Question

At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random. Navidi, William,Navidi, William. Statistics for Engineers and Scientists (p. 85). McGraw-Hill Higher Education. Kindle Edition.

Answer 1

At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random.

a) What is the probability that the student is majoring in engineering?

b) What is the probability that the student plays club sports?

c) Given that the student is majoring in engineering, what is the probability that the

student plays club sports?

d) Given that the student plays club sports, what is the probability that the student is

majoring in engineering?

e) Given that the student is majoring in engineering, what is the probability that the

student does not play club sports?

f) Given that the student plays club sports, what is the probability that the student is not

majoring in engineering?

Answer and Explanation

The venn diagram for the question is in the attachment.

Percentage majoring in engineering = 30% = 0.3

Percentage that plays club sport = 20% = 0.2

Percentage that major in engineering and play sport = 10% = 0.1

Percentage majoring in engineering & do not play club sport = 30 - 10 = 20% = 0.2

Percentage that plays club sport & do not major in engineering = 20 - 10 = 10% = 0.1

Total percentage = 100%

a) probability that student is majoring in Engineering P(E) = 30/100 = 0.3

b) probability that student plays club sport = 20/100 P(S) = 0.2

c) probability that the

student plays club sport given that the student is majoring in engineering, P(S|E) = (P(E and S))/(P(E))

P(E and S) = 10/100 = 0.1, P(E) = 0.3

P(S|E) = 0.1/0.3 = 0.3333

d) probability that the

student is majoring in engineering given that the student plays club sport, P(E|S) = (P(S and E))/(P(S))

P(S and E) = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

e) probability that the

student does not play club sport given that the student is majoring in engineering, P(S'|E) = (P(E and S'))/(P(E))

P(E and S') = 20/100 = 0.2, P(E) = 0.3

P(S'|E) = 0.2/0.3 = 0.667

f) probability that the

student is not majoring in engineering given that the student plays club sport, P(E'|S) = (P(S and E'))/(P(S))

P(S and E') = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

Answer 2

Answer:

The probability that Student both majors in engineering and plays club sport is 0.4.

Explanation:

• The probability that a student selected at random majors in engineering is 30% which is 0.3.
• The probability  that the student both majors in engineering and play club sports is 10% which is 0.1.

There are two possible ways were a student is selected at random in Engineering and Plays club.

The Probabilty that the student majors in Enginnering $=0.3$  

The probability that the student majors in plays club $=0.1$  

The probability that Student both majors in engineering and plays club

= Probability of engineering + probability of plays club

$=0.3+0.1 =0.4$

To learn more:

• https://brainly.com/question/20004122