The total number of flaws is given by:

`(0 × 4) + (1 × 3)` `+ (2 × 5) + (3 × 2)` `+ (4 × 4) + (5 × 1)` `+ (6 × 1)` `= 46`

So the average number of flaws for the 20 sheets is given by:

`mu=46/20=2.3`

The required probability is:

`"Probability"=P(X>=3)`

`=1-(P(x_0)+P(x_1)+P(x_2))`

`=1-((e^-2.3 2.3^0)/(0!)+(e^-2.3 2.3^1)/(1!)+(e^-2.3 2.3^2)/(2!))`

`=0.40396`

We can see the predicted probabilities for each of "No flaws", "`1` flaw", "`2` flaws", etc on this histogram.

[The histogram was obtained by graphing the following function for integer values of *x* only.

`(e^-2.3 2.3^x)/(x!)`

Then the horizontal axis was modified appropriately.]

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