Throw a dice 5 times. what is the probability that you get 4 or 6 in two throws, and 1 or 3 in 3 throws. combinations like 13461 are included.

Question

Now normally, I have 8 (111,113,133,131,311,331,333,313) combinations along with 3 ways to obtain it(4 4, 4 6, 6 6) , so 3/24 would be the answer according to my logic, but I am not sure that this is correct. I could really use some help.

Answer 1

Usually statements like "probability that you get a ____" mean "get at least one ___", so that is the interpretation I will adopt. Let $A = \{\text{At least a 4 or a 6 in first 2}\}$ and $B = \{\text{at least 1 or 3 in last 3}\}$. Then $\bar A = \{\text{ No 4 and no 6 in first two}\}$ and $\bar B = \{\text{No 1 and no 3 in last 3}\}$, and \begin{align*} P(AB) &= 1 - P(\bar A\cup \bar B)\\ &= 1- [P(\bar A)+P(\bar B)-P(\bar A \bar B)]\tag 1\\ &= 1- [P(\bar A)+P(\bar B)-P(\bar A)P(\bar B)]\tag 2\\ &=1-\left[\left(\frac{4}{6}\right)^2+\left(\frac{4}{6}\right)^3-\left(\frac{4}{6}\right)^2\left(\frac{4}{6}\right)^3\right]\\ &=0.3909465 \end{align*} where in $(1)$ I invoke inclusion exclusion, and in $(2)$ I recognize independence.

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If you want $A = \{\text{4 or 6 in two}\}, B = \{\text{1 or 3 in three}\}$, then $\bar A = \{\text{No 4 or 6 in two}\}, \bar B = \{\text{1 or 3 in three}\}$, and \begin{align*} P(A B)&= P(A)P(B)\tag 1\\ &=[1-P(\bar A)][1-P(\bar B)]\\ &=\left[1-\binom{2}{0}\left(\frac{2}{6}\right)^0\left(\frac{4}{6}\right)^2\right]\left[1-\binom{3}{0}\left(\frac{2}{6}\right)^0\left(\frac{4}{6}\right)^3\right]\tag 2\\ &= 0.3909465 \end{align*} where $(1)$ is true by independence and in $(2)$ I recognize the binomial distribution.