In the first part we would use W ~ B(40,0.05), where W denotes the number of defective washers in each packet of 40 washers.

From ii) , we know that P(W > 1)
= 1 - P(W ≤ 1)
≈ 1 - 0.39906406512
= 0.6009353488
= 0.601 (3s.f)

(Using GC's Binomcdf function, you can do manually as well by using the formula in MF26 and doing 1 - P(W = 1) - P(W = 0) )


Now in part ii), you are asked to find the greatest value of N such that P(at most 1 packet contains more than 1 defective washer) > 0.01


So now we have to form a new binomial distribution with N trials, and use P(W > 1) as our probability for each trial.


Letting X be the number of packets out of N randomly selected ones, whereby there is more than 1 defective washer in each of these X packets, we have :

X ~ B(N,0.601)


So what ii) is asking you to solve is : P(X ≤ 1) > 0.01

From here, we do the Iteration method.

Guess a value of N and do P(X ≤ 1) in the GC. Keep trying until you get P(X ≤ 1) > 0.01


(We can jump to different values of N if it's too far away from the desired 0.01 eg. If you're trying N = 18,19,20, and it's still too high, switch to something like N = 10)

When N = 9, P(X ≤ 1) ≈ 0.003735 < 0.01
When N = 8, P(X ≤ 1) ≈ 0.008392 < 0.01
When N = 7, P(X ≤ 1) ≈ 0.01860 > 0.01
When N = 6, P(X ≤ 1) ≈ 0.04053 > 0.01
When N = 5, P(X ≤ 1) ≈ 0.08632 > 0.01


From here we can confirm that N's greatest value to satisfy the inequality is 7

Alternative to the iteration method :

P(X ≤ 1) > 0.01
P(X = 1) + P(X = 0) > 0.01

Using the formula in MF26,

(ᴺ₀) (0.601)⁰ (1 - 0.601)ᴺ⁻⁰ + (ᴺ₁) (0.601)¹ (1 - 0.601)ᴺ⁻¹ > 0.01

(0.399)ᴺ + 0.601N(0.399)ᴺ⁻¹ > 0.01

(0.399)ᴺ + (0.601N/0.399) (0.399)ᴺ > 0.01

(0.399)ᴺ (1 + 1.50627N) > 0.01


Next, plot y = (0.399)ᴺ (1 + 1.50627N) and y = 0.01 into GC. (Use x instead of N)

The intersection points are N = -0.660 and N = 7.78

For -0.660 < N < 7.78,

(0.399)ᴺ (1 + 1.50627N) > 0.01

So largest value of N(a positive integer) = 7