Probability

[studybrat]

Hi Someone Help, please.

This is new to me; I'm trying to prepare myself for the next chapter, but I don't think the book provides good examples.

Here are two questions that seem hard:

A manufacturer of engine parts knows tht the probability of part A malfunctioning is .25 and the prob. of part B is .45. The manufacturer also knows that the two parts can malfunction together. What is the prob. of part A or part B malfunctioning?

Do I mult. both numbers together and divide by two? Help!

Also:

A company is hiring four people. What is the prob. that those four people were born in different months.

I don't even know how to start this one. Thanks.

 

[Unco]

A manufacturer of engine parts knows tht the probability of part A malfunctioning is .25 and the prob. of part B is .45. The manufacturer also knows that the two parts can malfunction together. What is the prob. of part A or part B malfunctioning?

Assuming the two events are independent, the probability of them both malfunctioning is 0.45 * 0.25 = 0.1125

Now you can use the formula:
P(A or B) = P(A) + P(B) - P(A and B)

 

[Unco]

A company is hiring four people. What is the prob. that those four people were born in different months.

P(an event occurs) = (no. of ways for that even to occur) / (total choices)

There are 12 places to allocate 4 people. The first person has 12 places to be allocated, the 2nd person has 11, the 3rd 10 and the 4th has 9.

Thus the total number of ways of the four people being born in different months is 12*11*10*9.

The total choices will come from each person being able to be allocated to all 12 months. So it is 12 * 12 * 12 * 12.

 

[studybrat]

That was Fast

I was getting on to see if I worked the 2nd prob. question right & I did, yeah!

Thanks for your help, you rock!

 

[soroban]

Hello, studybrat!

Here's another approach to #2 . . .

I must assume you are familiar with a basic theorem . . .

If A and B are independent events, then: . P(A and B) .= .P(A) × P(B)
. . (The probability that both happen is the product of their probabilities.)

A company is hiring four people.
What is the prob. that those four people were born in different months?

Call the four people: A, B, C, D.

Person A can be born in any month: he/she has 12 choices of months.
. . P(A has a birthday) = 12/12

Person B must not have the same birthmonth as A: he/she has 11 choices of months.
. . P(B has a different birthmonth) = 11/12

Person C must not have the same month as A or B: 10 choices.
. . P(C has a different birthmonth) = 10/12

Person D must not have the same month as A, B, or C: 9 choices.
. . P(D has a different birthmonth) = 9/12

The probability of all four of these events is the product of their probabilities:
. . 12/12 × 11/12 × 10/12 × 9/12 .= .55/96 .≈ .61%

 

[studybrat]

Yup, that's what I got...groovy!