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.

M7MBA wrote: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. 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.

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

M7MBA wrote: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

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