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Since C could selected for either of two remaining spaces, (5C1 + 4C1)/5C2 = 9/20 = .45 and rounding it up gives .5.
Where did I go wrong in this?
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DUMDUM21
For a drama, 3 students need to be selected out of 6 students A, B, C, D, E and F. If A is already selected, what is the probability of selecting C also?

A. 0.5
B. 0.2
C. 0.3
D. 0.4
E. 0.1

Since C could selected for either of two remaining spaces, (5C1 + 4C1)/5C2 = 9/20 = .45 and rounding it up gives .5.
Where did I go wrong in this?
­I'm not clear with your approach, so I'll offer another one involving combinations. Maybe that helps.

We need C to be among 2 chosen students out of the remaining 5:

1C1*4C1/5C2 = 4/10 = 0.4. Here, 1C1 represents choosing C, 4C1 represents choosing any from the remaining 4, and 5C2 represents the total number of ways to choose 2 out of 5.

Answer: D.­
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Bunuel
For a drama, 3 students need to be selected out of 6 students A, B, C, D, E and F. If A is already selected, what is the probability of selecting C also?

A. 0.5
B. 0.2
C. 0.3
D. 0.4
E. 0.1


Are You Up For the Challenge: 700 Level Questions­
As A is already selected. We are left with B,C,D,E and F for the remaining 2 spots.

Amongst the 5 members, selecting two gives us 5C2 options = 10.

This is the total number of cases = 10

The following are the possible cases for selecting 2 out of 5 : BC, BD, BE, BF, CD, CE, CF, DE,DF, EF

with each letter appearing 4 times,

Probability of selecting C = 4/10 = 0.4

option D
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Bunuel
For a drama, 3 students need to be selected out of 6 students A, B, C, D, E and F. If A is already selected, what is the probability of selecting C also?

A. 0.5
B. 0.2
C. 0.3
D. 0.4
E. 0.1


Are You Up For the Challenge: 700 Level Questions­
Probability of selecting C also = 1 - (Probability of not selecting C)

Probability of not selecting C = 4C2/5C2

because we will be selecting the remaining 2 members from 4 (excluding A and C)

1- (6/10) = 1- (3/5)

=2/5


1/5 is 0.2, so 2/5 will be 0.4
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