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.

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