Math Problem

[firstcitizen]

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To win a lottery, a person must select 4 numbers in any order from 20 numbers. Repetitions are not allowed. probability person wins is ?

a: 0.0002

 

[predentlsu1]

Use the combination formula to find out how many diff ways they can pick their numbers n!/r!(n-r)! Using this you will find they can pick their card 4,845 ways.......Since only one set of lottery numbers will win they have a 1/4,845 chance of winning.....which is 0.0002

Use the combination formula for no repeats
Use the permutation forumla for repeats

 

[firstcitizen]

one more 😀:

A survey found that one in five Americans visited a doctor in any [SIZE=-1]g[/SIZE]iven month. If 10 people are selected at random, the probability that exactly 3 visited a doctor last month is:

a: 0.201

 

[firstcitizen]

nvm i got it

 

[sacjumpman]

what's the fast way to do this problem? the way I do it would never work on the test...... i'm so screwed on probability.

Thanks

 

[smarterray]

How'd you answer that second one, firstcitizen?

 

[firstcitizen]

10C3 x (.2 raised to 3) x (.8 raised to 7)

 

[Mstoothlady2012]

i couldnt do the second one either...there are gonna be 4 probability questions...hopefully i will be able to answer 2 🙄

 

[DDSelin2mori]

nvm i got it

how did you get it?

 

[Mstoothlady2012]

10C3 x (.2 raised to 3) x (.8 raised to 7)

I understand why you did 10C3 x (.2 ^ 3)...but i would have never thought of doing .8^7

.2 is the probability of visiting a doctor & .8 is the prob of people not visiting a doctor right? since we are only looking for probability of people visiting doctor...y do we care about .8?

 

[predentlsu1]

Somethin is weird about that question.......I sat there for like 15 minutes trying to figure that one out and never got it. It still doesnt make since. I dont think you will see a problem that in depth on the test.

 

[Mstoothlady2012]

Somethin is weird about that question.......I sat there for like 15 minutes trying to figure that one out and never got it. It still doesnt make since. I dont think you will see a problem that in depth on the test.

yea i hope so...

 

[Streetwolf]

You need 0.8^7 (or 4/5 raised to the 7th power) because you need to account for all 10 people.

You say you understand 10c3, that's good.

You put in 0.2^3 because thats the odds of 3 people visiting a doctor. But that's only 3 people. What happened to the other 7?

You need 0.8^7 to show that the other 7 people are not visiting a doctor.

And since the three people visiting a doctor could be any three out of 10, you multiply by 10c3.

 

[z3u2]

I understand why you did 10C3 x (.2 ^ 3)...but i would have never thought of doing .8^7

.2 is the probability of visiting a doctor & .8 is the prob of people not visiting a doctor right? since we are only looking for probability of people visiting doctor...y do we care about .8?

Think of 0.8^7 * 0.2^3 is the probability of ONE combination of having exactly 3 people visiting doctors and exactly 7 people not visiting doctors.

(for example, what is the probility of the first three people visiting doc and the rest seven people not visiting doc? u'd do 0.2^3 *0.8^7 )

10 choose 3 is the total # of combinations

total # of combinations * probability per combination = total probability

 

[Mstoothlady2012]

thanks guys! i guess I kind of understand...not totally but its ok😕. I would have never thought of those 7 people. I suck at probability questions like this. I can do the easy ones though 😀..so hopefully real thing will have atleast 2/4 easy ones

 

[sacjumpman]

thanks guys! i guess I kind of understand...not totally but its ok😕. I would have never thought of those 7 people. I suck at probability questions like this. I can do the easy ones though 😀..so hopefully real thing will have atleast 2/4 easy ones

I'm right there with you.

I just don't know if its worth investing any more time in trying to figure out trig and probability for the exam. I've wasted soooooo much time with it already.

I got up and read this problem and could not do it and got so flustered for the whole morning.... at least I'm not the only one.

 

[sacjumpman]

one more 😀:

A survey found that one in five Americans visited a doctor in any [SIZE=-1]g[/SIZE]iven month. If 10 people are selected at random, the probability that exactly 3 visited a doctor last month is:

a: 0.201

What about if we changed this to AT LEAST 3 people visited the doctor (as opposed to exactly 3)?

I'm gonna see if I can figure it out, but want to see what the answer is....

edit to add:

so my answer: 10c3 * (1/5)^3

Since the last 7 people, just have to be people and not specifically those that haven't seen a doctor, the probability is 1?

Don't know if this is right or not...

 

[z3u2]

What about if we changed this to AT LEAST 3 people visited the doctor (as opposed to exactly 3)?

I'm gonna see if I can figure it out, but want to see what the answer is....

edit to add:

so my answer: 10c3 * (1/5)^3

Since the last 7 people, just have to be people and not specifically those that haven't seen a doctor, the probability is 1?

Don't know if this is right or not...

Two ways to do this.

Longer way: P of exactly 3 + P of exactly 4 + P of exactly 5.....+ P of exactly 10 = P of at least 3

Shorter way: 1 - P of exactly 0 - P of exactly 1 - P of exactly 2

 

[sacjumpman]

Two ways to do this.

Longer way: P of exactly 3 + P of exactly 4 + P of exactly 5.....+ P of exactly 10 = P of at least 3

Shorter way: 1 - P of exactly 0 - P of exactly 1 - P of exactly 2

Thanks. I'd never have time to do that. Mark and guess.......