Coin is tossed 5 times, so
n=5Let,
p= probability of getting heads
q= probability of getting tails.
∴p+q=1......(1)∴ Probability of getting 4 heads
=5C4⋅p4⋅qAnd probability of getting 5 heads
=5C5⋅p5Given,
5C4⋅p4⋅q=5C5⋅p5⇒5q=p...... (2) From equation (1) and (2), we get,
5q+q=1⇒6q=1 ⇒q=∴p=1−=Now, probability of getting atmost two heads
=p(x=0)+p(x=1)+p(x=2)p(x=0)= Getting zero head in 5 trials
=5C0⋅p0⋅q5p(x=1)= Getting one head in 5 trials
=5C1⋅p1⋅q4p(x=2)= Getting two heads in 5 trials
=5C2⋅p2⋅q3=5C0⋅q5+5C1⋅pq4+5C2⋅p2q3=()5+5⋅⋅()4+10⋅()2⋅()3===