Answer:
[tex]\dfrac{1}{16}[/tex]
Step-by-step explanation:
Probability Distribution Table for X
Where X is the score on a fair, six-sided dice:
[tex]\begin{array}{|c|c|c|c|c|c|c|}\cline{1-7} x & 1 & 2 & 3 & 4 & 5 & 6\\\cline{1-7} \text{P}(X=x) \phantom{\dfrac11}&\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\\cline{1-7}\end{array}[/tex]
Probability Distribution Table for Y
Where Y is the score on an unfair, six-sided dice, and the probabilities of rolling numbers 2, 3, 4, 5 and 6 are equal, and the probability of rolling a 1 has a different value:
[tex]\begin{array}{|c|c|c|c|c|c|c|}\cline{1-7} y & 1 & 2 & 3 & 4 & 5 & 6\\\cline{1-7} \text{P}(Y=y) \phantom{\dfrac11}&1-5k&k&k&k&k&k\\\cline{1-7}\end{array}[/tex]
Note: The probabilities of all the possible values that a discrete random variable can take add up to 1.
The outcomes of rolling one 3 and one 4 are:
- Rolling a 3 with the first die and a 4 with the second die, OR
- Rolling a 4 with the first die and a 3 with the second die.
Given the probability of rolling one 3 and one 4 is ¹/₂₄:
[tex]\implies \text{P}(X=3)\; \text{and} \; \text{P}(Y=4) \; \text{or} \; \; \text{P}(X=4) \; \; \text{and} \; \; \text{P}(Y=3)=\dfrac{1}{24}[/tex]
[tex]\implies \dfrac{1}{6} \times k +\dfrac{1}{6} \times k=\dfrac{1}{24}[/tex]
[tex]\implies \dfrac{k}{6} + \dfrac{k}{6}=\dfrac{1}{24}[/tex]
[tex]\implies \dfrac{2k}{6}=\dfrac{1}{24}[/tex]
[tex]\implies 24(2k)=6[/tex]
[tex]\implies 48k=6[/tex]
[tex]\implies k=\dfrac{1}{8}[/tex]
Therefore:
[tex]\implies 1-5k=1-\dfrac{5}{8}=\dfrac{3}{8}[/tex]
Hence, the probability distribution table for the unfair 6-sided dice is:
[tex]\begin{array}{|c|c|c|c|c|c|c|}\cline{1-7} y & 1 & 2 & 3 & 4 & 5 & 6\\\cline{1-7} \text{P}(Y=y) \phantom{\dfrac11}&\frac{3}{8} & \frac{1}{8} & \frac{1}{8}& \frac{1}{8} & \frac{1}{8}& \frac{1}{8} \\\cline{1-7}\end{array}[/tex]
There is only one possible outcome of rolling two 1's, so the probability of rolling two 1's is:
[tex]\implies \text{P$(X=1)$ and P$(Y=1)$}=\dfrac{1}{6}\times\dfrac{3}{8}=\dfrac{3}{48}=\dfrac{1}{16}[/tex]
Note: Please see attachment for the sample-space diagram showing the possible outcomes of rolling two dice. The possible outcomes of rolling one 3 and one 4 are highlighted in yellow. The possible outcome of rolling two 1's is highlighted in blue.
