Some (Easy) Probability Questions

[scem0]

I thought I'd get a little help with my math homework. It's easy stuff, but I'm having a little trouble.

Here's some problems (with problems I now understand whited out)

1) A box contains 6 white balls and 5 red balls. A sample of 4 balls is selected at random.
What’s the probability that the sample will have
a) all white balls
b) 2 white balls and 2 red balls

I can do a. It's 6/11 * 5/10 * 4/9 * 3/8 = 360/7920 = .0455

As for b, I'm not completely sure. It's the same concept right? 6/11 * 5/10 * 5/9 * 4/8 = 600/7920 = .0758

2. If P(a) = 0.3 P(b’) = 0.6 and P(a or b) = 0.5. What is P(a and b)?

I drew a Venn Diagram, but I'm not sure of my answer at all. My logic gets me a probability of 0, but that could definitely be wrong

Don't we need to know if he events are mutually exclusive or not to do P(a or b) = p(a) + p(b) vs. P(a or b) = p(a) + p(b) - p(a and b) ?

The rest are based on the given data:

Find probability that:
4. people live in Brooklyn and have a job

P(live in brooklyn) = 400/1000 = 2/5
P(have a job) <-- this is where I get lost. Do I do the probability of them having a job in brooklyn, 3/4, or the probability of them having a job in the whole set, 7/10?

5. people live in Queens or are unemployed

P(Live in Queens) = 3/5
P(Unemployed) = 3/10
P(Live in Queens and Unemployed) = 1/5

3/5 + 3/10 - 1/5 = 9/10 - 2/10 = 7/10

6. people are employed if they live in Brooklyn
P(unemployed and live in brooklyn) = 1/10
P(live in brooklyn) = 4/10
(1/10 )/( 4/10 ) = 1/4

7. people are unemployed if they live in Queens

P(Unemployed and live in queens) = 2/10
P( Live in Queens) = 6/10
(2/10)/(6/10) = 1/3


Thanks for the help!

e

 

[cait-sith]

Two events are mutually exclusive if P(A or B) = P(A) + P(B).

For #4, employed people in Brooklyn is given as 300. So P(Employed Brooklyn) = 300/1000.

An easier way to do #5 is to think, what conditions don't satisfy Unemployed or Queens. E.g., 1 - P(Employed and not in Queens) = 1 - P(Employed, Brooklyn).

 

[Veldek]

I thought I'd get a little help with my math homework. It's easy stuff, but I'm having a little trouble.

Here's some problems

1) A box contains 6 white balls and 5 red balls. A sample of 4 balls is selected at random.
What’s the probability that the sample will have
a) all white balls
b) 2 white balls and 2 red balls

I can do a. It's 6/11 * 5/10 * 4/9 * 3/8 = 360/7920 = .0455

As for b, I'm not completely sure. It's the same concept right? 6/11 * 5/10 * 5/9 * 4/8 = 600/7920 = .0758

Your way for b) doesn't include all possible cases.

You can also calculate it using combinations (you know what a over b means?). I'm using the calculator syntax aCb.

a) 6C4 / 11C4 = 0.045
This means the number of combinations taking 4 balls out of six (the white ones) divided by the number of combinations taking 4 balls out of 11.

You might now be able to solve b). In case you aren't:

b) 5C2 * 6C2 / 11C4 = 0.4545

 

[scem0]

Thanks, both of you.

One thing I don't get is why you use a combinations and not permutations for these ball in urn problems. It seems as though I've never used a permutation for any similar problem, but I use them all the time for other problems.

Thanks so much,

e

 

[Atlasland]

2. You don't need to know if the events are mutually exclusive.

P(a or b) = P(a) + P(b) - P(a and b)

is true irrespective of whether a & b are mutually exclusive. Think about it - it's a key concept. Draw a Venn Diagram to help you understand.

If a & b are mutually exclusive, then all that happens is P(a and b) becomes zero, and the term disappears. The original equation still holds. In fact, what you are left with is the blatantly obvious:

P(a or b) = P(a) + P(b)

*******************************************************

You have been given:

P(a) = 0.3
P(b') = 0.6, hence P(b) = 0.4
P(a or b) = 0.5


Inputing the data into that equation you have:

0.5 = 0.3 + 0.4 - P(a and b)

Hence:

P(a and b) = 0.3 + 0.4 - 0.5 = 0.2

 

[Veldek]

Thanks, both of you.

One thing I don't get is why you use a combinations and not permutations for these ball in urn problems. It seems as though I've never used a permutation for any similar problem, but I use them all the time for other problems.

Thanks so much,

e

You use permutations if you put the ball back into the urn right after taking one out IIRC.

 

[mkrishnan]

6. people are employed if they live in Brooklyn
P(unemployed and live in brooklyn) = 1/10
P(live in brooklyn) = 4/10
(1/10 )/( 4/10 ) = 1/4

You were close -- you got the next one right, but not this one. Read the question again and look at the chart again.