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

Redeem

Probability!

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 am
Thanked: 25 times
Followed by:7 members

Probability!

by Ozlemg » Tue Jul 05, 2011 4:54 am
A box contains ten assemblies of which two are defective. A sample of three assemblies is selected at random. What is the probability that the two defective parts will be selected?

A.0.0667
B.0.0776
C.0.667
D.0.433
E.0.20

Why is my colculation not correct? Where am I wrong?
Probability of Defective : 0.2
Probability of Non Defective : 0.8

Selecting 2 defective ones:
1/5*1/5*4/5 =4/125

OA is A
The more you suffer before the test, the less you will do so in the test! :)
Top
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Tue Jul 05, 2011 5:10 am
Hi,
Both the defensive ones are to be picked.This can be done in 1 way
From the remaining 8 one has to be picked in 8C1 ways.
Total number of ways of picked 3 out of 10 is 10C3
So, probability = 1*8C1/10C3 = 8/120 = 1/15

Hence, A

Coming to your method.
1st defective is selected in 2/10 ways
After it is picked 2nd one is picked in 1/9 ways
Third one is picked in 8/8 ways
So, probability is 1/5*1/9*1 = 1/45
But, here we are considering the order. In this case, the non-defective one is picked 3rd. It can be picked first or second as well
When non-defective one is picked first, probability is (8/10)*(2/9)*(1/8) = 1/45
When non-defective one is picked second, probability is (2/10)*(8/9)*(1/8) = 1/45
So, probability is 3(1/45) = 1/15.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pm
Location: Arlington, MA.
Thanked: 27 times
Followed by:2 members

by winniethepooh » Tue Jul 05, 2011 5:14 am
Answer is A.
The answer should be solved in this way:
You have in all 10C3 choices that is 120 choices, so this is your denominator.
In the Numerator you have 2 fixed choices to get the 2 defectives so 2C2 * 8C1(which is any assembly out of the remaining 8)
Hence, 8/120 = 0.0667
Top
Post new topic Post Reply
• Page 1 of 1