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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
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HHH
HHT
HTH
THH
TTT
TTH
THT
HTT

so there are 8 possible outcomes. 3 outcomes have only 2 heads. hence 3/8
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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
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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

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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
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hazelnut wrote:
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. \(\frac{7}{8}\)


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
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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
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BillyZ wrote:
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. \(\frac{7}{8}\)


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

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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
BillyZ wrote:
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. \(\frac{7}{8}\)


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.
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Re: A fair coin is flipped three times. What is the probability that the [#permalink]
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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!

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

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