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4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

Could anybody please clarify Thanks Cheers J

We have 4 dice (A, B, C, D).

\(C^2_4\) is the ways to select which two dice out of 4, will provide the same face: (A, B), (A, C), (A, D), (B, C), (B, D), or (C, D) . Next, these two dice can take 6 values, the remaining two 5 and 4, respectively. So, the # of favorable outcomes is \(C^2_4*6*5*4\).

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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06 Oct 2014, 05:11

Although I did get the answer right,I could not figure out the breakup of other possibilities.And it has been bugging me. Total possible outcomes = 6 * 6 * 6 * 6 = 1296

Possibility of having no.s on all dices same = 6 * 1 = 6

Possibility of having 3 same no.s = 6 * 5 * 4!/3! = 120

Possibility of having 2 same no.s and 2 different= 6 * 5C2 * 4!/2! = 6 * 10 * 12 = 720

Possibility of having 2 pairs of same no.s = 6 * 5 =30

Possibility of having all different no.s = 6 * 5 * 4* 3 = 360 Now these possibilities don't add up 6 + 120 + 720 + 30 + 360 = 1236(total is 1296 i.e. 60 more than this) Can anybody please tell which possibilities have i missed??? Thanks

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06 Oct 2014, 08:08

1

This post received KUDOS

Expert's post

arpitsharms wrote:

Although I did get the answer right,I could not figure out the breakup of other possibilities.And it has been bugging me. Total possible outcomes = 6 * 6 * 6 * 6 = 1296

Possibility of having no.s on all dices same = 6 * 1 = 6

Possibility of having 3 same no.s = 6 * 5 * 4!/3! = 120

Possibility of having 2 same no.s and 2 different= 6 * 5C2 * 4!/2! = 6 * 10 * 12 = 720

Possibility of having 2 pairs of same no.s = 6 * 5 =30

Possibility of having all different no.s = 6 * 5 * 4* 3 = 360 Now these possibilities don't add up 6 + 120 + 720 + 30 + 360 = 1236(total is 1296 i.e. 60 more than this) Can anybody please tell which possibilities have i missed??? Thanks

This should be 90: 3*(6*5). The number of ways to split 4 dice into 2 pairs is 3: AB - CD AC - BD AD - BC _________________

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11 Oct 2015, 21:28

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Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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18 Nov 2015, 05:06

Hello Bunuel

In such a question, is it imperative to assume that the dice will be different (not identical)?

Thank you

Bunuel wrote:

roshanaslam wrote:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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18 Nov 2015, 11:54

Expert's post

rsaahil90 wrote:

Hello Bunuel

In such a question, is it imperative to assume that the dice will be different (not identical)?

Thank you

Bunuel wrote:

roshanaslam wrote:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

Yes. 3 on the first die, say on blue one, and 4 on the second die, say on red one, is a different case from 3 on the second die, on red die, and 4 on the first die, on blue one. _________________

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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24 Jun 2016, 04:10

chetan2u wrote:

my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is 1/6 . third dice can have any of remaining 5 so probab is 5/6... fourth dice can have any of remaining 4 so probab is 4/6... so prob=1*1/6*5/6*4/6....but there are six different values so it becomes =1*1/6*5/6*4/6*6=5/9.. anyone if i have gone wrong

we multiply by 6 because there are 6 possibilities of having same numbers {1,1}.. {6,6} etc. we multiply by 4! for possible combinations in between the four dices, by 2! for two same numbers and again by 2! for possibilities between same & not-same numbers.

Can someone please explain why we are dividing again by 2!? Thanks!

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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24 Jun 2016, 23:52

This is a good question. Earlier I guessed the right answer and later understood why that must be right. Step 1. Choose a side : let us say 2. First we need to choose to put 2 in any of the four places. That is 4C2 = 6. To each of these there is 5*4 possibilities of other two sides not exactly same as 2. Now to each of the possibilities above 6*5*4 we need to multiply by 6. Since we chose 2 above, we can choose any of the six. So now required possibilities are 6*6*5*4. Prob = 6*6*5*4/6*6*6*6. = 20/36 = 5/9. Choice B

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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25 Jun 2016, 03:04

chetan2u wrote:

my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is 1/6 . third dice can have any of remaining 5 so probab is 5/6... fourth dice can have any of remaining 4 so probab is 4/6... so prob=1*1/6*5/6*4/6....but there are six different values so it becomes =1*1/6*5/6*4/6*6=5/9.. anyone if i have gone wrong

Nice explanation....I tried the same concept but faltered while calculating the probability of last dice (i considered 5/6 for that as well)... Your explanation just restored my faiths in my concepts (of course I made one of the classic error ....).

Thanks ..

gmatclubot

Re: 4 dices are thrown at the same time. What is the probability
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25 Jun 2016, 03:04

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