Thu 26 Jul, 2018 04:07 am - I can work out that p(at least 2 of same size) = 1-p(all different)
p(all different) = 80/80 * 79/80...55/80 = (80!/54!)/80^26 = 0.0103
So P(at least 2 of same size) = 1-0.0103 = 0.9897
But... (view)
Wed 25 Jul, 2018 08:06 am -
Thanks Engineer.
Is there an easy way to scale this, for example to:
A shoe shop has a shoe in 80 sizes and they keep 3 of each size in stock.
They expect 26 customers per week of this... (view)
Wed 25 Jul, 2018 02:08 am - A shoe shop has a shoe in 9 sizes and they keep 6 of each size in stock.
They expect 9 customers per week of this shoe - these can be any size.
What is the probability they will run out of stock? (view)
