The total number of possibilities 2^5 = 32

Out of which we need the ones with at least two heads

Reducing the failure cases from 32 will give us the happy cases.

The failure cases are TTTTT, HTTTT, THTTT, TTHTT, TTTHT, TTTTH (6 cases in total)

Reducing the failure cases from 32 gives us 26

The probability is 26/32 or 13/16 (D)

In such questions, we can use combination to quickly solve:

H occurs at-least twice, so 2H, 3H, 4H, or all 5H

Number of Favorable outcomes = 5C2 + 5C3 + 5C4 + 5C5 = 26
Total Outcomes = 2x2x2x2x2 = 32

hence, 13/16

Given that A fair coin is tossed 5 times and we need to find What is the probability that it lands heads up at least twice

P(At least twice heads) = P(2H) + P(3H) + P(4H) + P(5H) = 1 - P(0H) - P(1H)

P(0H)

Total number of cases = \(2^5\) = 32
Case in which we get 0H or 5T is TTTTT => 1

=> P(0H) = \(\frac{1}{32}\)

P(1H)

Case in which we get 1H can be found by putting H in any of the 5 slots _ _ _ _ _ => HTTTT, THTTT, TTHTT, TTTHT, TTTTH => 5

=> P(1H) = \(\frac{5}{32}\)

P(At least twice heads) = 1 - P(0H) - P(1H) = 1 - \(\frac{1}{32}\) - \(\frac{5}{32}\) = \(\frac{32 - 6}{32}\) = \(\frac{26}{32}\) = \(\frac{13}{16}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems