# 8: 7. Probability without Equally Likely Outcomes

- Page ID
- 25693

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

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

#### Prework:

- 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?
- 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 \(Pr[O_5]\).
- 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:

- 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 \(Pr(\text{orange or purple})=Pr(\text{orange})+Pr(\text{purple})=0.25+0.38=0.63.\)
- 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 \(Pr(O_5)=3Pr(O_2)=3\cdot 0.13=0.39.\)
- 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 \(Pr(A^c\cap B)=0.36\).