#Binomial distribution
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,rccw
$(0.999+0.001)^{25}=\sum_{k=0}^{25}\binom{25}{k}0.999^k\cdot 0.0001^{25-k}$
ooaa
take it as 1-P(none are defective)
WYM
eh not rly
think again
alternatively you can just sum from k=1 to 25 since that includes defective products from 1 to 25
but it does same thing
yes
,rccw
sick thanks
welcome
it should not
it is just rly rly rly small
well what should i do
put it in a better calculator lol
just whe whole equasion
yes
0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000249751
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