A fair coin is flipped three times. What is the probability that the

Question

A fair coin is flipped three times. What is the probability that the coin lands on heads exactly twice?

A.  \frac{1}{8}

B.  \frac{3}{8}

C.  \frac{1}{2}

D.  \frac{5}{8}

E.  \frac{7}{8}

chetan2u · Accepted Answer

Hi,

Ways of getting TWO heads in three flips..
3C2=3!/2!=3..

Total possible outcomes..
2*2*2=8..

Probability=3/8
B

shashankism · Answer

Probability of getting a head when a coin is flipped = 1/2
Probability of getting a tail when a coin is flipped = 1/2

Now coin is flipped thrice. The events in which coin lands in exactly 2 heads are THH, HTH, HHT

So required probability = 1/2*1/2*1/2 + 1/2*1/2*1/2 + 1/2*1/2*1/2 = 3/8

Answer B

pra1785 · Answer

HHH
HHT
HTH
THH
TTT
TTH
THT
HTT

so there are 8 possible outcomes. 3 outcomes have only 2 heads. hence 3/8

anwesha1691 · Answer

We should remember the rule:
Probability (Complex Event) = Probability (Individual Event) * Arrangements

By that logic, we have to find the probability of heads appearing exactly twice i.e., 
P(HHT) - this becomes our complex event.

We can break this down to individual events such as, 
P(H) * P(H) * P(T) which is 1/2 here for each case

Arrangement = 3!/2! 
This is because the combination (HHT) can be arranged in 3 ways - (HHT), (HTH), (THH). Now we can’t always make cases if the sample space is huge so the formula for that is (total outcome)!/(repetitions)!

Thus,
Probability of heads appearing exactly twice = 1/2 * 1/2 * 1/2 * 3!/2! = 3/8

Answer: 3/8

Posted from my mobile device

ScottTargetTestPrep · Answer

Let’s first considered the probability of H-H-T (i.e., two heads followed by a tail):

P(H-H-T) = (1/2)^3 = 1/8

However, two heads and one tail can occur in 3!/2! = 3 ways, so the total probability is 3 x 1/8 = 3/8.

Answer: B

GMATinsight · Answer

Method-1:

The number of ways of choosing 2 out of 3 heads  = ^3C_2 = 3

Total outcomes of 3 tosses  = 2*2*2 = 8  (every toss has two outcomes H/T)

Required Probability  = ^3C_2/2^3 = 3/8

Answer: Option B

Method-2:

Manually count outcomes of 3 tosses with two heads
(HHT), (HTH), (THH)

Total Outcomes = HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 8

Required Probability  = 3/8

Answer: Option B

minustark · Answer

Exactly 2 heads can happen when HHT = 3!/2! = 3 ways to get this composition
Total ways = 2*2*2=8
Probability = 3/8
B is our answer.

BrushMyQuant · Answer

Given that A fair coin is flipped three times and we need to find What is the probability that the coin lands on heads exactly twice?

Coin is tossed 3 times => Total number of cases =  2^3  = 8

To find the cases in which the coin lands on heads exactly twice we need to select two places out of three _ _ _ in which we will get Heads.

This can be done in 3C2 ways =  \frac{3!}{2!*(3-2)!}  =  \frac{3*2!}{2!*1!}  = 3 ways

=> P(2H) =  \frac{3}{8}

So,  Answer will be B 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Probability with Coin Toss Problems

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

jassonadder · Answer

Let's list out all the possible outcomes:

HHT

HTH

THH

TTH

THT

HTT

HHH

TTT

There are 8 total outcomes (23 = 8 outcomes when you flip 3 coins).

Now, how many of these have exactly two heads?

HHT ✅

HTH ✅

THH ✅

That's 3 outcomes with exactly two heads.

So, the probability = (Number of favorable outcomes) ÷ (Total outcomes) = 3/8.

If you use a Flip 3 coins simulator, you can also run a few trials and see that getting exactly two heads happens about 3 out of 8 times!

Thus, the correct answer is  B. 3/8.

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