
8: 7. Probability without Equally Likely Outcomes


The following topics are included in this series of eight videos.

1. Not Equally Likely, Example 1
2. Not Equally Likely, Example 2
3. Not Equally Likely, Example 3
4. Disjoint Events
5. Not Equally Likely, Example 4
6. Not Equally Likely, Example 5
7. Not Equally Likely, Example 6
8. Not Equally Likely, Example 7

Prework:

1. Suppose there are some orange, red, and purple balls in a bag, but you don’t know how many of each. You reach in and grab one. If the chance of picking an orange is $$25\%$$ and the chance of picking a red is $$37\%$$, what is the probability that you pick either an orange or purple ball?
2. Let $$S$$ be the sample space for an experiment. Let $$S = \{O_1, O_2, O_3, O_4, O_5\}$$. Assume that $$Pr[O_1] = 0.1$$ and $$Pr[O_4] = 0.25$$. Assume that $$O_2$$ and $$O_3$$ are equally likely and that $$O_5$$ is three times as likely as $$O_2$$. Find $$Pr[O_2]$$ and $$Pr[O_5]$$.
3. Suppose that $$Pr[A] = 0.4$$, $$Pr[B] = 0.6$$, and $$Pr[A \cap B^c] = 0.16$$. Determine $$Pr[A^c], Pr[A\cup B], Pr[A^c\cap B]$$.

Solutions:

1. Since the only three outcomes of this experiment are picking an orange, red, or purple ball, we have a $$100-25-37=38\%$$ chance of picking a purple ball. Therefore $$Pr(\text{orange or purple})=Pr(\text{orange})+Pr(\text{purple})=0.25+0.38=0.63.$$
2. Because there are only these 5 outcomes, we know $$Pr(O_1)+Pr(O_2)+Pr(O_3)+Pr(O_4)+Pr(O_5)=1$$. Since $$O_2$$ and $$O_3$$ are equally likely, we know that $$Pr(O_3)=Pr(O_2)$$, and since $$O_5$$ is three times as likely as $$O_2$$, we know $$Pr(O_5)=3Pr(O_2)$$. Substituting this information and the given probabilities into our initial equation, we get $$0.1+Pr(O_2)+Pr(O_2)+0.25+3Pr(O_2)=1$$. Combining like terms we see that $$0.35+5Pr(O_2)=1$$. We subtract $$0.35$$ and divide by $$5$$ to get that $$Pr(O_2)=0.13$$, and therefore $$Pr(O_5)=3Pr(O_2)=3\cdot 0.13=0.39.$$
3. The answer to the first question is that $$Pr(A^c)=0.6$$ since, for any set $$A$$, $$Pr(A)+Pr(A^c)=1$$. We next make the Venn diagram shown below by first using $$Pr(A\cap B^c)=0.16.$$ By subtraction, $$Pr(A\cap B)=0.24$$ and hence $$Pr(A^c\cap B)=0.36.$$ Therefore, $$Pr(A^c\cap B^c)=0.24.$$ Using the diagram we see that $$Pr(A\cup B)=0.16+0.24+0.36=0.76$$ and $$Pr(A^c\cap B)=0.36$$.