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What is the probability of flipping a fair coin three

tagged by: VJesus12

Top Member

What is the probability of flipping a fair coin three

What is the probability of flipping a fair coin three times and the coin landing on heads on exactly two flips?

A. 3/8
B. 5/8
C. 7/8
D. 1/8
E. 1/4

The OA is the option A.

Experts, what are the formulas that I should use here? Can you help me, please? I don't know how to start?

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Hi VJesus12,

We're asked for the probability of flipping a fair coin three times and the coin landing on heads on exactly two flips. Since the number of possible outcomes is so small, we can solve this question by simply listing out the possibilities and answering the exact question that is asked.

Since we're flipping a coin 3 times, there are only (2)(2)(2) = 8 possible outcomes. They are...

HHH
HHT
HTH
THH

TTT
TTH
THT
HTT

We're asked for the probability of getting exactly 2 HEADS from those 3 tosses. There are three ways (out of 8 total) to get exactly 2 heads (HHT, HTH and THH).

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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VJesus12 wrote:
What is the probability of flipping a fair coin three times and the coin landing on heads on exactly two flips?

A. 3/8
B. 5/8
C. 7/8
D. 1/8
E. 1/4
We can assume the first 2 flips are heads (H) and the last flip is tails (T). Thus:

P(H-H-T) = 1/2 x 1/2 x 1/2 = 1/8

However, we need to determine in how many ways we can get 2 heads and 1 tail. That number will be equivalent to how many ways we can organize the letters H-H-T.

We use the indistinguishable permutations formula to determine the number of ways to arrange H-H-T: 3!/2! = 3 ways. (Note: The 3 ways are H-H-T, H-T-H, and T-H-H.)

Each of these 3 ways has the same probability of occurring. Thus, the total probability is:

1/8 x 3 = 3/8

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Scott Woodbury-Stewart Founder and CEO

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A coin has two parts; head(H) and a tail(T). Flipping a coin will give 2 outcome H or T
Let's assume the first flip of the coin gives us T and the last two flip gives us H.
therefore, Pr(T-H-H)= 1/2 * 1/2 * 1/2 = 1/8
So, let's find the number of times we can have two head H and a tail T. This can only be done on the number of possible ways we can arrange the T-H-H.
We have T-H-H, H-T-H, H-H-T = 3ways
Each ways has same probabilty of 1/3.
Therefore the total probaility= 1/3 * 1/3 * 1/3 = 3/8
Hence the correct answer is a

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