Let `X =` number of rejected pistons

(In this case, "success" means rejection!)

Here, `n = 10`, `p = 0.12`, `q = 0.88`.

(a)

No rejects. That is, when `x=0`:

`P(X)` `=C_x^np^xq^(n-x)` `=C_0^10(0.12)^0(0.88)^10` `=0.2785`

One reject. That is, when `x=1`

`P(X)` `=C_1^10(0.12)^1(0.88)^9` `=0.37977`

Two rejects. That is, when `x=2`:

`P(X)` `=C_2^10(0.12)^2(0.88)^8` `=0.23304`

So the probability of getting no more than 2 rejects is:

`"Probability"=P(X<=2)`

`=0.2785+` `0.37977+` `0.23304`

`=0.89131`

(b) We could work out all the cases for `X = 2, 3, 4, ..., 10`, but it is much easier to proceed as follows:

`"Probablity of at least 2 rejects"`

`=1-P(X<=1)`

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

` =1-(0.2785+0.37977)`

`=0.34173`

Using Scientic Notebook , we can **define** the function `g(x)=C(10,x)(0.12)^x(0.88)^(10-x)` and then find the values at `0, 1, 2, ...`, which gives us the histogram:

Alternatively, using SNB :

`C(10,x)(0.12)^x(0.88)^(10-x)`

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