If John Throws a Coin Until a Series of Three Consecutive Heads GMAT Problem Solving

Question: If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

1. 1/16
2. 2/16
3. 3/16
4. 4/16
5. 6/16

“If John throws a coin until a series of three consecutive heads or three consecutive tails appears" - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been borrowed from the book “GMAT Official Guide Quantitative Review”.

To understand GMAT Club tests questions, applicants must possess fundamental qualitative skills. Quant tests a candidate's aptitude for reasoning and mathematics. The GMAT Quantitative Test's Club Test Phase consists of a question and a list of possible responses. By using mathematics to answer the question, the candidate must select the appropriate response. The problem-solving section of the GMAT Quant topic is made up of very complicated math problems that must be solved by using the right math facts.

Answer: B

Solution and Explanation:

Approach Solution 1:

It is given that john throws a coin until a series of three consecutive heads or three consecutive tails are obtained. It has asked to find out the probability that the game will end in the fourth round.
There are only two ways in which we can end the game in the fourth round.
Determine the likelihood that the two sequences, HTTT or THHH, will occur.The probability of THHH occurring = (1/2)^4 = 1/16. The probability of HTTT occurring = (1/2)^4 = 1/16.
The probability of either THHH or HTTT occurring = 1/16 + 1/16 = 2/16.

Hence 2/16 is the correct answer.

Approach Solution 2:

It is given that John throws a coin until a series of three consecutive heads or three consecutive tails is obtained. It has asked to find out the probability that the game will end in the fourth round.
It is a slightly different approach.

1. Determine the likelihood that the two sequences, HTTT or THHH, will occur. How likely is it that the first two throws will produce different results (which is required): 1/2.
2. What are the chances that the next two throws will both equal the last throw? 1/2 * 1/2.

The total probability will be the product of the two cases.

Answer = 1/2 * 1/2* 1/2 = 1/8.
We can also write it as 2/16
Hence, B is the correct answer.

Approach Solution 3:

It is a given that John throws a coin until a series of three consecutive heads or three consecutive tails is obtained. It has asked to find out the probability that the game will end in the fourth round.

Since there are two options, Beroulli's formula can also be applied:

2!/(1!1!) ∗ (1/2)^(4−1) ∗ (½)^(4-3)
2∗ ⅛ ∗ 1/ 2
⅛ = 2/16
Hence, B is the correct answer.

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