GMAT Problem Solving - Kate and David Each Have $10.

Question: Kate and David each have $10. Together they flip a coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?

1. \(\frac{5}{16}\)

2. \(\frac{15}{32}\)

100. \(\frac{1}{2}\)

500. \(\frac{21}{32}\)

5. \(\frac{11}{16}\)

“Kate and David each have $10. Together they flip a coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Answer

Approach Solution 1

Find the total number of combinations \(2^5=32\). 

Find the number of combinations when Kate wins: out of 5 games, she can win 3 or 4 times only as 5 victories would put her over the $14 mark and less than 3 victories, below the $10.

The number of combinations for winning 3 times:

\(C^3_5\)= 10

Number of combinations for winning 4 times:

\(C^4_5\)= 5

Remember, \(C^k_n=\frac{n!}{k!(n-k)!}\)

Probability = \(\frac{5+10}{32}=\frac{15}{32}\)

Correct option: B

Approach Solution 2

Write out the combinations:

[3]: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345

[4]: 1234, 1235, 1245, 1345, 2345

Total = \(\frac{15}{32}\)

The combinations can be summed because they have equal probabilities of \(\frac{1}{32}\) each.

Correct option: B

Approach Solution 3

There is equal probability that Kate or David will end up with more money by the end (and note there is no situation where Kate and David both end up with $10).

Therefore, we can subtract the probability that Kate wins all 5 flips from \(\frac{1}{2}\) to get the probability:

Combinations: \(2^5=32\)

Probability of winning all 5 flips = \(\frac{1}{32}\)

Probability Kate ends up with $11 - $14 = \(\frac{1}{2}-\frac{1}{32}=\frac{15}{32}\)

Correct option: B

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