Probability - I don't get what the question asks! - The Beat The GMAT Forum

Probability - I don't get what the question asks!

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Nailya: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Sat Feb 21, 2009 3:46 pm; Location: New York City; Thanked: 2 times

by Nailya » Wed Mar 18, 2009 5:33 pm

A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

What does this mean? What do I need to find?

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Source: Beat The GMAT — Problem Solving | Wed Mar 18, 2009 5:33 pm

Reader: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sat Mar 07, 2009 4:15 pm; Thanked: 2 times

by Reader » Wed Mar 18, 2009 7:09 pm

I am not sure, but I will try to answer it.

I think it's asking what's probability that the 3 item he picked, contains the two random chose item.

So it's (number of ways to pick 3 item that contain certain 2 items) /(10c3)

= 8 / (10!/(3!7!))

=8/120

=1/15

OA?

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Wed Mar 18, 2009 7:30 pm

Total number of items = 10
Let S be the sample space, then

n(S) = Numbers of drawing 2 items at random out of 10 = 10C2 = 45

Let E = Event of drawing 2 items at random none of which is the item chosen by that person.

So, n(E) = 7C2 = 21

So P(E) = n(E)/n(S) = 21/45 = 7/15

So answer should be 7/15

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PAB2706: Master | Next Rank: 500 Posts; Posts: 260; Joined: Sun Oct 12, 2008 8:10 pm; Thanked: 4 times

by PAB2706 » Wed Mar 18, 2009 8:27 pm

@ naylia pls give the answer choices.... or post the original answer.

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lilu: Master | Next Rank: 500 Posts; Posts: 148; Joined: Wed Dec 10, 2008 5:13 pm; Location: SF, CA; Thanked: 12 times

by lilu » Wed Mar 18, 2009 8:42 pm

I also think it's 7/15

Number of ways to choose 2 items out of 10: 10!/2!8!=45
Number of ways to choose 2 items when the 3 items that were bought are out of question: 7!/2!5!=21

So, the P is P of the 2 items that would be chosen out of 7 (when 3 other items are not an option) / P of the 2 items that would be chosen out of all 10 items ---> 21/45=7/15

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avenus: Master | Next Rank: 500 Posts; Posts: 238; Joined: Tue Feb 10, 2009 8:44 am; Thanked: 9 times

by avenus » Thu Mar 19, 2009 2:43 am

Here is my take:

the question is asking for the probability that the buyer does not have both the chosen items

The solution posted by codesnooker and Reader is the probability that the buyer does not have any of the chosen items

Not both includes none of the objects but also only one object:

7C2+3*7C1 = 42

So 42/45 = 14/15

It might be easier to obtain the probability that both items have been selected by the buyer p(both) and then take 1 - p(both)

1 - (3C2/10C3) = 1 - (3/45) = 14/15

Or even easier, p(both) without using combinatorics:

Chosen by the buyer: 3 items
Not chosen by the buyer: 7 items

p(both) = p(first object drawn's been chosen)*p(second object drawn's been chosen) = (3/10)*(2/9) = 1/15

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Nailya: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Sat Feb 21, 2009 3:46 pm; Location: New York City; Thanked: 2 times

by Nailya » Thu Mar 19, 2009 7:10 am

Guys, thank you all!!!
This is the OA and explanation:

Number of ways 2 can be selected without taking the three in question. : 7C2 Number of ways 2 can be selected out of 10 = 10C2 Prob. = 7C2/10C2 = 21/45 = 7/15

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avenus: Master | Next Rank: 500 Posts; Posts: 238; Joined: Tue Feb 10, 2009 8:44 am; Thanked: 9 times

by avenus » Thu Mar 19, 2009 8:04 am

Hi Nailya,

what's the source? As I said before, I think the cases in which only one of the two selected objects has been chosen by the buyer should also be considered. If the buyer has only one of the objects, it does not have both...

anyone any thoughts?

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Nailya: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Sat Feb 21, 2009 3:46 pm; Location: New York City; Thanked: 2 times

by Nailya » Thu Mar 19, 2009 8:40 am

These is the probability and combinatorics questions set posted by Eric in the Resourses section.

https://www.beatthegmat.com/probability- ... t6206.html

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