A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?A. 1/16B. 5/16C. 2/5D. 13/16E. 27/32This problem can be done in 2 ways
Method IThis approach does not need a pen and a paper.
We know the basic of probability is P(asked)=P(favourable outcome)/P(total outcome)
Because we are given a fair coin to deal with, P(total outcome)= 2^5 = 32 as each coin has 2 sides and both are equally likely to land.
Just visualize,
P(favourable outcome) = P(at least 2 H) = 1 - {P(no Head) or P(1 Head)}
P(no Head)={5!/5!}/32 = 1/32
P(1 Head)={5!/4!}/32 = 5/32
P(favourable outcome) = 1 - (1/32 + 5/32) = 1 - 6/32 = 26/32 = 13/16 (Option D)
Method IIP(landing heads at least twice) = P(heads twice) + P(heads thrice) + P(heads four times) + P(heads five times)
P(heads twice) = (5!/2!3!)/32 = 10/32
P(heads thrice) = (5!/3!2!)/32 = 10/32
P(heads four times) = (5!/4!)/32 = 5/32
P(heads five times) = (5!/5!)/32 = 1/32
P(landing heads at least twice) = 10/32 + 10/32 + 5/32 + 1/32 = (10+10+5+1)/32 = 26/32 = 13/16 (Option D)