total outcomes 5C2=10
Amir can be in 4 two-group combination sets
the required probability is 4/10=0.4
e
probability
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pemdas
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your original question modified below and responded for better grasp of concept
So the denominator is 5C2=10
The numerator for the three-person selections including Amir will be Amir=1*4C2 <- fix Amir as one (1) and variate the other two people selected (simple combination without order) and for the four-person selections including Amir will be Amir=1*4C3 <- fix Amir as one (1) and variate the other three people selected (find combination of 3 out of 4 remaining)
Hence answer to the first question will be (1*4C2)/5C2=6/10 or 0.6
The answer to the second question will be (1*4C3)/5C2=4/10 or 0.4
You could notice that two different questions - your originally posted and the last question with the four-person selections - have the same answers. It's mere coincidence
Three people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir is selected?
The total outcomes for both questions broached is the same. The question left open is in how many of the three-person selections and four-person selections Amir will be?Four people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir is selected?
So the denominator is 5C2=10
The numerator for the three-person selections including Amir will be Amir=1*4C2 <- fix Amir as one (1) and variate the other two people selected (simple combination without order) and for the four-person selections including Amir will be Amir=1*4C3 <- fix Amir as one (1) and variate the other three people selected (find combination of 3 out of 4 remaining)
Hence answer to the first question will be (1*4C2)/5C2=6/10 or 0.6
The answer to the second question will be (1*4C3)/5C2=4/10 or 0.4
You could notice that two different questions - your originally posted and the last question with the four-person selections - have the same answers. It's mere coincidence
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mayuragrawal2008
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How this formula will work for below mentioned question?
Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian is selected?
Please tell me where I'm wrong in above mentioned formula?
2 people are fixed, so nominator would be 2x3C0 = 2
and denominator would be 5C2.
Answer would be 2/10. Which is wrong I know.
Where I'm wrong to understand this formula? $$$$
Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian is selected?
Please tell me where I'm wrong in above mentioned formula?
2 people are fixed, so nominator would be 2x3C0 = 2
and denominator would be 5C2.
Answer would be 2/10. Which is wrong I know.
Where I'm wrong to understand this formula? $$$$
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GMATGuruNY
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The portion in red is incorrect.mayuragrawal2008 wrote:How this formula will work for below mentioned question?
Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian is selected?
Please tell me where I'm wrong in above mentioned formula?
2 people are fixed, so nominator would be 2x3C0 = 2
and denominator would be 5C2.
Answer would be 2/10. Which is wrong I know.
Where I'm wrong to understand this formula? $$$$
From the 5 people, the number of ways to choose 2 = 5C2 = (5*4)/(2*1) = 10.
Of the10 possible pairs, only 1 includes Amir and Brian.
Thus, P(Amir and Brian are both selected) = 1/10.
Alternate approach:
P(Amir or Brian is selected on the first pick) = 2/5. (Of the 5 people, 2 are Amir or Brian.)
P(Amir or Brian is selected on the second pick) = 1/4. (Of the 4 remaining people, only 1 is Amir or Brian, whoever of the two was not selected on the first pick).
To combine these probabilities, we multiply:
2/5 * 1/4 = 1/10.
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Brent@GMATPrepNow
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Your new question: Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian are BOTH selected?mayuragrawal2008 wrote:How this formula will work for below mentioned question?
Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian is selected?
Please tell me where I'm wrong in above mentioned formula?
2 people are fixed, so nominator would be 2x3C0 = 2
and denominator would be 5C2.
Answer would be 2/10. Which is wrong I know.
Where I'm wrong to understand this formula? $$$$
The correct answer to that question is 1/10. You're getting 2/10 as your answer, because you're treating the numerator different from the way you're treating the denominator.
For the denominator, you say there are 10 ways to select 2 people from 5 people. You used COMBINATIONS (5C2 to be exact), because you are saying that the ORDER in which we select the two people does not matter. For example, selecting Dhana first and Ebo 2nd, is the SAME as selecting Ebo first and Dhana 2nd. This is a perfectly fine approach.
However, for the denominator, you are used the Fundamental Counting Principle, in which you say there are 2 ways to select BOTH Amir and Brian. This means you are saying that the ORDER DOES matter. That is, you are saying that selecting Amir first and Brian 2nd, is DIFFERENT FROM selecting Brian first and Amir 2nd.
To get the correct answer, we must EITHER decide that order matters OR order does not matter, and then apply that strategy to the numerator AND the denominator.
If we go with your original solution (where the denominator is calculated as though order does NOT matter), then we must also treat the numerator the same way. If order does NOT matter, then we must determine the number of ways to select 2 people from a group of two people (Amir and Brian). Since order does not matter, we'll use combinations. We can select 2 people from 2 people in 2C2 ways. 2C2 = 1
So, P(both Amir and Brian are selected) = 1/10
-----------------------------------
ALTERNATIVELY, we can treat BOTH the numerator and denominator as though order DOES matter.
So, for the denominator, in how many ways can we select 2 people from 5 people?
We can select the 1st person in 5 ways
Then we can select the 2nd person in 4 ways
So, the TOTAL number of ways to select 2 people (if order DOES matter) = (5)(4) = 20
Then, for the numerator, in how many ways can we select 2 people from 2 people (Amir and Brian) if order DOES matter?
We can select the 1st person in 2 ways
Then we can select the 2nd person in 1 way
So, the TOTAL number of ways to select 2 people (if order DOES matter) = (2)(1) = 2
So, P(both Amir and Brian are selected) = 2/20 = 1/10
Cheers,
Brent
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Scott@TargetTestPrep
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The number of ways to select two people with Amir selected is 4C1 = 4. To clarify, the listing of these 4 ways is: AB, AC, AD, and AE (noting that the order of selection is not important).klaud wrote:
The number of ways to select 2 people from 5 is 5C2 = 5!/[2! x (5 - 2)!]! = (5 x 4)/2! = 10.
So, the probability that Amir is selected is 4/10 = 2/5 = 0.4.
Answer: E
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