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? $$$$

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?

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

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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

klaud wrote:

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).

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