Question

A company has a total of 30 employees.

17 employeers graduated from Berkeley.

Out of 30 employees, 18 are native Californians. Furthermore, the company has 12 mechanical
engineers. Still, only 1 employee can claim the highest honor of being a native Californian, Berkely graduate mechanical engineer.

(a)If an employee is selected at random from this company, what is the probability
that this employee is a Berkely graduate?

(b)A survey reveals that, out of 12 mechanical engineering working at this company,
half are native Californians. How many Californians are not mechanical engineers?

(c)Suppose that a native Californian is selected at random. The probability that this
employee is an Berkeley graduate is 1/2. If a Berkeley graduate is chosen at random, what is the probability that this employee is a native Californian?

(d)When the Berkeley graduates come together for a basketball game, there are only 3
mechanical engineers in the room. If an employee is selected at random from this company,
what is the probability that this employee is an mechanical engineer but neither a Berkeley graduate
nor a native Californian?

Please show work

Answer 1

Total number of employees = n(E) = 30

Employers graduated from berkley = n(B) 17

Employees who are native californians = n(C)18

Employees who are mechanical engineers = n(M) =12

Now from the given data ;

n(B^C^M) = 1

a) The probablility of an employee being berkley graduate = n(B)/n(E)

= 17/30

= 0.56

b) Also, from 12 mechanical engineers half are native californian

This means n(M^C) = 12/2 =6

Now , number of californian who are not mechanical engineers = n(C) - n(M^C) = 18-6 = 12

c) Given , when a native californian is selected , the probability that it is a berkley graduate is half.

This implies that n(B^C)/n(C) = 1/2

n(B^C) = 1/2 * n(C) =1/2 * 18 = 9

when a berkley graduate is selected , the probability that it is a native californian = n(B^C)/n(B)

= 9/17

= 0.529

d) When all the Berkley graduates came for a basketball match , there are only 3 Mechaical engineers ;

This implies that , n(B^M) = 3

Number of students who are mechanical engineers but not a berkley graduate or californian = n(M')

n(M') = n(M) - n(M^C) - n(M^B) + n(M^C^B)

= 12 - 6 - 3 + 1

= 4

Probability of students chosen at random who are mechanical engineers but not a berkley graduate or californian

P(M') = n(M')/ n(E) = 4/30 = 0.13

Hence, Probability of students chosen at random who are mechanical engineers but not a berkley graduate or californian is 0.13.