Math Destroyer Qs

[hope_to_match]

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Need some help with Math Destroyer (2013) Test 1

Q23: It is know that 20% of the population of Smallville has blue eyes. If 4 people are chosen at random from this population, what is the probability that at least one of the four has blue eyes?

I need help understanding how to go about solving this problem as I don't understand the explanation in the back. The answer is D btw (1-(4/5)^4=369/625

Q28: Can someone tell me what that little inverted sign represents between G and H in the question?

update: nvm i figured out number 28

Thanks 🙂

 

[SafetyDog]

I don't know the question for #28, but I can help with #23.

Okay, the reason why its (1-4/5) is that the 4/5 represents the chance of NOT getting a blue eyed person. So to not get 4 blue eyed people in a row is (4/5)^4. Then you subtract that from 1, which represents everything else. Including if you get 1, 2, 3, or 4 blue eyed people in your group.

I don't know if I explained it well enough, but let me know!

 

[hope_to_match]

I don't know the question for #28, but I can help with #23.

Okay, the reason why its (1-4/5) is that the 4/5 represents the chance of NOT getting a blue eyed person. So to not get 4 blue eyed people in a row is (4/5)^4. Then you subtract that from 1, which represents everything else. Including if you get 1, 2, 3, or 4 blue eyed people in your group.

I don't know if I explained it well enough, but let me know!

when we subtract 4/5^4 from 1, 1 represents the entire population or 80% of population? or something else?

 

[SafetyDog]

1 represents the entire population. But I don't like to think of it as a population, but rather a 1 is representative of a 100% chance.

Sorry, it's really difficult to explain for me! Maybe someone else can chime in.

 

[hope_to_match]

1 represents the entire population. But I don't like to think of it as a population, but rather a chance.

Sorry, it's really difficult to explain for me! Maybe someone else can chime in.

alright thanks i understand 🙂

 

[hope_to_match]

basically we find the probability of NOT having a blue eyed individual in the set of 4 ( by 4/5^4) and then subtract from entire population (which is one) to get the probability of having an individual with blue eyes, correct?

 

[SafetyDog]

Well, it's the probability of getting AT LEAST 1 blue eyed individual. So this can be 1, 2, 3, or 4 people that you got with blue eyes.

 

[hope_to_match]

Well, it's the probability of getting AT LEAST 1 blue eyed individual. So this can be 1, 2, 3, or 4 people that you got with blue eyes.

yup thanks!

 

[dent777]

Q28. It is a sign called intersection and what it means the numbers shared by both samples, i.e what is common b/t the two. Say sample A (1,2,3) sample B (2,4,5) A∩B is 2.

P.S there is also a "U" sign which is called a union and this one represents what is in either of the samples. Say sample A (1,2,3,4) and sample B (4,5,6). AUB would be (1,2,3,4,5,6). Note that even though 4 is in both samples, it is only listed once.

 

[hope_to_match]

Q28. It is a sign called intersection and what it means the numbers shared by both samples, i.e what is common b/t the two. Say sample A (1,2,3) sample B (2,4,5) A∩B is 2.

P.S there is also a "U" sign which is called a union and this one represents what is in either of the samples. Say sample A (1,2,3,4) and sample B (4,5,6). AUB would be (1,2,3,4,5,6). Note that even though 4 is in both samples, it is only listed once.

thanks a lot i was actually looking online for what the regular U means. And you answered that question as well cheers 👍

 

[Jordwin]

I don't know the question for #28, but I can help with #23.

Okay, the reason why its (1-4/5) is that the 4/5 represents the chance of NOT getting a blue eyed person. So to not get 4 blue eyed people in a row is (4/5)^4. Then you subtract that from 1, which represents everything else. Including if you get 1, 2, 3, or 4 blue eyed people in your group.

I don't know if I explained it well enough, but let me know!

So what would the probability be if we wanted:

1) at least 2
2) at least 3
3) 4
? I'm not sure i followed the whole conversation and am going over the same problem right now. Would 4 just be (1/4)^4?

 

[SafetyDog]

Yea, 4 should be (1/4)^4, but I have no idea how to do the other 2.

 

[dent777]

So what would the probability be if we wanted:

1) at least 2
2) at least 3
3) 4
? I'm not sure i followed the whole conversation and am going over the same problem right now. Would 4 just be (1/4)^4?

I don't think they would ask something like at least 2 or at least 3. It is too much to calculate in those time constraints.

 

[Jordwin]

I don't think they would ask something like at least 2 or at least 3. It is too much to calculate in those time constraints.

Okay thanks!