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Answer :
Final answer:
(a) Probability of exactly four defective facial tissues in a 45-second period: 0.0404.
(b) Probability of time between two defects being 20-50 seconds: 0.1841.
Explanation:
(a) To find the probability of exactly four defective facial tissues being removed in a 45-second period, we can use the Poisson distribution with a rate (λ) of 3.6 defects per minute. We need to convert the time to minutes to match the rate. The formula for the Poisson distribution is:
P(X = k) = [tex](e^(-λ) * λ^k) / k![/tex]
Where:
P(X = k) is the probability of k events occurring.
e is the base of the natural logarithm (approximately 2.71828).
λ is the rate of events (in this case, 3.6 defective tissues per minute).
k is the number of events (in this case, 4).
Substituting the values into the formula and rounding to four decimal places, we find that the probability is approximately 0.0404.
(b) To find the probability that the time between two successive removals of defective facial tissues is between 20 and 50 seconds, we need to find the cumulative probability of the Poisson distribution with rate λ = 3.6, for the interval [20/60, 50/60] minutes. This involves finding the probability of zero events in the first 20/60 minutes and exactly one event in the next 30/60 minutes, then summing these probabilities. The formula for the Poisson distribution is the same as in part (a).
Performing the calculations and rounding to four decimal places, we find the probability to be approximately 0.1841.
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