Given that When a certain coin is flipped, the probability of heads is 0.5 and the coin is flipped 6 times. And we need to find what is the probability that there are exactly 3 heads?

Now there are six places to fill as shown below

_ _ _ _ _ _

We need to get 3 Heads and 3 Tails.
Now lets find out the slots out of these 6 in which 3 heads will go.

We can find that using 6C3 = \(\frac{6!}{3!*(6-3)!}\) = \(\frac{6!}{3!*3!}\) = \(\frac{6*5*4*3*2*1}{3*2*1*3*2*1}\) = 20 ways

Now, in the remaining slots we will have Tails. So we can get 3H and 3T in 20 ways

We know that probability of getting a head, P(H), = Probability of getting a Tail, P(T) = \(\frac{1}{2}\)

=> Probability of getting 3H and 3T = Number of ways * P(H) * P(H) * P(H) * P(T) * P(T) * P(T) = 20 * \(\frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2}\) = \(\frac{5}{16}\)

So, Answer will be C
Hope it helps!

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

No. of total outcomes when a coin is tossed = 2^n , here n=6 = 64 outcomes
we want something like this (H,H,H,T,T,T) or (H,T,T,H,T,H)
but we can't count all so use combination
HHHTTT total letter is 6
H and T repeated 3 times
6!/3!3!= 20
20/64 answer