High School

On average, 5 conferences occur annually in a university. What is the probability that in any given year a) Exactly 9 conferences will occur in the university? b) Fewer than 6 conferences will occur in the university? c) At least 4 conferences will occur in the university?

Answer :

a) The probability of exactly 9 conferences occurring in a given year is approximately 0.104.

b) The probability of fewer than 6 conferences occurring in a given year is about 0.265.

c) The probability of at least 4 conferences occurring in a given year is roughly 0.924.

*a) Probability of Exactly 9 Conferences:*

The probability of exactly 9 conferences occurring in a given year can be calculated using the Poisson distribution formula:

P(X = k) = (λ^k * e^(-λ)) / k!

Where λ (lambda) is the average number of conferences, which is 5 in this case, and k is the desired number of conferences (9).

P(X = 9) = (5^9 * e^(-5)) / 9! ≈ 0.104

*b) Probability of Fewer than 6 Conferences:*

To calculate the probability of fewer than 6 conferences, we need to find the probabilities for 0 to 5 conferences and sum them up.

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

≈ (5^0 * e^(-5)) / 0! + (5^1 * e^(-5)) / 1! + (5^2 * e^(-5)) / 2! + (5^3 * e^(-5)) / 3! + (5^4 * e^(-5)) / 4! + (5^5 * e^(-5)) / 5!

≈ 0.265

*c) Probability of At Least 4 Conferences:*

To calculate the probability of at least 4 conferences, we'll find the probabilities for 4 to 9 conferences and sum them up.

P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

≈ (5^4 * e^(-5)) / 4! + (5^5 * e^(-5)) / 5! + (5^6 * e^(-5)) / 6! + (5^7 * e^(-5)) / 7! + (5^8 * e^(-5)) / 8! + (5^9 * e^(-5)) / 9!

≈ 0.924

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