probabilitystatistics
Midterm Question Help!!
A fair die is rolled 2 times. Let
A = the event that the sum of two numbers shown equals 3.
B = the event that the sum of two numbers equals to 7.
C = the event that at least one of the numbers shown is a “1”.
The probability P(A|C) is equal to?
The probability P(B|C)is equal to?
Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the same number is rolled six times. Thus, the probability of exactly two numbers appearing in six rolls of the die is $$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$
Exactly three numbers appear:
There are $\binom{6}{3}$ ways for three of the six numbers to appear. On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers. However, we have counted the sequences in which not all three of the numbers appear. There are $\binom{3}{2}$ ways for two of the numbers to appear. For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times. Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear. The number of sequences in which exactly one of the three numbers appears is $3$. Hence, the probability that exactly three of the numbers appear is $$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$
My thanks to @bof for clarifying my thinking about the first problem. Any errors that remain are entirely my responsibility.
Per @BGM's Comment, $X \sim \mathsf{Binom}(5, 1/36).$ You can apply the formula for the PDF of a binomial distribution to obtain the probabilities in the given table.
In R statistical software, the computation looks like this:
x = 0:5; pdf = dbinom(x, 5, 1/36)
cbind(x, pdf)
# x pdf
# 0 8.686158e-01
# 1 1.240880e-01
# 2 7.090741e-03
# 3 2.025926e-04
# 4 2.894180e-06
# 5 1.653817e-08
Now you can use the binomial PDF formula to obtain $P(X=4)$ and perhaps a few of the other probabilities in this distribution.
Thank you for showing what you have tried. That always makes it easier to be helpful. Here are some comments on your proposed methods:
Of course, all of these probabilities must add to $1$ as you say, so you could also use that method to verify the missing entry. But your computation $(1/36)^4 = 5.953742e-07$ is incorrect because it does not account for the one roll out of five that doesn't have a total of 12--or the fact that the 'non-12' could happen on any of the five rolls. Compare that computation with the binomial formula, and maybe you'll see what I mean.
Best Answer
Using Bayes Theorem, $$P(A|C)=\frac{P(C|A)P(A)}{P(C)}$$ $$P(B|C)=\frac{P(C|B)P(B)}{P(C)}$$
$$P(C)=1-\frac{5}{6}^2=\frac{11}{36}$$ for both
Note that the only two sequences that add to $3$ are $1,2$ and $2,1$ therefore $$P(C|A)=1$$ also using the information above $$P(A)=2\frac{1}{36}=\frac{1}{18}$$
all together $$P(A|C)=\frac{P(C|A)P(A)}{P(C)}=\frac{\frac{2}{36}}{\frac{11}{36}}=\frac{2}{11}$$
for $$P(B|C)=\frac{P(C|B)P(B)}{P(C)}$$
there are $6$ sequences that add to $7$, $(1,6),(6,1),(2,5),(5,2)(3,4),(4,3)$ therefore
$$P(C|B)=\frac{1}{3}$$ and $$P(B)=6\frac{1}{36}$$
altogether $$P(B|C)=\frac{P(C|B)P(B)}{P(C)}=\frac{\frac{1}{3}\frac{1}{6}}{\frac{11}{36}}=\frac{2}{11}$$