Question

At a certain university, 42% of the students are women, and 18% are engineering majors. Of the engineers, 22% are women. If a student at this university is selected at random, what is the probability that the selected person is either a woman or an engineering major?0.60000.03960.5604

Answer 1

Answer: 0.5604

Step-by-step explanation:

Let A represents the students who are women and B represents the students who are engineering majors.

Then , we have given that $P(A)=0.42\ ;\ P(B)=0.18\ ;\ P(A|B)=0.22$

By formula , $P(A\cap B)=P(A|B)\times P(B)=0.22\times0.18=0.0396$

Using formula, $\text{P(A or B)=P(A)+P(B) - P(A and B)}$, we have

$\text{P(A or B)}=0.42+0.18-0.0396=0.5604$

Hence, the probability that the selected person is either a woman or an engineering major = 0.5604

Answer 2

Answer:

The answer to this question is 0.5604

Step-by-step explanation:

From the question given we solve for the probability that the selected person is either a woman or an engineering major.

Recall that,

At a certain university, 18% are engineering majors, 42% of the students are women, and for the engineers 22% are women

so,

P( women ) = 0.42

P( engineering majors ) = 0.18

P( engineers who are women ) = 22% of 18 = (22/100) x 18 = 3.96% = 0.0396

Therefore,

P( woman or an engineering major ) = P(woman) + P( engineering major ) - P( Women and engineering major ) = 0.42 + 0.18 - 0.0396 = 0.5604