The booklets produced by a certain publisher contain, on average, 1 incorrect letter per 30 000 letters, and these errors occur randomly. A randomly chosen booklet from this publisher contains 12 500 letters.
- Use a suitable approximating distribution to find the probability that this booklet contains at least 2 errors.[3]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [B1] (λ=) 1/30000 * 12500 = 0.417 or better
- [M1] 1 - e^(-λ)(1 + λ); 1 – P(X= 0 or 1), by Poisson, using any λ, allow 1 – P(X = 0 or 1 or 2) for M1
- [A1] = 0.0661 or 0.0662 (3 sf); Final answer; SC use of Binomial (from 0.06607...) B1 only
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