math question

[shareshab]

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

i have no clue how mat destroyer solve this question please explain me in simplest way possible thanks alot 👍

 

[pk62281]

I think it's 1-(4/5)^4

(4/5)^4 is the chance that no one will have blue eyes from the chosen 4.
1 minus that is the chances of the rest of the scenarios which include getting 1, 2, 3, or 4.

 

[dentalWorks]

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

i have no clue how mat destroyer solve this question please explain me in simplest way possible thanks alot 👍

If I remember this stuff correctly...

"and" you multiply
"or" you add

Your going to pick 4 people... each one has a probability of being blue-eyed = 1/5 so...

1/5 OR 1/5 OR 1/5 OR 1/5
1/5 + 1/5 + 1/5 + 1/5 = 4/5 = answer

Is answer 4/5?

 

[chance8]

That's the same answer I got using this approach.

Probability that one person will get blue eyes = P(X<1) = 0.20.

P(X>1) = 1-P(X<1) = 1-0.2 = 0.8 or 4/5

If I remember this stuff correctly...

"and" you multiply
"or" you add

Your going to pick 4 people... each one has a probability of being blue-eyed = 1/5 so...

1/5 OR 1/5 OR 1/5 OR 1/5
1/5 + 1/5 + 1/5 + 1/5 = 4/5 = answer

Is answer 4/5?

 

[pk62281]

IMHO, I'm about 95% my method is correct. I got a question like this on topscore and it was calculated by (1 - chances that it wont happen%^n)

It makes a lot of sense if you think about it.
What are the chances that if you pick 4 random people from smallville, that none of them will have blue eyes?

The answer is (0.80)^4.

The question asks for "at least one" so that includes all the other possible scenarios other than getting no blue eyes, which is getting 1 or 2 or 3 or 4 blue eyes. Therefore, all you have to do is take 1 and minus it by (0.8)^4

 

[yankeefan86]

IMHO, I'm about 95% my method is correct. I got a question like this on topscore and it was calculated by (1 - chances that it wont happen%^n)

It makes a lot of sense if you think about it.
What are the chances that if you pick 4 random people from smallville, that none of them will have blue eyes?

The answer is (0.80)^4.

The question asks for "at least one" so that includes all the other possible scenarios other than getting no blue eyes, which is getting 1 or 2 or 3 or 4 blue eyes. Therefore, all you have to do is take 1 and minus it by (0.8)^4

Your method is right dude. The way you set up and solved the problem is how ,math destroyer answered it

 

[dentalWorks]

IMHO, I'm about 95% my method is correct. I got a question like this on topscore and it was calculated by (1 - chances that it wont happen%^n)

It makes a lot of sense if you think about it.
What are the chances that if you pick 4 random people from smallville, that none of them will have blue eyes?

The answer is (0.80)^4.

The question asks for "at least one" so that includes all the other possible scenarios other than getting no blue eyes, which is getting 1 or 2 or 3 or 4 blue eyes. Therefore, all you have to do is take 1 and minus it by (0.8)^4

you know what... I think you are right....

Its gotta be one of the 4 events bellow:
blue and non and non and non = ?
non and blue and non and non =
non and non and blue and non =
non and non and non and blue =

So now... do you say event 1 OR even 2 OR even 3 OR event 4?

can someone look up the answer please

 

[NewBee12]

the answer is : 1-(4/5)^4
the question says at least one which means 1 or 2 or 3 or 4 have blue eyes.
to do it the long way you need to find the probability of 1 having blue eyes, the probability of 2 having blue eyes etc.... then add them up.( see explanation below )
the easy way is 1- the probability of none having blue eyes which give:
1-(4/5)^4

the long way:
for 1 having blue eyes:
1C4 x (1/5)^1 x (4/5)^3
for 2 having blue eyes :
2C4 x(1/5)^2 x (4/5)^2
for 3
3C4 x (1/5)^3 x(4/5)^1
for 4
4C4 x(1/5)^4 x (4/5)^0
then add up all these probabilities. this is the long way to do it

 

[dentalWorks]

makes sense... thnx newbee

 

[chance8]

Ah you're right. I misinterpreted the equation.

P(x>1) = 1 -P(X<1) = 1-(.2)^4

Thanks!

the answer is : 1-(4/5)^4
the question says at least one which means 1 or 2 or 3 or 4 have blue eyes.
to do it the long way you need to find the probability of 1 having blue eyes, the probability of 2 having blue eyes etc.... then add them up.( see explanation below )
the easy way is 1- the probability of none having blue eyes which give:
1-(4/5)^4

the long way:
for 1 having blue eyes:
1C4 x (1/5)^1 x (4/5)^3
for 2 having blue eyes :
2C4 x(1/5)^2 x (4/5)^2
for 3
3C4 x (1/5)^3 x(4/5)^1
for 4
4C4 x(1/5)^4 x (4/5)^0
then add up all these probabilities. this is the long way to do it

 

[chance8]

"or" is add
"and" is multiply

In this case, I think it is event 1 and event 2 and event 3 and event 4 do not have blue eyes. Therefore, .2x.2x.2x.2 or .2^4 = probability that 0 blue eyes.

you know what... I think you are right....

Its gotta be one of the 4 events bellow:
blue and non and non and non = ?
non and blue and non and non =
non and non and blue and non =
non and non and non and blue =

So now... do you say event 1 OR even 2 OR even 3 OR event 4?

can someone look up the answer please