Probability!

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Bethxx: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Wed Dec 15, 2010 1:41 pm

by Bethxx » Wed Dec 15, 2010 1:43 pm

HELP NEED RIGHT NOW!

'If a class contains 12 boys and 8 girls, what is the probability that the teacher will pick at least 2 boys?'

Hellpp? <3

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Source: Beat The GMAT — Problem Solving | Wed Dec 15, 2010 1:43 pm

naremnaresh: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Tue Dec 14, 2010 8:53 pm; Location: India; Thanked: 2 times

by naremnaresh » Wed Dec 15, 2010 8:04 pm

evaluate the expression

12c2/20c2 + 12c3/20c3 +......+12c12/20c12

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sachin2411: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Dec 03, 2010 11:25 am

by sachin2411 » Wed Dec 15, 2010 8:13 pm

You forget that

"If a teacher chhoses 3, then it can be 2 girl and a boy... similarly if 4 then it can be 1 boy, 3 girl....... and so on......"

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naremnaresh: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Tue Dec 14, 2010 8:53 pm; Location: India; Thanked: 2 times

by naremnaresh » Wed Dec 15, 2010 8:16 pm

thanks sachin.
It will be a big equation to solve

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sachin2411: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Dec 03, 2010 11:25 am

by sachin2411 » Wed Dec 15, 2010 8:18 pm

yeah, its really big.... i don't think its a GMAT qns. Please specify the source.

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Thu Dec 16, 2010 12:41 am

Bethxx wrote:HELP NEED RIGHT NOW!

'If a class contains 12 boys and 8 girls, what is the probability that the teacher will pick at least 2 boys?'

Hellpp? <3

note: total outcomes = 20! our probability depends on the number of favorable trials possible out of 20! the number of outcomes was not specified in the problem therefore chose 20! ---- extra data, not useful in this problem

for class of probabilities containing combination of multiple probabilities find the complement probabilities 12 boys & 8 girls => select at least 2 boys => P(2 boys)+P(3 boys)+P(4 boys)... P(12 boys)
the compliment probabilities => 1) select one boy at any time, 2) select all girls at any time

P(1 boy)=12/20
P(all girls)=8!/(20*19*18*17*16*15*14*13)
P(at least 2 boys)=1-12/20 - 8!/(20*19*18*17*16*15*14*13)
1- 3/5 - 8*7*6*5*4*3*2*1/(20*19*18*17*16*15*14*13)
1-3/5 - 1/(2*19*15*13)= (2*19*15*13 - 2*19*9*13 - 1)/(2*19*15*13)
2*19*13*(15-9) - 1/(2*19*15*13)= 12*19*13 - 1/(19*15*26)=
cancel 1 above => 12*19*13/(19*15*26)= 12*13/(15*26)= 4*1/(5*2)=4/10

that's very Manhattanish, is the source MGMAT 700-800 or challenge series?

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IAN_DUDLA: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Dec 24, 2009 7:42 am

by IAN_DUDLA » Fri Dec 17, 2010 11:47 am

forbidden events (no boys)= 8/20*7/19=14/95

probability of good events =1-14/95=81/95

is it the official answer?

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tomada: Master | Next Rank: 500 Posts; Posts: 406; Joined: Mon Jan 25, 2010 11:36 amFollowed by: 4 members; Location: Syracuse, NY; Thanked: 23 times; GMAT Score:740

by tomada » Fri Dec 17, 2010 12:52 pm

To properly answer this question, don't we need to know how many students the teacher is selecting? If the teacher selects only 1 student, then the probability of selecting 2 boys = 0. If the teacher is selecting two students (without replacement), the probability that the first student is a boy = 12/20, and the probability that the second student is a boy = 11/19, for an overall probability of 132/380.

I'm really old, but I'll never be too old to become more educated.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Fri Dec 17, 2010 2:16 pm

Bethxx wrote:HELP NEED RIGHT NOW!

'If a class contains 12 boys and 8 girls, what is the probability that the teacher will pick at least 2 boys?'

Hellpp? <3

Hi,

Let me try in a different way, and hopefully if I don't make any basic mistake, this should be solvable within the time limits.

Lets assume (i,j) denotes i boys and j girls were selected :

So, total ways of selections :

(0,0), (0,1), (0,2)......,(0,8) -> 9 ways
(1,0),(1,1), (1,2),.......(1,8) -> 9 ways

(2,0),(2,1), (2,2),.......(2,8) -> 9 ways
(3,0),(3,1), (3,2),.......(3,8) -> 9 ways
.............................................................

(11,0),(11,1)...........(11,8) -> 9 ways
(12,0),(12,1),............(12,8) -> 9 ways

All the bold combinations are the favorable sections and the italicized are unfavorable ones.

So, the required probability = favorable/total = 11/13
or the required probability = 1- unfavorable/total = 1- 2/13 = 11/13.

Thanks

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 17, 2010 6:42 pm

anshumishra wrote: So, total ways of selections :

(0,0), (0,1), (0,2)......,(0,8) -> 9 ways
(1,0),(1,1), (1,2),.......(1,8) -> 9 ways

(2,0),(2,1), (2,2),.......(2,8) -> 9 ways
(3,0),(3,1), (3,2),.......(3,8) -> 9 ways
.............................................................

(11,0),(11,1)...........(11,8) -> 9 ways
(12,0),(12,1),............(12,8) -> 9 ways

All the bold combinations are the favorable sections and the italicized are unfavorable ones.

So, the required probability = favorable/total = 11/13
or the required probability = 1- unfavorable/total = 1- 2/13 = 11/13.

Thanks

Hey anshumishra, you make the selection of three boys and one girl (3,1) in one way?

can't we have more than one way of selecting three boys and one girl?
boy-girl-boy-boy; boy-boy-girl-boy; boy-boy-boy-girl; girl-boy-boy and so on...

how you do counting of the ways in your selection above, please explain.

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 17, 2010 6:59 pm

tomada wrote:To properly answer this question, don't we need to know how many students the teacher is selecting? If the teacher selects only 1 student, then the probability of selecting 2 boys = 0. If the teacher is selecting two students (without replacement), the probability that the first student is a boy = 12/20, and the probability that the second student is a boy = 11/19, for an overall probability of 132/380.

yes tomada, exactly this will happen as we try to account for each and every probability. We end with the very lengthy statements and an awkward solution. I have tried to simplify a solution process with alternative - complement probabilities.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Fri Dec 17, 2010 7:25 pm

Night reader wrote:
anshumishra wrote: So, total ways of selections :

(0,0), (0,1), (0,2)......,(0,8) -> 9 ways
(1,0),(1,1), (1,2),.......(1,8) -> 9 ways

(2,0),(2,1), (2,2),.......(2,8) -> 9 ways
(3,0),(3,1), (3,2),.......(3,8) -> 9 ways
.............................................................

(11,0),(11,1)...........(11,8) -> 9 ways
(12,0),(12,1),............(12,8) -> 9 ways

All the bold combinations are the favorable sections and the italicized are unfavorable ones.

So, the required probability = favorable/total = 11/13
or the required probability = 1- unfavorable/total = 1- 2/13 = 11/13.

Thanks
Hey anshumishra, you make the selection of three boys and one girl (3,1) in one way?

can't we have more than one way of selecting three boys and one girl?
boy-girl-boy-boy; boy-boy-girl-boy; boy-boy-boy-girl; girl-boy-boy and so on...

how you do counting of the ways in your selection above, please explain.

Night reader,

That is not the combinatorial selection. So here (3,1) doesn't mean ways of selecting 3 men and 1 woman.
They are just one of the possible outcomes of several selection.

Since here we don't know anything about the ways of selections (like he will choose 1, 2 , or 5 with 3 men , 2 women, etc..), it is best to use the symmetry. So likelihood of each of the selections as shown above are equal.
Why should one selection say (2,1) will have higher/lower probability than (3,0) (when there is no rule/restrictions) ?

Hope that makes sense. [Additionally I have included the case when his selection is (0,0) -> which means he selected o boys and 0 girls -> that might be debatable)]. But, important thing is you know the idea.

A question based on similar concept could be :

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary
scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
A. 24/64
B. 32/64
C. 36/64
D. 40/64
E. 42/64

I am sure, you may use the other ways to solve this(using finding the combinations of scores which will produce scores more than 10 e.g. 1,4,6 or 2,4,6 etc..and in how many ways we can achieve them), however lets see the above concept in use :

Total possible scores of Mary (unfavorable -italicized , favorable - bold) and since the average is 10.5 (i.e 3+18/2), the distribution lesser than this and more than this should occur equally.

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Hence the required probability = 1/2

Thanks

Last edited by anshumishra on Sun Dec 19, 2010 9:54 pm, edited 1 time in total.

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 17, 2010 7:43 pm

and with this in mind you solved the probability problem? connection of symmetry within the discrete probability distribution?
symmetry considerations do not suffice to justify or explain probability assignments! GMAT is not the math experiment, neither it is an empirical research.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Fri Dec 17, 2010 7:57 pm

Night reader wrote:and with this in mind you solved the probability problem? connection of symmetry within the discrete probability distribution?
symmetry considerations do not suffice to justify or explain probability assignments! GMAT is not the math experiment, neither it is an empirical research.

Night reader,

It would be better if we discuss the flaws in our assumptions. We are participating here to learn something new as there is hardly any chance that the same question would be repeated. I always try to find a new way if I could, and if someone finds any flaw in that, I am more than happy to get one of my flawed concept right . Forum is for that only

.

And believe me, symmetry helps ! I showed you using the dice example. Give it a shot without that and find the solution.
It is relatively an easier example however lets discuss the approach.

And by the way, "All the best to improve your quant score (i read in some other topic here) " !

Thanks

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