Probability of success `p = 0.8`, so `q = 0.2`.

`X =` success in getting through.

Probability of `7` successes in `10` attempts:

`text[Probability]=P(X=7)`

`=C_7^10(0.8)^7(0.2)^[10-7]`

`=0.20133`

Using the following function in SNB,

`C(10,x)(0.8)^x(0.2)^[10-x]`

we have:

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