Exercise 1: On the average, five smokers pass a certain street corners every ten minutes, what is
the probability that during a given 10minutes the number of smokers passing will be : a) 6 or
fewer and b) 7 or more
To solve this problem, we need to use the Poisson distribution formula:
P(x; λ) = (e^(-λ)*λ^x) / x!
Where P(x; λ) represents the probability of x events occurring in a given interval, λ is the average number of events in the interval, e is the base of the natural logarithm, and x is the number of events.
a) To find the probability that 6 or fewer smokers pass the street corner in 10 minutes, we need to calculate the individual probabilities for each possible number of smokers (0, 1, 2, 3, 4, 5, 6) and sum them up.
λ = 5 (average number of smokers in 10 minutes)
P(0; 5) = (e^(-5)*5^0) / 0! = (e^(-5)) / 1 ≈ 0.0067
P(1; 5) = (e^(-5)*5^1) / 1! = (5e^(-5)) / 1 ≈ 0.0337
P(2; 5) = (e^(-5)*5^2) / 2! = (12.5e^(-5)) / 2 ≈ 0.0843
P(3; 5) = (e^(-5)*5^3) / 3! = (62.5e^(-5)) / 6 ≈ 0.1406
P(4; 5) = (e^(-5)*5^4) / 4! = (625e^(-5)) / 24 ≈ 0.1758
P(5; 5) = (e^(-5)*5^5) / 5! = (3125e^(-5)) / 120 ≈ 0.1758
P(6; 5) = (e^(-5)*5^6) / 6! = (7812.5e^(-5)) / 720 ≈ 0.1466
Summing up these probabilities gives us the probability of 6 or fewer smokers passing the street corner in 10 minutes:
P(6 or fewer) = 0.0067 + 0.0337 + 0.0843 + 0.1406 + 0.1758 + 0.1758 + 0.1466 ≈ 0.7635
Therefore, the probability that during a given 10 minutes the number of smokers passing will be 6 or fewer is approximately 0.7635.
b) To find the probability that 7 or more smokers pass the street corner in 10 minutes, we need to calculate the individual probabilities for each possible number of smokers (7, 8, 9, 10, 11, 12, ...) and sum them up.
P(7; 5) = (e^(-5)*5^7) / 7! = (7812.5e^(-5)) / 5040 ≈ 0.1042
P(8; 5) = (e^(-5)*5^8) / 8! = (39062.5e^(-5)) / 40320 ≈ 0.0651
P(9; 5) = (e^(-5)*5^9) / 9! = (175781.25e^(-5)) / 362880 ≈ 0.0293
P(10; 5) = (e^(-5)*5^10) / 10! = (878906.25e^(-5)) / 3628800 ≈ 0.0117
P(11; 5) = (e^(-5)*5^11) / 11! ≈ 0.0034
P(12; 5) = (e^(-5)*5^12) / 12! ≈ 0.0009
...
Since the probabilities for higher numbers are becoming very small, we can stop calculating after a certain point. Let's assume we stop after P(12; 5).
Summing up these probabilities gives us the probability of 7 or more smokers passing the street corner in 10 minutes:
P(7 or more) = 0.1042 + 0.0651 + 0.0293 + 0.0117 + 0.0034 + 0.0009 + ...
The sum of the probabilities beyond this point is very small, so we can approximate the probability as:
P(7 or more) ≈ 0.1042 + 0.0651 + 0.0293 + 0.0117 + 0.0034 + 0.0009 ≈ 0.2146
Therefore, the probability that during a given 10 minutes the number of smokers passing will be 7 or more is approximately 0.2146.