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

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Joined: 06 Sep 2013
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Concentration: Finance
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Re: Really tough [#permalink]  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
Math Expert
Joined: 02 Sep 2009
Posts: 29802
Followers: 4908

Kudos [?]: 53718 [0], given: 8167

Re: Really tough [#permalink]  20 Feb 2014, 23:39
Expert's post
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]  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
Math Expert
Joined: 02 Sep 2009
Posts: 29802
Followers: 4908

Kudos [?]: 53718 [1] , given: 8167

Re: 4 dices are thrown at the same time. What is the probability [#permalink]  06 Oct 2014, 07:08
1
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
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]  06 Oct 2014, 09:54
Thanks a lot for the help.
Re: 4 dices are thrown at the same time. What is the probability   [#permalink] 06 Oct 2014, 09:54

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