Question

multiple events probability and statistics find the probability that event a or b will occur. \\(\frac{5}{19}\\) \\(\frac{6}{19}\\) \\(\frac{4}{19}\\) \\(\frac{4}{19}\\) \\(p(a\text{ or }b)=?\\)

Explanation:

Step1: Recall the formula for \( P(A \text{ or } B) \)

The formula for the probability of \( A \) or \( B \) occurring is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \). From the Venn diagram, we identify the probabilities: the part only in \( A \) is \( \frac{5}{19} \), the intersection \( A \cap B \) is \( \frac{6}{19} \), and the part only in \( B \) is \( \frac{4}{19} \). First, find \( P(A) \) which is the sum of the part only in \( A \) and the intersection: \( P(A)=\frac{5}{19}+\frac{6}{19}=\frac{11}{19} \). Then find \( P(B) \) which is the sum of the part only in \( B \) and the intersection: \( P(B)=\frac{4}{19}+\frac{6}{19}=\frac{10}{19} \). The intersection \( P(A \cap B)=\frac{6}{19} \).

Step2: Apply the formula

Substitute into the formula: \( P(A \cup B)=\frac{11}{19}+\frac{10}{19}-\frac{6}{19} \). First, add the numerators of the first two fractions: \( 11 + 10 = 21 \), so \( \frac{21}{19}-\frac{6}{19} \). Then subtract the numerators: \( 21-6 = 15 \), so \( \frac{15}{19} \). Alternatively, we can also calculate it by adding the three regions (only \( A \), intersection, only \( B \)): \( \frac{5}{19}+\frac{6}{19}+\frac{4}{19}=\frac{5 + 6+4}{19}=\frac{15}{19} \).

Answer:

\(\frac{15}{19}\)