Do not understand why my calculation for this die roll question is wrong

Question

You roll a fair six-sided die (all 6 of the possible results of a die roll are equally likely) 5 times, independently. Let $X$ (a random variable) be the number of times that the roll results in 2 or 3. Find $p_X(1)$.

This is a question from 6.431x at MITx, and I have exhausted my attempts so I know the correct way to compute this. I think I understand why the solution works and I am refraining from discussing it because that may violate the honour code.

What I would like to know is why my calculation was wrong.

$$\left(\binom{5}{1}\times \frac{1}{6}\left(1-\frac{1}{6}\right)^4\right)\times 2$$

$p_X(1)$ means the probability of there being a sequence that has either a 2 or a 3. In other words, out of the 5 slots in a 5-rolls sequence, there is one slot which is a 2, or a 3. Hence the $\binom{5}{1}$.

So we calculate the probability of getting a sequence which has a 2 (thus $1/6 \times (1-1/6)^4)$, and add that to the probability of getting a sequence which has a 3 (thus multiplying $\binom{5}{1}((1/6) \times (1-1/6)^4$ by $2$).

My answer is considerably larger than the correct answer (0.804 vs 0.329), so I suspect I have double counted somewhere; but where exactly did I go wrong?

Answer 1

So we calculate the probability of getting a sequence which has a 2 (thus $1/6 \times (1-1/6)^4)$, and add that to the probability of getting a sequence which has a 3 (thus multiplying $\binom{5}{1}((1/6) \times (1-1/6)^4$ by $2$).

My answer is considerably larger than the correct answer (0.804 vs 0.329), so I suspect I have double counted somewhere; but where exactly did I go wrong?

You had the right idea, however you added probabilities for a disjoint union of events at the wrong point; it should be done for the probability of the results for a particular roll of the die.

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You actually gave the probability that exactly one among five die show a 2 and exactly one among another five die show a 3.

Note: nothing prohibits the first five from showing one or more 3 or the other five from showing one or more 2.

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Instead, you seek the probability that exactly one among five die show a 2 or 3.

Now, a particular die will show that with probability of $2/6$.   Four other die will not show either with probability $(1-(2/6))^4$.

The rest of your logic was applied okay: there are $\tbinom 51$ ways to select which die among the five might show the favoured result, so the probability we seek is:$$\binom 51 \frac 26\left(1-\frac 26\right)^4$$

or simply $\dfrac{5\cdot 2^4}{3^5}$

Answer 2

You forgot that, once you know which of the throws gave you $2$ (or $3$), the remaining throws mustn't be either $2$ or $3$, so you should multiply by $\left(1-\frac{2}{6}\right)^4$ rather than by $\left(1-\frac{1}{6}\right)^4$.

Answer 3

The probability of a 2 or 3 is $\frac{2}{6} = \frac{1}{3}$

Therefore,

$px(1) = {5 \choose 1}\times\frac{1}{3}\times(1-\frac{1}{3})^4 \approx 0.329$