Probability Function Solution Clarification

Question

In the solution in the screenshot linked below, could somebody explain why P({4,6}) + P({7,8,9}) = 1/3 and clarify on how the probabilities are derived for each number.

Screenshot of solution

Thanks in advance!

Answer 1

We are tasked with finding a probability distribution on the set $S=\{0,1,2,3,4,5,6,7,8,9\}$ such that $P(\{0,1,2\})=P(\{3,5\})=\frac{1}{3}$, such that $P(\{0,1,4,6\})=\frac{1}{6}$, such that $P(\{8\})=0$, and such that the events $\{0,1,4,6\}$ and $\{4,6,7,8,9\}$ are independent of one another.

In your post, you specifically ask about the step of why we can from this information alone learn that $P(\{4,6\})+P(\{7,8,9\})=\frac{1}{3}$

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To begin, remember the following:

• Letting $S$ represent the sample space, one has $P(S)=1$

• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

  • In particular this implies when $A\cap B=\emptyset$ that $P(A\cup B)=P(A)+P(B)$

• $P(A) + P(S\setminus A) = 1$, or phrased differently $P(A^c)=1-P(A)$

• $A$ and $B$ are independent events if and only if $P(A\cap B)=P(A)\times P(B)$ (Not important for this specific step, but important for later steps)

Since we are told in the problem that we want our probability distribution to satisfy $P(\{0,1,2\})=P(\{3,5\})=\frac{1}{3}$ and since $\{0,1,2\}\cap \{3,5\}=\emptyset$ this implies that $P(\{0,1,2,3,5\}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$.

Further, since $S\setminus\{0,1,2,3,5\}=\{4,6,7,8,9\}$ we learn that $P(\{4,6,7,8,9\})=1-P(\{0,1,2,3,5\})=1-\frac{2}{3}=\frac{1}{3}$

Finally, since $\{4,6,7,8,9\}=\{4,6\}\cup\{7,8,9\}$ and since $\{4,6\}\cap \{7,8,9\}=\emptyset$ one has that $P(\{4,6,7,8,9\})=P(\{4,6\})+P(\{7,8,9\})$

Putting all of this together we have then $P(\{4,6\})+P(\{7,8,9\})=\frac{1}{3}$

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One can continue and find specific probabilities for several other of the numbers or at the least sets of numbers using these same properties. The pictured solution already does a fine job of going through each of these.