QR probablity question

[i4everpaki]

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it is known that 20% of the population of smallville has blue eyes. if four people are chosen at random from this population, what is the probability that at least one of the four has blue eyes?

a.(1/5)^4 = 1/625
b.(4/5)^4 = 256/625
c.1-(1/5)^4 = 624/625
d.1-(4/5)^4 = 369/625
e. (4/5)^1 + (1/5)^3 = 101/125

 

[Maygyver]

Is the answer 1-(4/5)^4? That would be what I would think; if it's right I'll tell you how.

 

[i4everpaki]

Is the answer 1-(4/5)^4? That would be what I would think; if it's right I'll tell you how.

yup, it is right. I just could not understand the explanation in destroyer. if someone can walk me through it, i would appreciate it. thanks

 

[wantVCUdental]

yup, it is right. I just could not understand the explanation in destroyer. if someone can walk me through it, i would appreciate it. thanks

What is the probability of NO blue eyes?
that's (4/5)^4

the total of all possibilities is 1 or 100%
Now if you know the probability of NO blue eyes is (4/5)^4
Then the rest must have at least 1 blue eye right?
and you express the rest as : 1-(4/5)^4

 

[i4everpaki]

The thing I do not understand is where do we get 4/5 from. I understand 1 bc that's comes from 100% and ^4 bc it's 4 people chosen. I am just really confuse where 4/5 is coming from. 1/5 = blue eyes, so does 4/5 represent that all 4 can have blue eyes and you raise it to 4 meaning it can be arrange in 4 ways? Sorry probablity is a really hard topic for me.

 

[Streetwolf]

The thing I do not understand is where do we get 4/5 from. I understand 1 bc that's comes from 100% and ^4 bc it's 4 people chosen. I am just really confuse where 4/5 is coming from. 1/5 = blue eyes, so does 4/5 represent that all 4 can have blue eyes and you raise it to 4 meaning it can be arrange in 4 ways? Sorry probablity is a really hard topic for me.

The above poster who solved it did so by first considering the opposite the question was asking. Instead of at least one person with blue eyes, they considered NO ONE with blue eyes.

When considering no one with blue eyes, you consider all the people who DON'T have blue eyes.

Since there's a 1/5 chance of HAVING blue eyes, it follows that there is a 4/5 chance of NOT HAVING blue eyes.

 

[Maygyver]

Yeah, basically you can solve this with a complement rule. Blue eyes = 1/5 and not blue eyes = 4/5. Hope that helps.

 

[i4everpaki]

The above poster who solved it did so by first considering the opposite the question was asking. Instead of at least one person with blue eyes, they considered NO ONE with blue eyes.

When considering no one with blue eyes, you consider all the people who DON'T have blue eyes.

Since there's a 1/5 chance of HAVING blue eyes, it follows that there is a 4/5 chance of NOT HAVING blue eyes.

ok so (4/5) dont have blue eyes raised by power of 4 because 4 random people are picked, minus it by 100% which is one 1. so this is why we get 1-(4/5)^4 it makes sense to me. so i am guessing this is the only way to solve this problem. we have to consider amount of people that dont have blue eyes and minus it with total to get the value we want.
THANKS a lot. for the help

 

[Streetwolf]

ok so (4/5) dont have blue eyes raised by power of 4 because 4 random people are picked, minus it by 100% which is one 1. so this is why we get 1-(4/5)^4 it makes sense to me. so i am guessing this is the only way to solve this problem. we have to consider amount of people that dont have blue eyes and minus it with total to get the value we want.
THANKS a lot. for the help

No. One other way.

At least 1 means the cases where 1, 2, 3, and 4 people have it.

Case where 1 has it: (4C1)(1/5)(4/5)^3 = 256/625
Case where 2 have it: (4C2)(1/5)^2 * (4/5)^2 = 96/625
Case where 3 have it: (4C3)(1/5)^3 * (4/5) = 16/625
Case where 4 have it: (1/5)^4 = 1/625

Add them up = 369/625

In each case you have to choose which people have it out of the 4. In the case of 2 people, there are (4C2) = 6 ways (combination - google it) to select 2 people from a group of 4 people. For each of those 6 ways there is a (1/5)^2 * (4/5)^2 = 16/625 chance that 2 of them have it. Multiply 16/625 by 6 to get 96/625.

Of course the other way is far easier.

 

[i4everpaki]

No. One other way.

At least 1 means the cases where 1, 2, 3, and 4 people have it.

Case where 1 has it: (4C1)(1/5)(4/5)^3 = 256/625
Case where 2 have it: (4C2)(1/5)^2 * (4/5)^2 = 96/625
Case where 3 have it: (4C3)(1/5)^3 * (4/5) = 16/625
Case where 4 have it: (1/5)^4 = 1/625

Add them up = 369/625

In each case you have to choose which people have it out of the 4. In the case of 2 people, there are (4C2) = 6 ways (combination - google it) to select 2 people from a group of 4 people. For each of those 6 ways there is a (1/5)^2 * (4/5)^2 = 16/625 chance that 2 of them have it. Multiply 16/625 by 6 to get 96/625.

Of course the other way is far easier.

this way is very time consuming, i will stick to the first way. thanks alot