Problem Solving

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..

I hope ans is B. I can try to explain if it is correct.

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...

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..

Ankit
i can do it from 1-P(A) formulae but not able to figur out what i did wrong ..

OA...B

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

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

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

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!

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

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

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

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

ankit1383

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

Thanks

Ket
I dnt remember the Source exactly...but rest assure the answer provided is absolutely correct......