The mean of defective blades supplied in packets of $10$ is $1$. in how many packets of this make out of $1000$ packages would you expect to find at least $4$ non defective blades.
Answer given in my book is $13$ whereas me and my friends are getting $999$
On
p = 0.1 q = 0.9 n = 10 N = 1000 (x is no. of defective items)
P(X=x) + 10Cx * (0.1)^x * (0.9)^(10-x)
P(X>=4) = 1 - P(X=0,1,2,3 non defective items)
0 non defective = 10 defective 1 non defective = 9 defective and so on
1 - P(X=0,1,2,3 non defective items) = 1 - (10C10 * (0.1)^10 * (0.9)^0 + 10C9 * (0.1)^9 * (0.9)^1 + 10C8 * (0.1)^8 * (0.9)^2 + 10C7 * (0.1)^7 * (0.9)^3) = 0.999
N*0.999 = 999
To get you started:
The mean number of defective blades in a pack of $10$ is $1$, meaning the probability that a given blade in a pack is defective is $0.1$.
What is the probability that in $10$ blades, at least $4$ of them are defective?
Hint: Think Binomial Distribution and Complement Rule.
Then use this answer, which will be the probability that a given pack will have at least $4$ defective blades, and multiply by $1000$ to find the expected number of packs that have at least $4$ defective blades.