Question: A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3, 2 will be red and 1 will be green ?

1. 7/40
2. 7/20
3. 49/100
4. 21/40
5. 7/10

Correct Answer: D
Solution and Explanation:
Approach Solution 1:
The problem statement informs that:
Given:

• A certain basket contains 10 apples, 7 of which are red and 3 are green.
• 3 different apples are randomly selected.

Find Out:

• The probability that out of those 3, 2 will be red and 1 will be green.

The number of ways to select 2 different red apples out of 7 is \(C_7^2={(7*2)\over2!} =21\)
The number of ways to choose 1 green apple out of \(3=C_3^1=3\)
The total number of ways to choose 3 different apples out of total 10 apples \(​​​​= C_{10}^3={(10*9*8)\over3*2}\)
= 5*3*8 = 120
As per the formula of Probability, we know that:
Probability = Number of favorable outcomes / total number of outcomes;
Hence, \(P = {C_7^2*C_3^1\over{C_{10}^3}}= {21 * 3\over120} = {63\over120} = {21\over40}\)
Therefore, the probability that out of those 3, 2 will be red and 1 will be green = \(21\over40\)

Approach Solution 2:
The problem statement states that:
Given:

• A certain basket contains 10 apples, 7 of which are red and 3 are green.
• 3 different apples are randomly selected.

Find Out:

• The probability that out of those 3, 2 will be red and 1 will be green.

We can solve the problem by using the probability approach.
That is we can say:
\(P(RRG)= {3!\over{2!}} * {7 \over10} * {6\over9} * {3\over8}={21\over40}\)
Here, we are multiplying by 3!/2! as the case of RRG can occur in 3 ways: RRG, RGR, GRR - the number of permutation of 3 letters out of which 2 are identical.
Therefore, the probability that out of those 3, 2 will be red and 1 will be green =\(21\over40\).

Approach Solution 3:
The problem statement states that:
Given:

• A certain basket contains 10 apples, 7 of which are red and 3 are green.
• 3 different apples are randomly selected.

Find Out:

• The probability that out of those 3, 2 will be red and 1 will be green.

We can solve the problem by applying certain rules of probability.
Let's find the probability of choosing a red apple 1st, a red apple 2nd, and a green apple 3rd (i.e RRG)
P(red apple 1st AND red apple 2nd AND green apple 3rd) = P(red apple 1st) x P(red apple 2nd) x P(green apple 3rd)
= 7/10 x 6/9 x 3/8
= 7/40
It is important to select a red apple 1st, a red apple 2nd, and a green apple 3rd (i.e RRG). It is just ONE WAY to get 2 red apples and 1 green apple. There are also other ways like RGR and GGR.
We already know that P(RRG) = 7/40, which implies that P(RGR) = 7/40 and P(GRR) = 7/40
Therefore, P(select 2 red apples and 1 green apple) = P(RRG or RGR or GRR)
= P(RRG) + P(RGR) + P(GRR)
= 7/40 + 7/40 + 7/40
= 21/40
Thus, the probability that out of those 3, 2 will be red and 1 will be green is = 21/40.

“A certain basket contains 10 apples, 7 of which are red and 3 are green” - is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “GMAT Official Guide 2021”. To solve the GMAT Problem Solving questions, the candidates must gain concrete knowledge of mathematics and calculations. The GMAT Quant practice papers further help the candidates to analyse varieties of questions that will enable them to improve their mathematical knowledge.

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