# Random Person Probability Problem

• thechicgeek
In summary: So the answer is .000064.Ok. Does he want you to round it off too? For c either the first one is 85 and the second one isn't or the second one is 85 and the first one isn't, or they are both 85. You could find the numbers for all three cases and add them. But the simple way to do it is to notice the answer is 1-(probability neither is 85). So the answer is .000064.In summary, the probability of picking an 85 year old is .008.
thechicgeek

## Homework Statement

The probability that a person picked at random is 85 years old is 0.008 . if two people are picked at random find the probability of that:
a. they are both 85
b. neither is 85
c. at least one is 85

I have two questions like this one and I am so confused as to how you would find the answer if you aren't given the pool size. could you show me how one would work this problem out?
Thanks :)

You have to try. Start with the first one. Take the pool size to be infinite. So the choice of the first person doesn't affect the probabilities in making the second choice.

A probability of 0.008 means 8 out of a 1000.

right so the second result isn't affected by the first result right? You could pick another 85. I'm just trying to understand how they know that the chance of picking an 85 year old is .008 like how do they know that? Am I thinking too hard about this?

chrisk said:
A probability of 0.008 means 8 out of a 1000.

Hah! okay that makes much more sense. :)

thechicgeek said:
Hah! okay that makes much more sense. :)

That doesn't mean the pool size is 1000. I.e. the probability of picking a second 85 year old isn't 7/999. It's still 8/1000.

Dick said:
That doesn't mean the pool size is 1000. I.e. the probability of picking a second 85 year old isn't 7/999. It's still 8/1000.

got it. Okay well I'm going to try and work a, b, and c out and see what i can come up with.

okay so would a. =.006%
b. =98%
am I on the right track here?

thechicgeek said:
okay so would a. =.006%
b. =98%
am I on the right track here?

You multiplied the probabilities, right? I would leave it as 0.000064 rather that changing to % and rounding.
Same with the second. But yes, it looks like you are on the right track.

Would c. be .8% or .0079
The professor wants the answer in percentage format.

thechicgeek said:
Would c. be .8% or .0079
The professor wants the answer in percentage format.

Ok. Does he want you to round it off too? For c either the first one is 85 and the second one isn't or the second one is 85 and the first one isn't, or they are both 85. You could find the numbers for all three cases and add them. But the simple way to do it is to notice the answer is 1-(probability neither is 85).

## 1. What is the "Random Person Probability Problem"?

The "Random Person Probability Problem" is a mathematical problem that asks, given a group of people, what is the probability that at least two people share the same birthday.

## 2. How is the probability calculated?

The probability is calculated using the formula: P = 1 - (365!/365^n) where n is the number of people in the group. This formula takes into account the number of possible combinations of birthdays and subtracts it from 1 to get the probability of at least two people sharing the same birthday.

## 3. Is this problem based on real data?

Yes, this problem is based on the mathematical concept of the Birthday Paradox, which has been proven with real data. In fact, with a group of 23 people, there is a 50% chance that two people will share the same birthday.

## 4. Can this problem be applied to other scenarios?

Yes, this problem can be applied to any scenario that involves randomly selecting items or individuals from a group. It can also be used in the field of cryptography to determine the probability of two people generating the same random number.

## 5. Are there any limitations to this problem?

Yes, this problem assumes that all birthdays are equally likely to occur and that the distribution of birthdays is random. It also assumes that the group of people being considered is large enough to make the probability significant. Additionally, this problem does not take into account leap years or the possibility of multiple people sharing the same birthday.

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