Final answer:
To find the probability that exactly two one-year-old children out of 9000 cannot recognize their mother's voice when the average likelihood is 0.001 per child, we use the Poisson distribution with λ = 9. The calculated probability using the Poisson probability mass function is approximately 0.0025 or 0.25%.
Step-by-step explanation:
The question involves calculating the probability of an event occurring in a given number of trials using the Poisson distribution. Specifically, we are interested in finding the probability that exactly two one-year-old children out of 9000 cannot recognize their mother's voice, given the likelihood of this happening is very low (0.001 per child on average). The Poisson distribution is appropriate here because we are dealing with rare events over a large number of trials.
First, we need to determine the average rate (λ) at which these events (children not recognizing their mother's voice) occur across the population. If on average, 1 out of 1000 children does not recognize their mother's voice, then for 9000 children, λ = 9000/1000 = 9 children are expected not to recognize their mother's voice.
To find the probability that exactly two children do not recognize their mother's voice, we use the Poisson probability mass function: P(x; λ) = (e^{-λ} * λ^x) / x!, where x is the number of successes (in our case, children not recognizing their mother's voice), λ is the average number of successes, and e is the base of the natural logarithm.
Substituting the values into the formula: P(2; 9) = (e^{-9} * 9^2) / 2! = (0.00012341 * 81) / 2 = 0.00498361 / 2 = 0.0024918055. Therefore, the probability that exactly two children out of 9000 do not recognize their mother's voice is approximately 0.0025 or 0.25%.