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number of accidents (n)| 0 | 1| 2| 3
Probability of n accidents| 0.75| 0.15| 0.08| 0.02

what kinda distribution do ya use?

since it is discrete
\[E(\text{ accidents in one week })=\sum_{n=0}^{3} n \cdot p(n)\]

probability mass function?

aka also known as mean value

Oh! That makes sense! I was over complicating it again...

have you found the sum of the products that I mentioned?

that is have you computed:
0*.75+1*.15+2*.08+3*.02

Yes, indeed I was working on that. cx

dayum probability magics

for my final answer I got 0.37

ok sounds awesome

that is what I have gotten too