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Four dice are thrown simultaneously. The probability that $4$ and $3$ appear on two of the dice given that $5$ and $6$ appear on the other two dice is:

a) $1/6$

b) $1/36$

c) $12/51$

d) None of these

Since the events are independent, I feel the probability is $1/6 times 1/6 = 1/36$

But answer is c. Why?

Even if there are only two dice, the probability of observing $4$ and $3$ is not $1/6 times 1/6$, but rather, $1/18$, since the sample space is the set of ordered pairs $(a,b)$ where each $a, b in {1, 2, 3, 4, 5, 6}$. Thus there are two desired outcomes $(4,3)$, $(3,4)$ out of $6^2 = 36$ possible outcomes.

When there are four dice, two of which you are told are $5$ and $6$, you must reason carefully and precisely. Given the set of all outcomes of four dice rolls, select those that show at least one $5$ and at least one $6$. Of these, how many show $3$ and $4$ on the other two dice?

Doing mathematics is not about "feelings." It is about showing and justifying your reasoning. Describing your calculation without providing a sound basis for why you are doing what you are doing, is not math.

The tricky part of the solution is to find the number of outcomes such that 2 dice land $5$ and $6$; it can be done using inclusions/exclusions.

Let $Omega$ be the set of all outcomes, of size $6^4$.
Let $S_5$ be the set of outcomes that contain no $5$'s, of size $5^4$.
Let $S_6$ be the set of outcomes that contain no $6$'s, of size $5^4$.
Let $S_{5,6}$ be the set of outcomes that contain no $5$'s no $6$'s, of size $4^4$.

Using inclusion/exclusion principle the number of outcomes that contain at least one $5$ and at least one $6$ is
$$6^4-5^4-5^4+4^4=302$$
The rest is simple, and the answer is $12/151$