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# 4 dices are thrown at the same time. What is the probability

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20 Feb 2014, 11:21
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.

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

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.

Thanks
Cheers
J
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20 Feb 2014, 23:39
jlgdr 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.

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

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.

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 $$C^2_4*6*5*4$$.

Hope it's clear.
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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
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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06 Oct 2014, 07:08
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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
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]

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06 Oct 2014, 09:54
Thanks a lot for the help.
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11 Oct 2015, 20:28
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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, 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.

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

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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, 10:54
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.

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

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.
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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

Hi,

Why have you multiplied by 6?

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

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24 Jun 2016, 21:44
MBAhereIcome wrote:
$$= (\frac{1}{6}*\frac{1}{6}*6)*\frac{5}{6}*\frac{4}{6}*\frac{4!}{2!*2!}$$ $$= \frac{5}{9}$$

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!
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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
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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 ....).

Thanks ..
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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.

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

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.?
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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] 04 Oct 2016, 07:06

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