When a certain coin is flipped, the probability of heads is

Out of 6 tosses, exactly 3 heads implies exactly 3 tails.
You can get 3 heads and 3 tails in various ways: HHHTTT, HHTHTT, TTTHHH etc
How many such ways are there? We just need to find the number of ways of arranging 3 Hs and 3 Ts in a row. This can be done in 6!/3!*3! = 20 ways (the 3 Hs are identical and the 3 Ts are identical)

Probability of obtaining HHHTTT = (0.5)*(0.5)*(0.5)*(0.5)*(0.5)*(0.5)

Probability of obtaining 3Hs and 3 Ts in any order = (0.5)^6 * 20 = 5/16

Check out this post for a discussion on binomial probability concept:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/02 ... obability/

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

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