4 Out Of 15 Apples Are Rotten. They Are Taken Out One By One At Random GMAT Problem Solving

Question: 4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replaced. What is the probability that the 9th one examined is the last rotten one?

A. \(\frac{4c_3*11c_5}{15c_8*7c_1}\)

​B. \(\frac{11c_5*4c_4}{15c_9}\)

C.\(\frac{4c_3*11c_5}{15c_9}\)

​D. \(\frac{11c_5}{15c_9*7c_1}\)

​E. None of these

“4 out of 15 apples are rotten in a bag”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review". GMAT Quant section consists of a total of 31 questions. To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

In order to answer this, consider up to the 9 apples, first 8 apples contain 3 rotten ones. Further, the remaining 5 out of the 8 apples are in good condition. Thus, this can be chosen as \(4c_3*11c_5\) ways. Thus, the total number of selections of 8 apples out of the total of 15 can be stated as  \(15c_8\) . Thus, the last apple is the only rotten apple that is left and which further can be selected in a single manner. This implies, total number of ways of selecting a single apple from remaining is to consider 15-8= 7 apples that states \(7c_1\)  approaches.

Correct Answer: A

Approach Solution 2:

Lets consider upto the 9th apple. The first 8 apples should have 3 rotten ones and remaining 5 good ones. This can be chosen in \(4c_3*11c_5\) ways.

Total number of selections of 8 apples out of 15 apples is \(15c_8\).
The last apple is the only rotten one left, which can be selected in 1 way.
Total number of ways of selecting that 1 apple from remaining 15 – 8 = 7 apples = \(7c_1\) ways

Total probability = \(4c_3*11c_5\)/ 1/\(7c_1\)

Correct Answer: A

Approach Solution 3:

So 8 have been picked. These 8 should be 5 normal and 3 rotten - ways = 11C5*4C3
Total ways = 15C8
Probability = 11C5∗4C3/15C8*1/7C1
BUT now the next can be any of the 7 remaining. However, it is the rotten one which we are looking at so probability is 1/7=1/7C1

Final answer = \(\frac{11c_5*4c_4}{15c_9}\)

Correct Answer: A

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