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Seeds: when a binomial model is an approximation

(i) In a small plot of land, there are 20 seeds being sown, of which only 4 will germinate. If 5 seeds are selected at random without replacement, find the probability that at least a seed that germinates would be selected. marks

(ii) In a huge plot of land where a large number of seeds are sown, 10% of the seeds would germinate. If 5 seeds are selected at random without replacement, find the probability that more than a seed that germinates would be selected. marks

State an assumption made in calculating the probability in (ii). mark

[2009/DHS/II/7]

CJC · Review Problems · Question 1

Answer

(i) 232/323232/323. (ii) 0.081460.08146. Assume a constant success probability and approximately independent selections in the huge population.

(i)

Use the complement, accounting for the changing population:

P(at least one)=1−P(all five do not germinate)P(\text{at least one})=1-P(\text{all five do not germinate})

=1−16201519141813171216=232323=1-\dfrac{16}{20}\dfrac{15}{19}\dfrac{14}{18}\dfrac{13}{17}\dfrac{12}{16}=\boxed{\frac{232}{323}}.

Equivalently, 1−(165)/(205)1-\binom{16}{5}/\binom{20}{5}. A fixed p=4/20 binomial model is not exact for this small without-replacement sample.

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(ii)

Let X count germinating seeds among the five selected. For the very large population, use the source model X∼B(5,0.10)X\sim B(5,0.10).

P(X>1)=1−P(X≤1)=1−0.95−5(0.1)(0.9)4=0.08146P(X>1)=1-P(X\leq1)=1-0.9^5-5(0.1)(0.9)^4=\boxed{0.08146}.

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Assumption

The source accepts either a constant germination probability of 0.1 or independence between selected seeds as the stated assumption. The binomial approximation uses both: selecting five seeds changes a huge population negligibly, so successive germination outcomes can be treated as approximately independent with constant success probability.

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Common mistakes

The small-plot calculation is exact; the huge-plot binomial calculation relies on a population-size approximation.

Read the hint