At a highschool the failure rate is 0.2 per year. Of 15 students registered in the freshman class, what is the probability?

Here we have a binomial probability distribution

so, P(pass) = 0.8 and P(fail) = 0.2

a) P(15 pass) = 15C15 x (0.8)¹⁵ x (0.2)⁰

i.e. 0.8¹⁵ = 0.035...to 3 d.p.

c) P(12 pass) = 15C12 x (0.8)¹² x (0.2)³

i.e. 0.250...to 3 d.p.

b) At least 12 passing means P(12) + P(13) + P(14) + P(15)

Now, P(13) = 15C13 x (0.8)¹³ x (0.2)²

i.e. 0.231

And P(14) = 15C14 x (0.8)¹⁴ x (0.2)¹

i.e. 0.132

Combining we get:

0.250 + 0.231 + 0.132 + 0.035 = 0.648

Note: Using 'cumulative' tables we get that P(pass ≤ 11) = 0.352

And P(pass ≥ 12) = 1 - P(pass ≤ 11)

i.e. 1 - 0.352 => 0.648...same result.

:)>  

a)  0.8^15 = 0.0351843721

b) 1 - P(0 Fail) - P(1 Fails) -  P(2 Fail) - P(3 Fail) =

    1 - 0.0351843721 -  (15C1)(0.2^1)(0.8^14) - (15C2)(0.2^2)(0.8^13) - (15C3)(0.2^3)(0.8^12) = about 0.35518

c)  (15C3)(0.2^3)(0.8^12) = about 0.2501