The probability that a male and has a favorable opinion is selected is 2/5
The probability of an event is defined as the possibility of an event occurring against sample space.
[tex]\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }[/tex]
Permutation is the number of ways to arrange objects.
[tex]\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }[/tex]
Combination is the number of ways to select objects.
[tex]\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }[/tex]
Let us tackle the problem.
This problem is about Probability.
Given:
Of the 62 favorable responses, there were 40 males and 22 females.
Of the 38 unfavorable responses, there were 15 males and 23 females.
There were total (40+15) males = 55 males
Let P(M) = probability a male is selected
[tex]P(M) = \frac{\texttt{Number of Male Citizens}}{\texttt{Total Number of Citizens}}[/tex]
[tex]P(M) = \boxed{\frac{55}{100}}[/tex]
There were 62 favorable responses
Let P(F) = probability citizens has a favorable opinion is selected
[tex]P(F) = \frac{\texttt{Number of Favorable Opinion Citizens}}{\texttt{Total Number of Citizens}}[/tex]
[tex]P(F) = \boxed{\frac{62}{100}}[/tex]
Let P(M ∪ F) = probability a male or citizens has a favorable opinion is selected
[tex]P(M \cup F) = \frac{\texttt{Number of Favorable Opinion Citizens or Male Citizens}}{\texttt{Total Number of Citizens}}[/tex]
[tex]P(M \cup F) = \frac{40 + 15 + 22}{100}[/tex]
[tex]P(M \cup F) = \boxed{\frac{77}{100}}[/tex]
At last , the probability a male and has a favorable opinion could be calculated as follows:
[tex]P(M \cap F) = P(M) + P(F) - P(M \cup F)[/tex]
[tex]P(M \cap F) = \frac{55}{100} + \frac{62}{100} - \frac{77}{100}[/tex]
[tex]P(M \cap F) = \frac{40}{100}[/tex]
[tex]P(M \cap F) = \large{\boxed {\frac{2}{5}} }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Probability
Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation
