Question

An engineering school reports that 52% of its students are male (M), 33% of its students are between the ages of 18 and 20 (A), and that 27% are both male and between the ages of 18 and 20. What is the probability of a random student being chosen who is a female and is not between the ages of 18 and 20?

Answer 1

Answer:

42%

Step-by-step explanation:

Given: P(M) = 0.52, P(A) = 0.33, and P(M and A) = 0.27.

Find: P(not M and not A).

P(not M and not A) = 1 − P(M or A)

P(not M and not A) = 1 − (P(M) + P(A) − P(M and A))

P(not M and not A) = 1 − (0.52 + 0.33 − 0.27)

P(not M and not A) = 1 − 0.58

P(not M and not A) = 0.42

Answer 2

Treating these probabilities as Venn probabilities, it is found that there is a 0.42 = 42% probability of a random student being chosen who is a female and is not between the ages of 18 and 20.

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The events are:

• Event A: Female.
• Event B: Not between the ages of 18 and 20.

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• 52% of the students are male, thus, 48% are female, and $P(A) = 0.48$.
• 33% are between the ages of 18 and 20, thus, 67% are not between these ages, which means that $P(B) = 0.67$
• 27% are both male and between these ages, which means that 73% are either female or not between these ages, thus $P(A \cup B) = 0.73$.

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The probability of a random student being chosen who is a female and is not between the ages of 18 and 20 is given by:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Inserting the probabilities we found:

$P(A \cap B) = 0.48 + 0.67 - 0.73 = 0.42$

0.42 = 42% probability of a random student being chosen who is a female and is not between the ages of 18 and 20.

A similar problem is given at https://brainly.com/question/21421475