PS - Probability - 2 - exactly one - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

PS - Probability - 2 - exactly one

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

karthikpandian19: Legendary Member; Posts: 1665; Joined: Thu Nov 03, 2011 7:04 pm; Thanked: 165 times; Followed by:70 members

PS - Probability - 2 - exactly one

by karthikpandian19 » Tue Jul 24, 2012 5:10 am

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

Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank"

---
---Never stop until cracking GMAT---

Quote

Source: Beat The GMAT — Problem Solving | Tue Jul 24, 2012 5:10 am

niketdoshi123: Master | Next Rank: 500 Posts; Posts: 210; Joined: Thu Mar 08, 2012 11:24 pm; Thanked: 62 times; Followed by:3 members

by niketdoshi123 » Tue Jul 24, 2012 5:49 am

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

Quote

CSASHISHPANDAY: Senior | Next Rank: 100 Posts; Posts: 84; Joined: Sat Jun 18, 2011 9:50 pm; Location: New Delhi; Thanked: 35 times; Followed by:3 members; GMAT Score:800

by CSASHISHPANDAY » Wed Jul 25, 2012 5:38 am

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.

Quote

NicoleWhite: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Wed Jul 25, 2012 1:13 pm; Thanked: 1 times

by NicoleWhite » Wed Jul 25, 2012 1:53 pm

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?

Quote

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pm; Thanked: 339 times; Followed by:49 members; GMAT Score:770

by eagleeye » Wed Jul 25, 2012 2:11 pm

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

Quote

NicoleWhite: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Wed Jul 25, 2012 1:13 pm; Thanked: 1 times

by NicoleWhite » Wed Jul 25, 2012 3:40 pm

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!

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Jul 25, 2012 4:52 pm

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.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Quote

CSASHISHPANDAY: Senior | Next Rank: 100 Posts; Posts: 84; Joined: Sat Jun 18, 2011 9:50 pm; Location: New Delhi; Thanked: 35 times; Followed by:3 members; GMAT Score:800

by CSASHISHPANDAY » Wed Jul 25, 2012 8:01 pm

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