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

Question

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

what i have done is:
total ways of getting values when 4 dices are thrown: 6^4
the 2 dices showing the same phase can be done in: 6 ways ............. (1)
total ways of arranging these 4 face value: 4!/2!.................................(2)
the remaing 2 places can be filled in 5c2 ways.....................................(3)

Bunuel · Accepted Answer

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.

chetan2u · Answer

my take on this is....
choosing first no will have possibility of 1.... 
choosing the same on any of other three dice is  \frac{1}{6} 
third dice can have any of remaining 5 so probability is  \frac{5}{6.} ..
fourth dice can have any of remaining 4 so probability is  \frac{4}{6} ...
 so prob= 1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6} ...
.but there are six different values so it becomes = 1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}*6=\frac{5}{9} ..

shahideh · Answer

Here is my approach:

If we name the dices: 1-2-3-4
there are 6 possibilities that just 2 of them are the same:  1&2-1&3-1&4-2&3-2&4-3&4 
and to calculate probability of the first one, 1&2, we know that the first dice can be every thing. but the second should be the same as first. the third should be something else than the first and second, and finally the forth should be something else than the three previous ones. So:
 1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}

and as i said, we have 6 possibilities. So total probability is:
 6*1*\frac{1}{6}*\frac{5}{6}*\frac{4}{6}=\frac{5}{9}

R2I4D · Answer

Did the exact same thing. I hope that having similar thought processes to Bunuel's continues...

krishnasty · Answer

Please explain why am i going wrong -

favourable outcomes -
1st dice can have any number..hence, 6 possible ways
2nd dice ( same number ) = 1 way
3rd dice = 5 ways
4th dice = 4 ways
total favourable outcome = 6*1*5*4 = 120 ways

total outcome = 6*6*6*6 ways = 1296.

Hence, probablity = 120/1296 = 5/54

prakhag · Answer

Done in the same manner:

Probability of getting two same faces: 1 and 1/6
For other two faces, probabilities: 5/6 and 4/6 respectively.
The dice combinations can be arranged in 4!/2! ways
Total probability = 1*1/6*5/6*4/6*4!/2! = 5/9

venmic · Answer

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

Answer: 5/9.

************

This was a solution posted by you on probability
can you please explain why is the 6 shown twice ??

why is the 6 given twce it shuld be only ONCE

becuase the only possibility of 2 dices getting the same number is

66
55 
44
33
22
11

that is 6 possibilities

so 6*5*4/6^4

why is the 6C2 in the picture

please explain

Bunuel · Answer

We have 4 dice (A, B, C, D), not 2. So,  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.

SVaidyaraman · Answer

1. The formula is nCr 
2. n is 6 as a dice has 6 values
3. r is 1 as only one value is shown
4. But there are 4 dices and there is a constraint on the value of n for 3 dices
5. Let the first dice show any value. Thus it can show 6 values and n is 6. r is 1
6. The second dice in consideration has to show the same value. Therefore n is 1. r is 1
7.The third dice cannot show the value on the first two dices and therefore n is 5. r is 1
8. Similarly for the fourth dice n is 4. r is 1
9. We can select 2 dices out of 4 dices in  4C2 ways 
10. The total number of ways of exactly 2 dices out of 4 dices showing the same face is  4C2 *6C1*1C1*5C1*4C1= 720 
11. The total number of possibilities with any value with 4 dices is 6^4=1296 
12. The probability is 720/1296 = 5/9

jlgdr · Answer

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

Bunuel · Answer

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.

arpitsharms · Answer

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

Bunuel · Answer

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

rsaahil90 · Answer

Hello Bunuel

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

Thank you

Bunuel · Answer

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.

HanoiGMATtutor · Answer

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.

IIMC · Answer

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.

chetan2u · Answer

you are wrong with your combinations  \frac{4!}{2!}  and that is why your answer gives you probability >1

now where do you use 4!/2!
say "how many words can be formed from BULB?" 
4!/2! is Ok because arrangement is important

But in our case ALL dices are similar, so selection is what is required and therefore 4C2

hope it helps

BrushMyQuant · Answer

Given that 4 dices are thrown at the same time and We need to find What is the probability of getting ONLY 2 dices showing the same face?

As we are rolling four dice => Number of cases =  6^4  = 1296

Now lets pick the two dice which will show the same number in 4C2 ways
=>  \frac{4!}{2!*(4-2)!}  =  \frac{4*3*2!}{2!*2!}  = 6 ways

Now, these two dice can get any of the six numbers in 6 ways
The third dice can get any of the remaining 5 numbers in 5 ways
The fourth dice can get any of the remaining 4 (excluding the two numbers used above) numbers in 4 ways

=> Total number of cases = 6 * 6 * 5 * 4

=>  probability of getting ONLY 2 dices showing the same face  =  \frac{6 * 6 * 5 * 4}{6^4}  =  \frac{5}{9}

So,  Answer will be B 
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

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

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