Reading Assignments

Probability for n>=4 is ~= 48.1%

@joe-shmo said
if you have a spreadsheet handy; using the general formula BigDoggProblem derived you can verify the following sum for the ways to read/not read set X :

P( n = 0) + P( n = 1) + P( n = 2) + ... + P( n = 99) + P( n = 100) = 1

[101 terms in total ]

Thanks Joe.I understand all that but I don't think google sheets has a probability function as such

@venda said
Thanks Joe.I understand all that but I don't think google sheets has a probability function as such

If you understand, that works for me!

@bigdoggproblem said
Probability for n>=4 is ~= 48.1%

Correct!

The probability of at least 4 people reading set X in class of 100 is nearly 50%

P( n ≥ 4 ) = 1 - [ P ( n = 0 ) + P( n= 1 ) + P( n = 2 ) + P( n = 3 ) ] ≈ 48.1%

@joe-shmo said
Correct!

The probability of at least 4 people reading set X in class of 100 is nearly 50%

P( n ≥ 4 ) = 1 - [ P ( n = 0 ) + P( n= 1 ) + P( n = 2 ) + P( n = 3 ) ] ≈ 48.1%

I found it a bit strange that the probability for n=3 is the highest of all n's.

n prob
0 2.63%
1 9.75%
2 17.88%
3 21.64%
4 19.43%
5 13.82%
6 8.10%
7 4.03%
8 1.74%
9 0.66%

@bigdoggproblem said
I found it a bit strange that the probability for n=3 is the highest of all n's.

n prob
0 2.63%
1 9.75%
2 17.88%
3 21.64%
4 19.43%
5 13.82%
6 8.10%
7 4.03%
8 1.74%
9 0.66%

I can’t think of any logical reason at the moment why that should be. It’s not odd that there is a max. The binomial distribution is symmetrical, so that factor would skew it toward the middle (50). But the remaining factor heavily skews it toward the beginning. So i would kind of expect a maximum for sure between 0 and 50. The fact that it’s so heavily skewed is interesting.

I've enjoyed your puzzles Joe and I'd like to think I've learned something form them and from the responses of the others so thank you to all.
This weeks newspaper puzzle was rubbish by the way so it's not worth the effort of posting it.