Computing Probability

[JW Smith]

Need some help...

I have a question that I'm researching and cannot for the life of me figure out the answer. I need the answer rather quickly.

According to records of a manufacturing firm, 70% of its employees are male, 50% work on the assembly line and 30% are males and work on the assembly line.

a. Find the probability that a randomly selected employee is either male or works on the assembly line.

b. Find the probability that a randomly selected employee is either male or works on the assembly line, but not both.

c. Find the probability that a randomly selected employee is neither a male nor works on the assembly line.

d. Find the probability that a randomly selected employee is a male and does not work on the assembly line.

Thanks.
JW

[PeterVincent]

Hi,
This is my attempt and needs checking.

If 70% are male then 30% are female.
If 30% (male AND assembly) then (male AND NOT assembly) = 70%-30%=40%
If 50% assembly and 30% (male AND assembly) then (female AND assembly) = 50% - 30% = 20%

Part a.
(male OR assembly) = male + (female AND assembly) = 70% + 20% = 90%

Part b.
male - (male AND assembly) + (female AND assembly)
= 70% - 30% +20% = 60%

Part c.
(neither a male nor assembly) = (NOT male and NOT assembly) = NOT(male OR assembly)

(male OR assembly) = 90% from part a, therefore NOT(male OR assembly) = 100% -90% = 10%

Part d.
male AND (NOT assembly) = male - (male AND assembly) = 70% - 30% = 40%

It would be great if someone could check this for me and post theri comments.

Thanks,
Peter

[JohnM]

Excellent!

Want to be a contributor?

[PeterVincent]

Thanks JohnM.

Would like to be a contribuitor but I will wait until my knoledge is better.

Peter