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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, 04:11

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

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

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18 Nov 2015, 04:06

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.

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, 03:10

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, 22:52

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, 02:04

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 ....).

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

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26 Jun 2016, 00:37

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.

The only thing I'd like to add, a potential shortcut, is that since the numerator has 5, and 5 is not divisible by 6, the final answer has to have a numerator divisible by 5, a condition that only answer choice B meets.
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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28 Aug 2016, 10:29

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

Hi Chetanu- I was attempting the question in same way. But I did not multiply it by 6. Why are we trying to multiply by 6.?

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

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04 Oct 2016, 07:06

I solved it in this way there are 4 positions to be filled --> _ * _ *_ * _ for 1st position we can choose any of 6 number --> 6 * _ *_ * _ 2nd position must match 1st position that can be done in 1/6 ways --> 6 * 1/6 *_ * _

3rd must not match 1st or 2nd that can be done in 5/6 ways --> 6 * 1/6 *5/6 * _ 4th must not match 1st or 2nd or 3rd that can be done in 4/6 ways --> 6 * 1/6 *5/6 * 4/6

so answer comes to be 5/9.

Though I got the answer ,I'm not sure if my logic is correct. Can someone please verify?
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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20 Nov 2017, 07:31

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.

Since we are only selecting 2 dice why isn't it 4C2*6*1. I am a little confused now

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.

Since we are only selecting 2 dice why isn't it 4C2*6*1. I am a little confused now

What is the logic behind 4C2*6*1? Check the logic behind 4C2*6*5*4 in the highlighted part above.
_________________

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

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20 Nov 2017, 22:04

kwesibabban wrote:

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.

Since we are only selecting 2 dice why isn't it 4C2*6*1. I am a little confused now

Well if I have a deck of cards comprising of 2 suits and 6 cards each with values from 1 to 6. The numbers of ways of selecting 4 cards with the same value is 6C4*2*2*2*2? Is there any scenario where this could be 6C4*2*2*2*2*2*2? And if I have deck of cards comprising of 3 suits and 6 cards each. The numbers of ways of selecting 2 cards with different values will be 6C2*3*2? So then if I am rolling 4 dice and I am selecting 2 out of the 4 with the same face why isn't it 4C2*6*1. Is it that in this case of the dice, all 4 were cast but we are only selecting 2? Where as in the case of the cards only some were drawn and not all being drawn and then us selecting some. I hope you understand my logic, there may be some flaws somewhere.

Well if I have a deck of cards comprising of 2 suits and 6 cards each with values from 1 to 6. The numbers of ways of selecting 4 cards with the same value is 6C4*2*2*2*2? Is there any scenario where this could be 6C4*2*2*2*2*2*2? And if I have deck of cards comprising of 3 suits and 6 cards each. The numbers of ways of selecting 2 cards with different values will be 6C2*3*2? So then if I am rolling 4 dice and I am selecting 2 out of the 4 with the same face why isn't it 4C2*6*1. Is it that in this case of the dice, all 4 were cast but we are only selecting 2? Where as in the case of the cards only some were drawn and not all being drawn and then us selecting some. I hope you understand my logic, there may be some flaws somewhere.

You did not explain the logic. The analogy with cards is not accurate.

I''l try to explain once more. The dice are different. Two of them must show the same number. 4C2 is the number of ways to choose which two will show the same number. These two can show any of the 6 numbers, so *6. The remaining two dice can show 5*4 different combination of number. So, 4C2*6*5*4.
_________________

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

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27 Nov 2017, 06:00

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.

I am a bit confused:

my solution is, as given there are 4 dice, A,B,C and D. now, let the number that appears on the 2 dice is 1, so I am left with 2 dice which will have 2 different numbers say 5 and 6 so the one set I will get will be 1,1,5,6. Now, since this re-arranged in 4!/2! ways. now to fill the 2 dice with same number I have 6C1 choices, and for 3rd dice I am left with 5 choices and for the 4th dice I am left with 4 choices so, 6*5*4 and since it can be rearranged in 4!/2! ways.it will be 6*5*4*4!/2! and total possible outcomes are 6^4, so it should be (4!/2!*6*5*4) /6^4....

but doing this I am getting wrong answer, please let me know where I am going wrong.

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

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27 Nov 2017, 07:22

I guess (B).

Reason: Assume u have 4 blanks _,_,_,_ The first blank can be filled in 6 ways, the second can be filled in 1 way (i.e with the same number as obtained during the first fill), the third blank in 5 ways (i.e any number other than first 2 blanks), and the fourth blank in 4 ways. Now these 4 blanks can re reshuffled i.e 3! I took 3! Coz 1 blank needs to be fixed with a repeated number. Now number of total outcomes is 6*6*6*6. So finally, Probability = (6*1*5*4*3!)/(6*6*6*6) = 5/9.