We are given that A fair coin is tossed six times and we need to find What is the probability that the result will be exactly three heads and three tails

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

This is an example of what I call number of distributions * probability/distribution:

The number of distributions, or ways to organize, three Hs and three Ts (HHHTTT, HHTHTT, etc.) is the factorial of the total number of terms (6) divided by the factorial of each of the sets of repeated terms (3 Hs and 3 Ts, naturally).

6!/(3!3!) = 20

The probability/distribution is a sample-case probability for ONE of these (not that it matters since the coin toss will always produce a probability of 1/2 per toss):

H H H T T T ==> (1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = (1/2)^6 = 1/64

So that means that...

number of distributions * probability/distribution = 20*1/64 = 20/64 = 5/16.

Answer C.

If you're interested in more on this technique, I've written an extensive article about it here: https://privategmattutor.london/gmat-probability-number-of-distributions-x-probability-per-distribution/

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