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Answer :
(i) The probability of at most 1 fracture occurring in a person's lifetime is approximately 0.1992.
(ii) The probability that Mr. Lim has at most 3 fractures in the second half of his lifetime is approximately 0.9343.
The number of fractures happening in a person's lifetime can be modeled by a Poisson distribution.
To find this probability, we can use the Poisson probability formula. Let's denote the average number of fractures occurring in a person's lifetime as λ (lambda). In this case, λ is given as 3 fractures.
The formula for the probability of k occurrences in a Poisson distribution is:
P(k; λ) = (e^(-λ) * λ^k) / k!
To find the probability of at most 1 fracture, we need to calculate the probabilities for 0 and 1 fractures and then add them together.
P(0; λ) = (e⁻³ * 3⁰) / 0!
= e⁻³
≈ 0.0498
P(1; λ) = (e⁻³ * 3¹) / 1!
= 3 * e⁻³
≈ 0.1494
Therefore, the probability of at most 1 fracture occurring in a person's lifetime is approximately 0.0498 + 0.1494 = 0.1992.
Now let's move on to the second part of the question. Mr. Lim is currently 43 years old, and he suffered his first fracture at the age of 42. We are asked to find the probability that Mr. Lim has at most 3 fractures in the second half of his lifetime.
Since the average number of fractures in a lifetime is 3, we can assume that the average number of fractures in the second half of a lifetime is 1.5 (half of 3). Let's denote this λ' (lambda prime).
To find the probability, we can again use the Poisson probability formula. This time, we need to calculate the probabilities for 0, 1, 2, and 3 fractures and add them together.
P(0; λ') = (e⁻¹.⁵ * 1.5⁰) / 0!
≈ 0.2231
P(1; λ') = (e⁻¹.⁵ * 1.5¹) / 1!
≈ 0.3347
P(2; λ') = (e⁻¹.⁵ * 1.5²) / 2!
≈ 0.2510
P(3; λ') = (e⁻¹.⁵ * 1.5³) / 3!
≈ 0.1255
Therefore, the probability that Mr. Lim has at most 3 fractures in the second half of his lifetime is approximately 0.2231 + 0.3347 + 0.2510 + 0.1255 = 0.9343.
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