Question

the accompanying table shows the numbers of male and female students in a certain region who received bachelors degrees in a certain field in a recent year. a student is selected at random. find the probability of each event listed in parts (a) through (c) below.
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(a) the student is male or received a degree in the field
the probability is 0.521.
(type an integer or a decimal. round to three decimal places as needed.)
(b) the student is female or received a degree outside of the field
the probability is 0.895.
(type an integer or a decimal. round to three decimal places as needed.)
(c) the student is not female or received a degree outside of the field
the probability is
(type an integer or a decimal. round to three decimal places as needed.)

Explanation:

Step1: Use the formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Let \(A\) be the event that the student is not - female (i.e., male) and \(B\) be the event that the student received a degree outside of the field.
We know that \(n(A) = 798025\) (number of male students), \(n(B)=1494277\) (number of students with degrees outside the field), and \(n(A\cap B)=602701\) (number of male students with degrees outside the field). The total number of students \(n = 1861136\).

Step2: Calculate \(P(A)\), \(P(B)\) and \(P(A\cap B)\)

\(P(A)=\frac{n(A)}{n}=\frac{798025}{1861136}\), \(P(B)=\frac{n(B)}{n}=\frac{1494277}{1861136}\), \(P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{602701}{1861136}\)

Step3: Calculate \(P(A\cup B)\)

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Answer:

\(0.908\)