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Worked solution

Find the smallest binomial distribution with two tied modes

The random variable XX is the number of successes in nn independent trials of an experiment in which the probability of a success at any trial is pp.

Denoting P(X=k)P(X=k) by pkp_k, it is given that pkpk−1=(n−k+1)pk(1−p)\dfrac{p_k}{p_{k-1}}=\dfrac{(n-k+1)p}{k(1-p)}, k=1,2,…,nk=1,2,\ldots,n.

A distribution is said to be bimodal if it has two modes.

Find the least value of nn, and the corresponding modes of XX, given that XX is bimodal and that p=1825p=\dfrac{18}{25}. marks

[2017/ACJC/MYE/I/8]

CJC · Review Problems · Question 14

Answer

Least n=24n=24; modes 1717 and 1818.

Tie condition

Two neighbouring binomial modes occur when pk/pk−1=1p_k/p_{k-1}=1. Using the given ratio, (n−k+1)p=k(1−p)(n-k+1)p=k(1-p), hence (n+1)p=k(n+1)p=k. With p=18/25p=18/25, this becomes 18(n+1)=25k18(n+1)=25k.

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Least n and both modes

Since 18 and 25 are coprime, n+1n+1 must be a positive multiple of 25. Its smallest possible value is 25, giving n=24n=24 and k=18k=18. The tied modes are k−1=17k-1=17 and k=18k=18. The explorer probabilities at the two peaks are authored checks, not an additional printed ask.

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Common mistakes

The tied counts are k−1 and k, not k and k+1.

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