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?
have 32 integers from 1 to 32. We will choose 4 from them.
probability of choosing one integer is its value divided by total of all
For example, if there is 4 integers, the total of them is 4+3+2+1
so probability of selecting 1 is 1/10
probability of selecting 2 is
probability of selecting 3 is 3/10
probability of selecting 4 is
Using the above method we will choose 4 integers out of 32. What is
the probability of 32 being chosen?
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 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 **
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.
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 **