What is the probability of getting only 1 head in a single

[M7MBA]

What is the probability of getting only 1 head in a single throw of three fair coins?

A) 1/2
B) 3/8
C) 1/4
D) 1/5
E) 5/8

The OA is the option B.

How can I determine the correct answer? Shouldn't it be (1/2)^3=1/8? I am confused.<i class="em em-confused"></i>

[EconomistGMATTutor]

What is the probability of getting only 1 head in a single throw of three fair coins?

A) 1/2
B) 3/8
C) 1/4
D) 1/5
E) 5/8

The OA is the option B.

How can I determine the correct answer? Shouldn't it be (1/2)^3=1/8? I am confused.<i class="em em-confused"></i>

Hi M7MBA.

Let's take a look at your question.

First, we have _ _ _ (3) places, and each place has 2 possibilities (H or T). Hence, there are a total of 2*2*2=8 possibilities.

Now, we want to obtain ONLY one head, this implies that we have the next cases:

HTT
THT
TTH

Therefore, there are 3 cases.

Finally, the answer is cases favorable / total of cases = 3/8. This is why the correct answer is B.

I hope this answer may help you.

Feel free to ask me again if you have any doubt.

Regards.

[[email protected]]

Hi M7MBA,

Since we're tossing just 3 coins, there aren't that many possible outcomes. This question can be solved simply by listing out the possibilities and answering the exact question that is asked. With 3 coins, there are (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 1 HEAD from those 3 tosses. There are three ways (out of 8 total) to get exactly 2 heads (HTT, THT and TTH).

Final Answer: B

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

[Jeff@TargetTestPrep]

What is the probability of getting only 1 head in a single throw of three fair coins?

A) 1/2
B) 3/8
C) 1/4
D) 1/5
E) 5/8

We need to determine the probability of H-T-T = 1/2 x 1/2 x 1/2 = 1/8.

Since H-T-T can be arranged in 3!/2! = 3 ways, the total probability is 3/8.

Answer: B