Part iii

P (at least one box out of n boxes which have > 2 defective toys) > 0.95

1 - P(no boxes have > 2 defective toys) > 0.95

1 - 0.95 > P(no boxes have > 2 defective toys)

0.05 > P (all boxes have ≤ 2 defective toys)

Recall that P(one box has > 2 defective toys) is 0.268.
So P(one box has ≤ 2 defective toys)
= 1 - 0.268 = 0.732

Since there are n boxes,

when 0.05 > P (all boxes have ≤ 2 defective toys)

0.05 > 0.732^n

(0.732 for first box, X 0.732 for second box, X 0.732 for third box, X 0.732 for fourth... .... X 0.732 for the nth box)

Take ln on both sides,(you can use log too whichever you prefer)

ln 0.05 > ln (0.732)^n

ln 0.05 > n ln(0.732)

ln 0.05/ln 0.732 < n

(note that sign changes as ln 0.732 is negative. Recall that sign changes when dividing by a negative value)

9.6024827 < n

n > 9.602487

So least value of n = 10

Part iv)

10 randomly chosen boxes.

Number of toys = 10 x 30 = 300

You can treat this as a binomial distribution with 300 trials (n=300),
p = 0.06.

Letting X be the number of defective toys,
X ∼ B(300,0.06)

P( < 17 toys are defective)

= P( ≤ 16 toys are defective)

= P ( X ≤ 16)

Key in binomcdf(300,0.06,16) into your GC.

You should get 0.36954287062

≈ 0.370

Thank you!!