Given that A fair coin is flipped 5 times.

Quantity A: The probability of getting more heads than tails

Method 1

As there are odd number of coin tosses so there are only two outcomes possible. Either the number of heads will be more or the number of tails will be more. So there is equal change of occurrence of both the outcomes.

=> P(Getting more heads than tails) = \(\frac{1}{2}\)

=> Quantity A = Quantity B = \(\frac{1}{2}\)

So, Answer will be C

Method 2

Coin is tossed 5 times => Total number of cases = \(2^5\) = 32

P(Getting more heads than tails) = (Cases in which we will get 3H + 4H + 5H) / 32

Cases in which we get 3H

Cases in which we get 3H can be found by putting 3 heads in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5C3 ways = \(\frac{5!}{3!*(5-3)!}\) = \(\frac{5*4*3!}{3!*2!}\) = 10 ways

Cases in which we get 4H

Cases in which we get 4H is same as getting 1 Tail and this can be found by putting 1 tail in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5 ways by putting T in any of the 5 slots => 5 ways

Cases in which we get 5H

This can be done in 1 way (HHHHH)

=> P(Getting more heads than tails) = \(\frac{10 + 5 + 1 }{ 32 }\) = \(\frac{16}{32}\) = \(\frac{1}{2}\)

=> Quantity A = Quantity B = \(\frac{1}{2}\)

So, Answer will be C
Hope it helps!

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