Thread: easy probability question

Hi, I have two questions which I think are easy but I am not sure as I am starting to learn probability.

1. We have 32 integers from 1 to 32. we choose 16 of them
randomly. what is the probability of the integer 32 being chosen?

2. We
have 32 integers from 1 to 32. We will choose 4 from them.

The
probability of choosing one integer is its value divided by total of all
integers.
For example, if there is 4 integers, the total of them is 4+3+2+1
= 10
so probability of selecting 1 is 1/10
probability of selecting 2 is
2/10
probability of selecting 3 is 3/10
probability of selecting 4 is
4/10

Using the above method we will choose 4 integers out of 32. What is
the probability of 32 being chosen?

Originally Posted by ruleworld

Hi, I have two questions which I think are easy but I am not sure as I am starting to learn probability.

1. We have 32 integers from 1 to 32. we choose 16 of them
randomly. what is the probability of the integer 32 being chosen?

Depends on whether the 16 numbers are chosen with replacement or without. If chosen without replacenment you end up with two piles of numbers - one pile of 16 with the chosen numbers and one pile of 16 of numbers not chosen. The probability of the number 32 being in one pile of the other is the same, right?

Originally Posted by ruleworld

2. We have 32 integers from 1 to 32. We will choose 4 from them.

The
probability of choosing one integer is its value divided by total of all
integers.
For example, if there is 4 integers, the total of them is 4+3+2+1
= 10
so probability of selecting 1 is 1/10
probability of selecting 2 is
2/10
probability of selecting 3 is 3/10
probability of selecting 4 is
4/10

Using the above method we will choose 4 integers out of 32. What is
the probability of 32 being chosen?

Not following you on this. If you have 4 integers the probability of selecting any one of the four integers is 1/4.

ebaines, in the second problem the OP is saying (because 1+ 2+ 3+ ...+ 31+ 32= 32(33)/2= 495) the probability of choosing 1 is 1/495, the probability of choosing 2 is 2/495, etc.

The probability of 32 being chosen is 32/495 so the probability of it NOT being chosen is 1- 32/495= 464/495. Assuming "sampling with replacement" (sampling without replacement would be very difficult- the probabilty of a specific number being chosen after another has been removed will depend stongly on what number had been removed) the probability 32 is not chosen in four trials is $\displaystyle (464/495)^4= 0.772$ approximately.

HallsOfIvy: It's a strange problem, and poorly worded. It would be as if we have a bin of 495 balls, with one of them having "1" printed on it, two of them having a "2" etc up to thirty-two balls having "32" printed on them, and from this bin of 495 balls we randomly select four. In this case the probability of choosing a 32 on any one draw is 32/495, and under those conditions then yes, I agree with your analysis.

**EDIT: the sum of 1 to 32 is 32x33/2 = 528, not 495, so the probability of choosing 32 on any draw with replacement is 32/528 **

Originally Posted by HallsofIvy

ebaines, in the second problem the OP is saying (because 1+ 2+ 3+ ...+ 31+ 32= 32(33)/2= 495) the probability of choosing 1 is 1/495, the probability of choosing 2 is 2/495, etc.

The probability of 32 being chosen is 32/495 so the probability of it NOT being chosen is 1- 32/495= 464/495. Assuming "sampling with replacement" (sampling without replacement would be very difficult- the probabilty of a specific number being chosen after another has been removed will depend stongly on what number had been removed) the probability 32 is not chosen in four trials is $\displaystyle (464/495)^4= 0.772$ approximately.

You are right. You cannot replace anything after choosing it. For example, if you choose first choose 10, then there will be 31 integers to choose from and you have to choose 9 integers from them.

But actually I don't need to choose from integers, I gave that example to explain easily but I didn't catch that trick that is if a specific number is removed the selection process would become complicated for integers.

What I need a rank based choice from items. We have 32 items ranked from 1 to 32. We choose each item according to the rules. Here are the rules revised now -

If there is 4 items, they will be ranked 1 to 4 (suppose the best rank is 4). the total of the ranks is 4+3+2+1 = 10
so probability of selecting item with rank 1 is 1/10
probability of selecting item with rank 2 is 2/10
probability of selecting item with rank 3 is 3/10
probability of selecting item with rank 4 is 4/10

suppose we choose the item ranked 10 first, then we remove the items from list. Then we re-rank the remaining 31 items from 1 to 31 and follow the selection process again. We do this 4 times to choose 4 items. So what is the probability of getting the highest ranked item(32) being chosen as one of those 4 items?

Thanks for help.

Originally Posted by ebaines

HallsOfIvy: It's a strange problem, and poorly worded. It would be as if we have a bin of 495 balls, with one of them having "1" printed on it, two of them having a "2" etc up to thirty-two balls having "32" printed on them, and from this bin of 495 balls we randomly select four. In this case the probability of choosing a 32 on any one draw is 32/495, and under those conditions then yes, I agree with your analysis.

Hi, It's my mistake. I have corrected it. Hope you don't find it very strange now.

If I understand the rules:

The probability of NOT choosing 32 on the first draw is 1-32/528 [note - the sum of 1 to 32 is 32x33/2 = 528, not 455 per the previous posts]. Assuming 32 is not drawn, then what was number 32 is now called 31. The sum of 1 to 31 is 31x32/2 = 496, and the probability of NOT choosing 31 on the second draw is 1-31/496. Continuing on - if 31 is not chosen it becomes number 30 and the probability of not choosing 30 on the third draw is 1-30/465, and then finally for the last draw the probability of not choosing 29 is 1-29/435. Putting it all together: the probability of NOTchoosing the original 32 in all four draws is (1-32/528)(1-31/496)(1-30/465)(1-29/435) = 0.873. Hence the probability of picking 32 in one of the four draws is 1- 0.873 = 0.127.

**EDIT: there's an error here - see post #9 **

Originally Posted by ebaines

If I understand the rules:

The probability of NOT choosing 32 on the first draw is 1-32/528 [note - the sum of 1 to 32 is 32x33/2 = 528, not 455 per the previous posts]. Assuming 32 is not drawn, then what was number 32 is now called 31. The sum of 1 to 31 is 31x32/2 = 496, and the probability of NOT choosing 31 on th second draw is 1-31/496. Continuing on - if 31 is not chosen it becomes number 30 and the probability of not choosing it on the third draw is 1-30/465, and fibally for the last draw the probability of not choosing 29 is 1-29/435. Putting it all together: the probability of NOTchoosing the original 32 in all four draws is (1-32/528)(1-31/496)(1-30/465)(1-29/435) = 0.873. Hence the probability of picking 32 in one of the four draws is 1- 0.873 = 0.127.

Yes! That should be it. Thank you very much.

Oops - math error:

Originally Posted by ebaines

Putting it all together: the probability of NOTchoosing the original 32 in all four draws is (1-32/528)(1-31/496)(1-30/465)(1-29/435) = 0.873. Hence the probability of picking 32 in one of the four draws is 1- 0.873 = 0.127.

This should be: (1-32/528)(1-31/496)(1-30/465)(1-29/435) = 0.769. Hence the probability of picking 32 in one of the four draws is 1- 0.769 = 0.231.

Also the first question where we pick 16 items out of 32 regardless of rank should be 16 / 32 = 0.5 then.

Originally Posted by ruleworld

Also the first question where we pick 16 items out of 32 regardless of rank should be 16 / 32 = 0.5 then.

Right.