Three dice are thrown together. What is the probability that exactly two of them show the same face?
I can get anything over first die…so sure event thus probability $1$.
Now in second have to get the very same number as of I got over the first die…so the probability would then be $\frac{1}{6}$ and then in the third one I have to get a number other than that on the first (…which is also same as second)..thus the probability then would be $\frac{5}{6}$.
Thus, total probability would be $\frac{5}{36}$ for one case …but we have such $^3C_2 = 3$ cases…
Thus, total probability now would be $=3.\frac{5}{36}=\frac{5}{12}$
So is my answer and arguments made right…?
If not please comment on my approach.
Best Answer
Let's first count how many ways are there for there to be at least $2$ of the same face, and then how many ways for $3$ of the same face.
For the first, we use complementary counting, as we can count the number of cases when all three dice show different values. This can happen in $_6P_3=6\cdot5\cdot4=120$ ways. So, the final result is $6^3-120=96$.
For the second, there are clearly only $6$ ways this can happen.
So, the final number of possibilities is $96-6=90$. We divide by the total number of possibilities, which is $216$, to get $$\frac5{12}$$
Hence, your answer is correct. I've just provided an alternate approach.