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Probability

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Probability

by ankit1383 » Thu Aug 06, 2009 8:40 pm
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. 1/12

B. 5/14

C. 4/9

D. 1/2

E. 2/3

Please point out my mistake

Let assume we have ball 1 2 3
In first try i can select any ball...so probability is 1
In second try i will select 5/8 (not to include the other ball)
third try 4/7
fourth try 3/6

Probability: 1x5/8x4/7x3x6= 5/28

I have the OE..will post it later..please correct my method..
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by real2008 » Thu Aug 06, 2009 9:24 pm
I hope ans is B. I can try to explain if it is correct.
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by aspiregmat » Thu Aug 06, 2009 9:26 pm
Ankit I didn't get your reasoning...

I tried to solve the same.. here is my approach..


Lets first find whats the probability of getting 3 different cups..

total posibilities = 9C4

favorable posibility = 3C2 * 3C1 * 3C1 * 3 ( last 3 because any of the cup can be 2)

probability = 3*3*3*3/ 9C4
= 9/14

therefore no getting 3 different cups = 1-9/14 = 5/14

IMO B

OA plz...
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Re: Probability

by ankitns » Thu Aug 06, 2009 9:37 pm
In my opinion answer should be [spoiler]5/14 --> B[/spoiler]

Think of this in terms of P(A) = 1 - P(A')

Consider what is the probability that the contestant will taste all three samples. And then subtract that from 1.

I think with your approach, you could considering the order in which the cups are selected. Which means that the cups selected in the order 1, 2, 3 and 4 will be different from cups selected in the order 4, 3, 2, and 1. I am not sure what you did wrong.


ankit1383 wrote:At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. 1/12

B. 5/14

C. 4/9

D. 1/2

E. 2/3

Please point out my mistake

Let assume we have ball 1 2 3
In first try i can select any ball...so probability is 1
In second try i will select 5/8 (not to include the other ball)
third try 4/7
fourth try 3/6

Probability: 1x5/8x4/7x3x6= 5/28

I have the OE..will post it later..please correct my method..
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by ankit1383 » Thu Aug 06, 2009 9:48 pm
Ankit
i can do it from 1-P(A) formulae but not able to figur out what i did wrong ..

OA...B
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by ankitns » Thu Aug 06, 2009 10:58 pm
Not sure as to what exactly you are doing...sorry cant help much.
ankit1383 wrote:Ankit
i can do it from 1-P(A) formulae but not able to figur out what i did wrong ..

OA...B
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by kaulnikhil » Fri Aug 07, 2009 6:42 am
can someone please tell me where i am goin wrong in selecting the favourable conditions
3c1 * 3c1 * 3c1* 6 c1

as we have 6 cups remaining so we can select any one of these
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by ManSab » Tue Aug 11, 2009 9:40 pm
favorable posibility = 3C2 * 3C1 * 3C1 * 3 ( last 3 because any of the cup can be 2)

probability = 3*3*3*3/ 9C4

==========================

Can some one please break down above calculation....???

Thanks
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Question

by GambitOS » Wed Aug 12, 2009 1:21 am
Hello guys!

I ask you to explain me where you take this
favorable posibility = 3C2 * 3C1 * 3C1
What's name of this formula, can you advice me some links!

Thanks a lot!
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by tom4lax » Thu Aug 13, 2009 5:07 am
Look at it this way:

AAABBBCCC

So for favorable condition, we want at least one of each

From AAA we get 3!/2!1! (choose 2 from 3)

From BBB we get 3!/1!2! (choose 1 from 3)

From CCC we get 3!/1!2! (choose 1 from 3)

Since it does not matter which one of the groups chooses 2, we multiply by 3

3*3*3*3
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by ket » Thu Aug 13, 2009 11:46 am
Can someone tell me what's wrong with my logic?

1. First we need to find the probability of contestant tastes all three samples

we have to choose 4 cups.

the probability that cup #1 is from any three - 3/9
the probability that cup #2 is from another three samples - 3/8
the probability that cup #3 is from another three samples - 3/7
the last cup can be any out of 6 therefore the probability is 1

so i get that the probability of choosing all three samples is

3/9X3/8X3/7 and the answer should be 1-3/9X3/8X3/7 = 53/56 - this answer does not exist :)

Someone please help me :)
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Re: Probability

by maihuna » Thu Aug 13, 2009 8:44 pm
ankit1383 wrote:At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. 1/12

B. 5/14

C. 4/9

D. 1/2

E. 2/3
Total selection :9*8*7*6

The question is how many times the four selection is not entirely of 3 of a kind:
XXXYYYZZZ

First selection any 9
Second selection any 8
Third four will go out, so remainig one is 3
Fourth one in 2 ways

so prob = 9*8*3*2/9*8*7*6 = 1/7

bUT THIS IS NOT ONE OF THE ANS
Charged up again to beat the beast :)
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by [email protected] » Thu Aug 13, 2009 9:01 pm
probability that a contestant does not taste all of the samples =
1 - probability that a contestant tastes all of the samples

now, lets calculate probability that a contestant tastes all of the samples.

there are 3 cases for this, as there are three samples i. e. 2,1,1(two cups of 1st sample, 1 cup of 2nd sample, 1 cup of 3rd sample), 1,2,1 and 1,1,2.

probability that a contestant tastes all of the samples = (3C2 * 3C1 * 3C1 + 3C1 * 3C2 * 3C1 + 3C1 * 3C1 * 3C2)/ 9C4
= (3*27)/(63*2)
= 9/14

therefore,
probability that a contestant does not taste all of the samples = 1 - 9/14
= 5/14
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by ket » Fri Aug 14, 2009 10:51 am
ankit1383

Could you please tell me what's the source of this problem? Is it OG?

Thanks
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by ankit1383 » Fri Aug 14, 2009 5:53 pm
Ket
I dnt remember the Source exactly...but rest assure the answer provided is absolutely correct......
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