×‌
51!49!
100!
⇒‌
p
1−p
=‌
51
50
‌ where ‌q=1−p ⇒50p=51−51p⇒101p=51 ⇒P=‌