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.