Question

Just prior to jury selection for O. J. Simpson’s murder trial in 1995, a poll found that about 18,4% of the adult population believed Simpson was innocent (after much of the physical evidence in the case had been revealed to the public). Ignore the fact that this 18,4% is an estimate based on a subsample from the population; for illustration, take it as the true percentage of people who thought Simpson was innocent prior to jury selection. Assume that the 10 jurors were selected randomly and independently from the population (although this turned out not to be true). Find the probability that the jury had at least two members who believed in Simpson’s innocence prior to jury selection. [Hint: Define the Binomial(10;18,4/100) random variable X to be the number of jurors believing in Simpson’s innocence.

Answer 1

Here, n = 10, p = 0.184, (1 - p) = 0.816 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 2).
P(X >= 2) = (10C2 * 0.184^2 * 0.816^8) + (10C3 * 0.184^3 * 0.816^7) + (10C4 * 0.184^4 * 0.816^6) + (10C5 * 0.184^5 * 0.816^5) + (10C6 * 0.184^6 * 0.816^4) + (10C7 * 0.184^7 * 0.816^3) + (10C8 * 0.184^8 * 0.816^2) + (10C9 * 0.184^9 * 0.816^1) + (10C10 * 0.184^10 * 0.816^0)
P(X >= 2) = 0.2995 + 0.1801 + 0.0711 + 0.0192 + 0.0036 + 0.0005 + 0 + 0 + 0
P(X >= 2) = 0.5740

answered by: anonymous