Question: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
- \(\frac{3}{14}\)
- \(\frac{19}{84}\)
- \(\frac{11}{42}\)
- \(\frac{15}{28}\)
- \(\frac{3}{4}\)
Correct Answer: B
Solution and Explanation:
Approach Solution 1:
The problem statement states that:
Given:
- Triplets Adam, Bruce, and Charlie enter a triathlon.
- If there are 9 competitors in the triathlon, medals are awarded for the first second, and third place.
Asked:
- To find out the probability that at least two of the triplets will win a medal.
According to the question it is asking to find out the probability that at least two of the triplets will win a medal. This implies exactly two of the triplets will win a medal or more than two will win a medal.
This is a combination and permutation problem.
Some general formulas are given below-
- No. of combinations of n objects taking r at a time =\(\left(\! \begin{array}{c} n \\ r \end{array} \!\right) = \frac{n!}{r!(n-r)!}\)
- No. of permutations of n objects taking r at a time = \(^nP_r\) =\(\frac{n!}{(n-r)!}\)
Firstly we’ll calculate the probability that exactly two of the triplets will win a medal.
Let it be p1.
The number of ways to choose 2 people from the group of 3 people is \(^3C_2\) = 3
The number of ways to choose 1 person from the remaining 6 person is \(^6C_2\) = 6
Total number of ways of choosing 3 people from the group of 9 people = \(^6C_3\) = 84
P1 =\(6*\frac{3}{84}=\frac{18}{84}\)
Now we’ll calculate the probability that exactly three of the triplets will win a medal.
Let it be p2.
The number of ways to choose 3 people from a group of 3 people is \(^3C_3\) = 1
Total number of ways of choosing 3 people from a group of 9 people = \(^9C_3\) = 84
P2 = \(\frac{1}{84}\)
The probability that at least two of the triplets will win a medal = p1 + p2
= \(\frac{1}{84}+\frac{18}{84}\)
= \(\frac{19}{84}\)
Hence, the probability that at least two of the triplets will win a medal is 19/84.
Approach Solution 2:
The problem statement informs that:
Given:
- Triplets Adam, Bruce, and Charlie enter a triathlon.
- If there are 9 competitors in the triathlon, medals are awarded for the first second, and third place.
Asked:
- To find out the probability that at least two of the triplets will win a medal.
It is required to note that all medals are equal which is not different for the 1, 2 and 3 positions.
There are two cases.
Case 1: only two get medals.
Let’s select 2 out of three A, B, and C ---3C2=3.
The third can be any of the remaining 6.
Therefore, total ways=3*6=18.
Case 2: all three get medals.
Therefore, it exhibits only 1 way.
Total ways at least to get medals= 18+1=19.
Therefore, total ways= 9C3 = 9!/6!3!= 84
Hence, the probability that at least two of the triplets will win a medal is 19/84.
Approach Solution 3:
The problem statement implies that:
Given:
- Triplets Adam, Bruce, and Charlie enter a triathlon.
- If there are 9 competitors in the triathlon, medals are awarded for the first second, and third place.
Asked:
- To find out the probability that at least two of the triplets will win a medal.
The total number of ways to choose 3 people from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.
The number of ways 2 of the triplets are among the 3 awarded is 3C2 x 6C1 = 3 x 6 = 18.
The number of ways all 3 triplets are the 3 awarded is 3C3 x 6C0 = 1 x 1 = 1.
Therefore, the probability is (18 + 1)/84 = 19/84.
Hence, the probability that at least two of the triplets will win a medal is 19/84.
“Triplets Adam, Bruce, and Charlie enter a triathlon”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “Advanced GMAT Quant”. GMAT Problem Solving questions enable the candidates to access information and solve quantitative problems. GMAT Quant practice papers help the candidates to practice several sorts of questions that will enable them to enhance their mathematical understanding.
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