PS - Probability - 2 - exactly one

karthikpandian19 wrote:Two of six students in a seminar are to be randomly assigned to read On Liberty. If Deringer and Hay are both in the section, what is the probability that exactly one of them is assigned to read On Liberty?


(A) 4/15

(B) 2/5

(C) 8/15

(D) 3/5

(E) 2/3

Total possible selections = 6C2 = 15
Selecting exactly one out of Deringer and Hay = 2 ways
Selecting others = 4 ways
Total ways = 2*4 = 8
Probability = [spoiler]8/15[/spoiler]
Ans is c

Let pick D AND not H:

(2/6)*(4/5)= 8/30



same way pick H AND not picking D:

(2/6)*(4/5)= 8/30



Probability of picking only one of them: sum both probabilities:

(8/30)+(8/30)=8/15


Let's say you're interested in picking D then picking Not H. The probability of picking D on the first draw is 1/6. Then there are 5 people left out of the original 6. Of those 5 people, you are interested in 4 of them, because the remaining person is H, and you don't want to pick him. Thus, the probability of picking Not H on the second draw is 4/5.

I am confused as to why the probability of picking D is 2/6.

Let pick D AND not H:

(2/6)*(4/5)= 8/30

D is 1 person among 6 people, so why are you using 2/6?

When I solve this, I do P(D)*P(not H) + P(H)*P(not D), which is (1/6)*(4/5) + (1/6)*(4/5) = (4/30) + (4/30) = (8/30) = (4/15)

Can you explain where I am going wrong?

NicoleWhite wrote:I am confused as to why the probability of picking D is 2/6.

Let pick D AND not H:

(2/6)*(4/5)= 8/30

D is 1 person among 6 people, so why are you using 2/6?

When I solve this, I do P(D)*P(not H) + P(H)*P(not D), which is (1/6)*(4/5) + (1/6)*(4/5) = (4/30) + (4/30) = (8/30) = (4/15)

Can you explain where I am going wrong?

I can't comment on the thought process CSASHISHPANDAY used to get 2/6, but I can tell you where you are going wrong.

The way you are calculating it is equivalent to doing this:

1. Pick D first. Pick one of the people other than H second.
2. Pick H first. Pick one of the people other than D second.

This is a permutational way of calculating probability. You are missing two cases. Those are:
1. Pick one of people other than D and H first. Pick D second.
2. Pick one of people other than D and H first. Pick H second.

This would give you the correct probability of 8/15.

Another way you could do it using 2/6 is:

1. (Pick either one of D or H first) * (Pick one of the other two people second)*2!
= 2/6*4/5*2!.
The 2! above is for arranging/considering the case where one of D or H is picked second.


The fastest way using selections of doing this question would be:

Required probability = (Choose one of D or H)*(Choose one of the other 4)/(Select 2 out of 6)
= 2C1*4C1/6C2 = 2*4*2/(6*5) = 8/15.

Let me know if this helps

Ahh, I did not account for the cases you described:

1. Pick one of people other than D and H first. Pick D second.
2. Pick one of people other than D and H first. Pick H second.

If I continued with my method, I would get P(not D or H)*P(D) = (4/6)*(1/5) = (4/30) and P(not D or H)*P(H) = (4/6)*(1/5) = (4/30)

Adding these to the other (4/30)'s would get me to the correct answer. Thank you!

karthikpandian19 wrote:Two of six students in a seminar are to be randomly assigned to read On Liberty. If Deringer and Hay are both in the section, what is the probability that exactly one of them is assigned to read On Liberty?


(A) 4/15

(B) 2/5

(C) 8/15

(D) 3/5

(E) 2/3

P(D or H on the first pick) = 2/6. (Of the 6 students, 2 are D or H).
P(not D or H on the second pick) = 4/5. (Of the 5 remaining students, 1 is D or H, so the remaining 4 are not D or H).
Since we want both events to happen, we multiply the fractions:
2/6 * 4/5.
Since the order the picks could be reversed to not D or H on the first pick followed by D or H on the second pick, we multiply by 2:
2/6 * 4/5 * 2 = 8/15.

The correct answer is C.

Alternate approach:
P(D and H) = 2/6 * 1/5 = 1/15.
P(neither D nor H) = 4/6 * 3/5 = 6/15.
P(D or H but not both) = 1 - 1/15 - 6/15 = 8/15.

Dear nicokle, read once again

Two of six students in a seminar are to be randomly assigned to read On Liberty. If Deringer and Hay are both in the section, what is the probability that exactly one of them is assigned to read On Liberty?

Here you have to pick two students which may be either D or H (two person) either way like D and not H/H but not D so pick should be 2/6
let it be simpler P(D or H be the 1st pick i.e. out of six 2 D or H)=2/6

OA is C.....thanks for all of your explanations guys...