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gcponte
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HELP with these problems!!! 07 Oct 2003, 01:22
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I am taking the GMAT on Friday.
I really don't know how to solve the following:

1) From7 teachers and 5 pupils, a random selection of 7 is make. what is the probability that is contains at least 4 teachers? [149/198]

2) A committee of 3 men and 4 women is to be chosen from 6 men and 7 women. what is the probability that is contains a particular manand a particular woman? [2/7]

Please, help me!!! Friday I'll let you know about the outcome.
Thanks a lot



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HELP with these problems!!! Updated on: 07 Oct 2003, 02:50
Originally posted by stolyar on 07 Oct 2003, 02:36.
Last edited by stolyar on 07 Oct 2003, 02:50, edited 1 time in total.
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(1) 'at least 4 teachers' means 4, 5, 6, 7 teachers in a selection. Therefore there will be 3, 2, 1, 0 pupils accordingly.

P=[7C4*5C3/12C7]+[7C5*5C2/12C7]+[7C6*5C1/12C7]+[7C7*5C0/12C7]

(2) My approach: let us call a particular man and a particular woman John and Mary.
So, we have 5 men+John and 6 women+Mary
Take 3 men and 4 women such then both John and Mary are taken
Favorable outcomes=1C1(to take John)*5C2(to take other two men)*1C1(to take Mary)*6C3(to take other three women)
Total outcomes=13C7=11*12*13

P=10*20/11*12*13=50/429 differs from your 2/7
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gcponte
I am taking the GMAT on Friday.
I really don't know how to solve the following:

1) From 7 teachers and 5 pupils, a random selection of 7 is make. what is the probability that is contains at least 4 teachers? [149/198]



Alteast 4 means your favorable outcomes may contain 4 ,5,6 or 7 teachers..

we have to select 7....so we have the following possibilities:

Select 4 teachers out of 7 and 3 pupils out of 5 : 7C4 * 5C3

Select 5 teachers out of 7 and 2 pupils out of 5 : 7C5 * 5C2

Select 6 teachers out of 7 and 1 pupil out of 5 : 7C6 * 5C1

Select 7 teachers out of 7 : 7C7

Add the four..thats your favorable outcomes

Now how many ways can we select 7 from the total of 12 ( 7 teachers + 5 pupils )

12C7 ...thats your total outcomes

Divide the favorable outcomes by the total outcomes..thats the required Probability.

{( 7C4 * 5C3) + (7C5 * 5C2) + (7C6 * 5C1) + (7C7 * 5C0 )} /( 12C7)

i doubt that you will be asked to calculate huge numbers...so for gmat purposes..the above form is ok..for smaller problems...you may be required to go a step further...

Good Enough?

Thanks
praetorian
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gcponte

2) A committee of 3 men and 4 women is to be chosen from 6 men and 7 women. what is the probability that is contains a particular manand a particular woman? [2/7]

Thanks a lot


Its as good as saying ...Select 2 men out of 5 and 3 women out of 6..why?

Because we consider that the particular man and particular woman are already selected...

Favorable Cases : 5C2 * 6C3

Now for the total cases, we do away with that restriction...that is ...select 3 men out of 6 and 4 women out of 7

Total Cases : 6C3 * 7C4


Required probability : ( 5C2 * 6C3 ) / ( 6C3 * 7C4) = 10/35 =2/7

Thanks
Praetorian
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what is my mistake?
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Thanks a lot praetorian. It is a great help for me. Thanks again!
I am a bit tense, in 3 days I have the test. The only thing I didn't cover properly is permutation/combination.
Where are u from? Any suggestion to give me for these last days??
I'll write other examples, so that I have more exercises solved as examples.

1. Five cards are drawn at random from a pack of 52 playing cards . what is the probability that they are all from the same suit? [33/16660]

2. Five cards are drawn from a pack of 52 playing carts. What is the probability of drawing at leas 3 aces? [19/10829]

3. From 7 teachers and 5 pupils a commitee of 7 is to be formed. How many commitees can be selected if both teachers and pupils are represented and the teachers are in a majority? [595]
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stolyar
what is my mistake?


Stoylar , you cannot club the 7 and 6 together..since they are two different groups.

We have to select 3 MEN and 4 Women... that specifically mentioned..

When you say 13C7 ... your 7 includes cases where there 6 men and 1 woman..and 1 man and 6 women..

That wouldnt be correct..

Good enough?

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1. Five cards are drawn at random from a pack of 52 playing cards . what is the probability that they are all from the same suit?

P=4(13/52*12/51*11/50*10/49*9/48)=33/16660

2. Five cards are drawn from a pack of 52 playing carts. What is the probability of drawing at least 3 aces?

P(3 or 4 aces) = [4C3*48C2+4C4*48C1]/52C2

3. From 7 teachers and 5 pupils a commitee of 7 is to be formed. How many commitees can be selected if both teachers and pupils are represented and the teachers are in a majority? [595]

variants are 1p+6t, 2p+5t, 3p+4t

N=[5C1*7C6]+[5C2*7C5]+[5C3*7C4]=35+210+350=595
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gcponte

1. Five cards are drawn at random from a pack of 52 playing cards . what is the probability that they are all from the same suit? [33/16660]


There are 4 suits ... clubs, spades, hearts and diamonds..13 cards of each

So our favorable cases are : pick 5 cards out of clubs OR pick 5 cards out of spades OR pick 5 cards out of hearts OR pick 5 cards out of diamonds

Favorable cases : 13C5 + 13C5 + 13C5 + 13C5 = 4 * 13C5

Total Cases are where we pick ANY 5 cards out of 52 cards

Total Cases : 52C5

Required Probability : (4 *13C5) / 52C5

Thanks
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gcponte wrote:
Quote:
2. Five cards are drawn from a pack of 52 playing carts. What is the probability of drawing at least 3 aces? [19/10829]


We have 4 aces in a pack of 52 cards

Atleast 3 aces means... 3 aces and 4 aces

Prob(3 aces)

Favorable cases ...pick 3 aces of 4 and pick 2 other cards out of the remaining 48

Favorable cases : 4C3 * 48C2

Prob ( 4aces )

Pick 4 aces out of 4 and 1 card out of the remaining 48

Favorable : 4C4 * 48C1

Total cases : 52C5

Required Probability : {(4C3 * 48C2) + (4C4 * 48C1)} /( 52C5)
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gcponte wrote:

Quote:
3. From 7 teachers and 5 pupils a commitee of 7 is to be formed. How many commitees can be selected if both teachers and pupils are represented and the teachers are in a majority? [595]


Teachers in a majority and the pupils are represented..

That means minimum of 4 teachers and maximum of 6 ..why?

Because BOTH teachers and pupils are represented...if we take 7 maximum for teachers...pupils will be 0 and we will violate the condition in the problem

using the same concept as we used earlier

Combinations : 7C4 * 5C3 + 7C5 * 5C2 + 7C6 * 5C1
= 35 *10 + 21 * 10 + 7 *5 = 595

thanks
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gcponte
Thanks a lot praetorian. It is a great help for me. Thanks again!
I am a bit tense, in 3 days I have the test. The only thing I didn't cover properly is permutation/combination.
Where are u from? Any suggestion to give me for these last days??
I'll write other examples, so that I have more exercises solved as examples.


Unless you start solving these problems on your own, there is absolutely no way you can solve a TOTALLY NEW problem in less than 2 minutes

you can only learn so much from others work..the real deal is trying to use all your knowledge to solve a particular problem...thats how i do it.

if you get it wrong...great..because you eliminated a potential mistake on your gmat and learnt a new thing too...

if you get it right.. great...move on. :)

last three days. work really really hard in the next two days on probability and counting methods... that wouldnt be enough..but hey..its always good to practice..

Post your probability or combination questions here...someone may help.

thanks
praetorian



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