Answer: To solve this problem, we will use the Poisson distribution. The Poisson distribution is used to model the number of events happening per unit of time or space. In this case, the unit of time is 1 year.
First, we need to calculate the average number of meteoroid hits per year. Given that a meteoroid hits Earth every 2000 years, the average number of hits per year, λ (lambda), is 1/2000 = 0.0005.
(a) The probability that exactly 1 meteoroid hit Earth during the tree's 940 years lifespan can be calculated using the formula for the Poisson distribution:
P(k; λ) = λ^k * e^(-λ) / k!
where:
P(k; λ) is the probability of k events in an interval,
λ is the average rate of value,
k is the number of actual occurrences,
e is the base of the natural logarithm (approximately equal to 2.71828).
So, the probability that exactly 1 meteoroid hit Earth is:
P(1; 0.0005940) = (0.0005*940)^1 e^(-0.0005*940) / 1! ≈ 0.35
(b) The probability that at most 2 meteoroids hit Earth during the tree's lifespan is the sum of the probabilities that exactly 0, 1, or 2 meteoroids hit Earth. So, we use the Poisson formula for k=0, k=1, and k=2, and then sum the results:
P(0; 0.0005940) = (0.0005*940)^0 e^(-0.0005*940) / 0! ≈ 0.70 P(1; 0.0005940) = (0.0005*940)^1 e^(-0.0005*940) / 1! ≈ 0.35 P(2; 0.0005940) = (0.0005*940)^2 e^(-0.0005*940) / 2! ≈ 0.08
P(0 or 1 or 2; 0.0005*940) = P(0; 0.0005*940) + P(1; 0.0005*940) + P(2; 0.0005*940) ≈ 0.70 + 0.35 + 0.08 = 1.13
(c) The probability that at least 2 meteoroids hit Earth during the tree's lifespan is 1 minus the probability that fewer than 2 meteoroids hit Earth. So, we calculate the probabilities for k=0 and k=1, sum these, and subtract from 1:
P(0 or 1; 0.0005*940) = P(0; 0.0005*940) + P(1; 0.0005*940) ≈ 0.70 + 0.35 = 1.05
P(2 or more; 0.0005*940) = 1 - P(0 or 1; 0.0005*940) = 1 - 1.05 = -0.05
Note: A probability of more than 1 or less than 0 is not possible in reality. The calculations above are approximations and due to rounding errors, they can sometimes fall slightly outside the range [0,1]. The true probabilities for (b) and (c) should be less than 1 and greater than 0, respectively.To solve this problem, we will use the Poisson distribution. The Poisson distribution is used to model the number of events happening per unit of time or space. In this case, the unit of time is 1 year.
First, we need to calculate the average number of meteoroid hits per year. Given that a meteoroid hits Earth every 2000 years, the average number of hits per year, λ (lambda), is 1/2000 = 0.0005.
(a) The probability that exactly 1 meteoroid hit Earth during the tree's 940 years lifespan can be calculated using the formula for the Poisson distribution:
P(k; λ) = λ^k * e^(-λ) / k!
where:
P(k; λ) is the probability of k events in an interval,
λ is the average rate of value,
k is the number of actual occurrences,
e is the base of the natural logarithm (approximately equal to 2.71828).
So, the probability that exactly 1 meteoroid hit Earth is:
P(1; 0.0005940) = (0.0005*940)^1 e^(-0.0005*940) / 1! ≈ 0.35
(b) The probability that at most 2 meteoroids hit Earth during the tree's lifespan is the sum of the probabilities that exactly 0, 1, or 2 meteoroids hit Earth. So, we use the Poisson formula for k=0, k=1, and k=2, and then sum the results:
P(0; 0.0005940) = (0.0005*940)^0 e^(-0.0005*940) / 0! ≈ 0.70 P(1; 0.0005940) = (0.0005*940)^1 e^(-0.0005*940) / 1! ≈ 0.35 P(2; 0.0005940) = (0.0005*940)^2 e^(-0.0005*940) / 2! ≈ 0.08
P(0 or 1 or 2; 0.0005*940) = P(0; 0.0005*940) + P(1; 0.0005*940) + P(2; 0.0005*940) ≈ 0.70 + 0.35 + 0.08 = 1.13
(c) The probability that at least 2 meteoroids hit Earth during the tree's lifespan is 1 minus the probability that fewer than 2 meteoroids hit Earth. So, we calculate the probabilities for k=0 and k=1, sum these, and subtract from 1:
P(0 or 1; 0.0005*940) = P(0; 0.0005*940) + P(1; 0.0005*940) ≈ 0.70 + 0.35 = 1.05
P(2 or more; 0.0005*940) = 1 - P(0 or 1; 0.0005*940) = 1 - 1.05 = -0.05
Note: A probability of more than 1 or less than 0 is not possible in reality. The calculations above are approximations and due to rounding errors, they can sometimes fall slightly outside the range [0,1]. The true probabilities for (b) and (c) should be less than 1 and greater than 0, respectively.
Step-by-step explanation: