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Please help...stuck on these 2 probability questions

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Please help...stuck on these 2 probability questions

by kandie_101 » Sun Jun 05, 2011 7:24 pm
1.a class consists of 14 men and 16 women. A group of 5 is randomly chosen
a) the probability of this group containing at least 2 women and at least 2 men?
b) probability the group contains the same gender.


2. a box contains 2 defective lights inadvertenly mixed with 8 nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly 3 trials
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by cans » Sun Jun 05, 2011 7:44 pm
q1. 14 m, 16w. total 30.
a)case1: 2 men, 3 women = 14C2*16C3 = ; case2: 3 men, 2 women = 14C3*16C2
total 5 = 30C5
Thus prob(atleast 2 men and atleast 2 women) =[ 14C2*16C3 + 14C3*16C2 ] /30C5 (you can calculate it)
b)prob(all m) = 14C5/30C5 ; prob (all w) = 16C5/30C5
prob(same gender) = (14C5 + 16C5) / 30C5
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by sjmit4 » Sun Jun 05, 2011 10:39 pm
kandie_101 wrote: 2. a box contains 2 defective lights inadvertenly mixed with 8 nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly 3 trials
number of ways of getting exactly 3 trails= 2
As 3rd trail should have second defective light so first defective light can be detected in first or second attempt.

Total number of ways of testing= 10 x 9/ 2! = 45 (both defective lights consider similar )

So probability is = 2/45

I am not sure if this is the right answer though .. :)
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by cans » Sun Jun 05, 2011 11:32 pm
2. a box contains 2 defective lights inadvertenly mixed with 8 nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly 3 trials
What does 'after exactly 3 trials' mean???
Both defective found in exactly 3 trials, or in the 4th trial?????
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by sohrabkalra » Mon Jun 06, 2011 2:44 am
1.a class consists of 14 men and 16 women. A group of 5 is randomly chosen
a) the probability of this group containing at least 2 women and at least 2 men?
b) probability the group contains the same gender.
Lets work this backwards.

Probability = favourable outcomes/ Total possible outcomes
So in this case it is = Ways of choosing five ppl in which there are atleast two men and two women/Total ways of choosing 5 people

Now lets calculate these
Total No of people = 14+16 = 30
Total Possible outcomes =Total Possible ways of choosing five ppl = 30C5 -----(a)

Now for favourable outcomes....

You need atleast 2 men - so they can be choosen in 14C2 ways
2 women can be choosen in --- 16C2 Ways
And the remaining one person can be a man or woman so that can be choose out of the remaining 26 ppl by - 26C1 ways....

So ANS = (16C2 x 14C2 x 26C1) / 30C5

b) For all same gender = Either all are women or all are men ..so as P(A or B) = P(A) + P(B)
= Probability that all are women + Probability that all are men
= 16C5/30C5 +14C5/30C5

Correct me if i am wrong :)
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by sjmit4 » Mon Jun 06, 2011 11:48 pm
cans wrote:
What does 'after exactly 3 trials' mean???
Both defective found in exactly 3 trials, or in the 4th trial?????
I think it means exactly 3 trials.
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by Ozlemg » Wed Jun 08, 2011 12:56 pm
Quote:
2. a box contains 2 defective lights inadvertenly mixed with 8 nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly 3 trials

I calculate this Q as following :

number of selectinng non defective ones for the 1st and 2 nd trial and 3rd trial and selecting defective ones for the 4th and 5th trial (cos it is mentioned "both")

number of non defective : 8
number of defective : 2
total : 10
*because we do not replace, denominator will reduce one by one
8/10*7/9*6/8*2/7*1/6 : 1/45
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by MM_Ed » Sun Jun 19, 2011 5:17 pm
If the defective (D) lights are found after exactly 3 trials, it means the 3rd one tried is defective. One of the first two tried is also defective, while the other is good (G). So the results are GDD or DGD.

The number of permutations possible for three bulbs chosen out of 10 is 10x9x8 = 720. Since only 2 permutations will work for us, the probability is 2/720 = 1/360.
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by ma127 » Sun Jun 19, 2011 5:55 pm
MM_Ed wrote:If the defective (D) lights are found after exactly 3 trials, it means the 3rd one tried is defective. One of the first two tried is also defective, while the other is good (G). So the results are GDD or DGD.

The number of permutations possible for three bulbs chosen out of 10 is 10x9x8 = 720. Since only 2 permutations will work for us, the probability is 2/720 = 1/360.
neat
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