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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]
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
Re: 4 dices are thrown at the same time. What is the probability [#permalink]
06 Oct 2014, 07:08
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 _________________
Re: 4 dices are thrown at the same time. What is the probability [#permalink]
11 Oct 2015, 20:28
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Re: 4 dices are thrown at the same time. What is the probability [#permalink]
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.
Re: 4 dices are thrown at the same time. What is the probability [#permalink]
18 Nov 2015, 10:54
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. _________________
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