Geometric Distribution Application

Question

The questions Alice throws a die 12 times then gives it to Bob, Bob throws the die 6 times. They then honestly report that the die landed 3 times on 1. What is the probability that Alice will get 1 on exactly 2 times rolling on 12 times rolling?

My attempt My intepretation for this question is the probability that Alice get the 1st success on 2nd trials. I think this is Geometric distribution, so i have p = 3/12 = 0.25 (probability of sucess). Applying pmf of geometric distribution, i have $$(1-0.25)^{2-1}0.25= 0.1875$$

I am not sure if it is correct since i am not confident

Any help would be appreciated. Thank you :)

Answer 1

We know that a 1 is obtained on exactly $3$ of the $18$ rolls of the die, which gives an experimental probability of obtaining a 1 of $$p = \frac{3}{18} = \frac{1}{6}$$ which happens to also be the probability of obtaining a 1 when a fair die is rolled.

We want to find the probability that a 1 is obtained on exactly two of the first twelve rolls. The geometric distribution is used to find the probability that it will take $X$ Bernoulli trials to obtain the first success for an event with probability $p$, which is not what we need here. Instead, we want to use the Binomial distribution, which gives the probability that exactly $k$ successes will occur in $n$ trials of an experiment if each trial has probability $p$ of success. The probability of exactly $k$ successes in $n$ trials is given by the formula $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $\binom{n}{k}$ is the number of ways exactly $k$ successes can occur in $n$ trials, $p^k$ is the probability of $k$ successes, and $(1 - p)^{n - k}$ is the probability of $n - k$ failures.

In this problem, $n = 12$ and $k = 2$. Hence, the probability that exactly two of the twelve rolls that Alice makes result in the outcome 1 is $$\Pr(X = 2) = \binom{12}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{10}$$

Answer 2

The probability that Alice will get 1 on exactly 2 times rolling on 12 times rolling.

$\frac{\binom{12}{2}\times5^{10}}{6^{12}}=\frac{66\times5^{10}}{6^{12}}=0.2961$

$\binom{12}{2}\times5^{10}$ is the number of possible outcomes of two ones if rolling a die 12 times

Answer 3

Huh... I interpret the wording of this question completely differently. I thought it was asking for this: Conditioned on there being three 1s among the 18 rolls, what is the probability that two of the three belonged to Alice? The answer to my question is simply ${12 \choose 2} {6 \choose 1} / {18 \choose 3}$.

I know this conflicts with the word "will" in the question... but other interpretations don't make too much sense either. E.g. if one ASSUMES the die is fair and six-sided, then the info of there being three 1s on 18 rolls is irrelevant. Whereas if one does NOT assume the die is fair, and wants to ESTIMATE the probability $p$ of rolling a 1, saying three 1s out of 18 rolls is insufficient without (IMHO) explicitly stating that one should use the obvious estimate (sample mean).