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

Question

When a certain coin is flipped, the probability of heads is 0.5. If the coin is flipped 6 times, what is the probability that there are exactly 3 heads?

A. 1/4 
B. 1/3 
C. 5/16 
D. 31/64 
E. 1/2

Bunuel · Accepted Answer

When a certain coin is flipped, the probability of heads is 0.5. If the coin is flipped 6 times, what is the probability that there are exactly 3 heads?

A. 1/4
B. 1/3
C. 5/16
D. 31/64
E. 1/2

We need the proability of HHHTTT (any combination of this): \(P=\frac{6!}{3!3!}*(\frac{1}{2})^6=\frac{20}{64}=\frac{5}{16}\). We are multiplying by \(\frac{6!}{3!3!}\) since HHHTTT scenario can occur in several ways: HHHTT, THHHTT, TTHHHT, ... basically the number of permutations of 6 letters HHHTTT, out of which 3 H's and T's are identical (\(\frac{6!}{3!3!}\));

Answer: C.

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Probability chapter of out Math Book:
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Hope it helps.

MacFauz · Answer

Required series is HHHTTT =  (\frac{1}{2})^6  =  1/64

This combination can occur in  \frac{6!}{3!*3!}  ways. ie 20 ways

So answer =  \frac{1}{64}*20   = \frac{5}{16}

Kudos Please... If my post helped.

KarishmaB · Answer

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/

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=LRURv38kZw8

bumpbot · Answer

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

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

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