Suppose we roll 10 fair six-sided dice. What is the probability that there are
exactly two 2’s showing?
Solution:
There are ${10\choose2} = 45$ ways of choosing which two dice will have 2 showing. Then the probability that those two dice show 2, and the other eight do not, is $(1/6)^{2}(5/6)^{8}$. So, the answer is $45(1/6)^2(5/6)^8 = 0.2907$
I don't get a lot of parts.
"There are ${10\choose2} = 45$ ways of choosing which two dice will have 2 showing."
What does above statement mean visually? For example the outcomes of one dice is
$$S = \{1,2,3,4,5,6\}$$
And we roll it 10 times so the sample space is $(1/6)^{10}$
So what is 10 choose 2 saying?
$\{(1,1,1,1,1,1,1,1,1,1), (1,1,1,1,1,1,1,1,1,2), \cdots , (6,6,6,6,6,6,6,6,6,6)\}$
Would 10 choose 2 be the number of posibilities for there to be two 2's like
$(1,1,1,1,1,1,1,1,2,2), (1,1,1,1,1,1,1,6,2,2)$, … , etc.
Best Answer
We have 10 outcomes. Lets call each outcome $X_i$ for $i \in (1,2,3,4,5,6,7,8,9,10)$ which is either a 2 or not a 2. Therefore the $P(X_i=2)= \frac 16$ since there is 1 in 6 ways to roll a 2 on a fair die AND $P(X_i \neq 2)= \frac 56$ since there are 5 out of 6 ways to NOT roll a 2.
Now it is clear that the probability of rolling a 2 on the first go and then a 2 on the second go is $$P(X_1=2) \times P(X_2=2)$$ Then we must roll NOT 2 on all the other rolls. That is: $$P(X_3 \neq 2) \times P(X_4 \neq 2) \times ... \times P(X_{10} \neq 2)$$. Multiplying these together gives:
$$P(Roll \space EXACTLY \space two 2s)$$ $$= P(X_1=2) \times P(X_2=2) \times P(X_3 \neq 2) \times P(X_4 \neq 2) \times ... \times P(X_{10} \neq 2)$$ $$=(\frac 16)^2 \times (\frac 56)^{10-2}=(\frac 16)^2 \times (\frac 56)^8$$
But... we are not done yet. This is only if the first roll and the second roll were 2 and the remaining 8 rolls being not 2. We could also have had the first roll and third roll being 2, and the remaining 8 rolls being not 2. Let us ask then: How many ways are there that we can arrange two 2s in 10 places? Turns out there are 2 objects and 10 places. The objects here is the key. The first object is our two 2s. The second object is all our other rolls that were not 2s as before. This is where the ${10\choose2} = 45$ comes from.
So in summary, what we are saying is that first, the probability of getting two 2s then 8 non-2s is $(\frac 16)^2 \times (\frac 56)^8$ and there are 45 ways to place exactly two 2s into 10 spots.
Let me know if I can clarify anything!