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Sagot :
Alright, let's work through these questions step-by-step.
### Part (a):
Question: One out of a thousand people reacted to a newly manufactured vaccine against an endemic disease. If 3000 people were treated with this vaccine, find the probability that at most two people reacted to the vaccine.
Solution:
We are dealing with a binomial distribution where the probability of reacting to the vaccine (p) is 1/1000, and the number of trials (n) is 3000. When n is large and p is small, we can approximate the binomial distribution with a Poisson distribution. The Poisson parameter λ (lambda) is given by [tex]\( \lambda = n \times p \)[/tex].
In this case:
[tex]\[ \lambda = 3000 \times \frac{1}{1000} = 3 \][/tex]
We need to find the probability that at most two people reacted, which is [tex]\( P(X \leq 2) \)[/tex]. Using the Poisson distribution, this is the sum of the probabilities of 0, 1, and 2 reactions.
[tex]\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \][/tex]
Using the Poisson probability mass function (pmf):
[tex]\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \][/tex]
Summing these probabilities for [tex]\( k = 0, 1, 2 \)[/tex]:
[tex]\[ P(X \leq 2) = \left( \frac{3^0 e^{-3}}{0!} \right) + \left( \frac{3^1 e^{-3}}{1!} \right) + \left( \frac{3^2 e^{-3}}{2!} \right) \][/tex]
Evaluating these terms gives:
[tex]\[ P(X \leq 2) \approx 0.423190 \][/tex]
### Part (b):
Question: Find the probability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin.
Solution:
This situation follows a binomial distribution with [tex]\( n = 100 \)[/tex] (the number of tosses) and [tex]\( p = 0.5 \)[/tex] (the probability of getting heads in each toss). We need to find the probability that the number of heads, [tex]\( X \)[/tex], falls between 40 and 60 inclusive: [tex]\( P(40 \leq X \leq 60) \)[/tex].
We sum the binomial probabilities for each [tex]\( k \)[/tex] from 40 to 60.
[tex]\[ P(40 \leq X \leq 60) = \sum_{k=40}^{60} \binom{100}{k} (0.5)^k (0.5)^{100-k} \][/tex]
This includes calculating the binomial probability mass function (pmf) at each value of [tex]\( k \)[/tex] from 40 to 60.
[tex]\[ P(X = k) = \binom{100}{k} (0.5)^{100} \][/tex]
Summing these probabilities gives:
[tex]\[ P(40 \leq X \leq 60) \approx 0.964800 \][/tex]
### Part (c):
Question: Find the probability that in a family of 4 children, there will be at least 1 boy.
Solution:
This also follows a binomial distribution where [tex]\( n = 4 \)[/tex] (the number of children) and [tex]\( p = 0.5 \)[/tex] (the probability of having a boy in each birth). We need to find the probability of having at least 1 boy, which is [tex]\( 1 - P(X = 0) \)[/tex], where [tex]\( X \)[/tex] is the number of boys.
[tex]\[ P(\text{at least 1 boy}) = 1 - P(X = 0) \][/tex]
The probability of having 0 boys (all girls) can be calculated using the binomial pmf:
[tex]\[ P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = (1) (0.5)^4 = 0.0625 \][/tex]
Thus, the probability of having at least 1 boy is:
[tex]\[ P(\text{at least 1 boy}) = 1 - 0.0625 = 0.9375 \][/tex]
### Summary of Results:
(a) Probability that at most two people reacted to the vaccine: [tex]\( \approx 0.423190 \)[/tex]
(b) Probability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin: [tex]\( \approx 0.964800 \)[/tex]
(c) Probability that in a family of 4 children, there will be at least 1 boy: [tex]\( \approx 0.9375 \)[/tex]
### Part (a):
Question: One out of a thousand people reacted to a newly manufactured vaccine against an endemic disease. If 3000 people were treated with this vaccine, find the probability that at most two people reacted to the vaccine.
Solution:
We are dealing with a binomial distribution where the probability of reacting to the vaccine (p) is 1/1000, and the number of trials (n) is 3000. When n is large and p is small, we can approximate the binomial distribution with a Poisson distribution. The Poisson parameter λ (lambda) is given by [tex]\( \lambda = n \times p \)[/tex].
In this case:
[tex]\[ \lambda = 3000 \times \frac{1}{1000} = 3 \][/tex]
We need to find the probability that at most two people reacted, which is [tex]\( P(X \leq 2) \)[/tex]. Using the Poisson distribution, this is the sum of the probabilities of 0, 1, and 2 reactions.
[tex]\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \][/tex]
Using the Poisson probability mass function (pmf):
[tex]\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \][/tex]
Summing these probabilities for [tex]\( k = 0, 1, 2 \)[/tex]:
[tex]\[ P(X \leq 2) = \left( \frac{3^0 e^{-3}}{0!} \right) + \left( \frac{3^1 e^{-3}}{1!} \right) + \left( \frac{3^2 e^{-3}}{2!} \right) \][/tex]
Evaluating these terms gives:
[tex]\[ P(X \leq 2) \approx 0.423190 \][/tex]
### Part (b):
Question: Find the probability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin.
Solution:
This situation follows a binomial distribution with [tex]\( n = 100 \)[/tex] (the number of tosses) and [tex]\( p = 0.5 \)[/tex] (the probability of getting heads in each toss). We need to find the probability that the number of heads, [tex]\( X \)[/tex], falls between 40 and 60 inclusive: [tex]\( P(40 \leq X \leq 60) \)[/tex].
We sum the binomial probabilities for each [tex]\( k \)[/tex] from 40 to 60.
[tex]\[ P(40 \leq X \leq 60) = \sum_{k=40}^{60} \binom{100}{k} (0.5)^k (0.5)^{100-k} \][/tex]
This includes calculating the binomial probability mass function (pmf) at each value of [tex]\( k \)[/tex] from 40 to 60.
[tex]\[ P(X = k) = \binom{100}{k} (0.5)^{100} \][/tex]
Summing these probabilities gives:
[tex]\[ P(40 \leq X \leq 60) \approx 0.964800 \][/tex]
### Part (c):
Question: Find the probability that in a family of 4 children, there will be at least 1 boy.
Solution:
This also follows a binomial distribution where [tex]\( n = 4 \)[/tex] (the number of children) and [tex]\( p = 0.5 \)[/tex] (the probability of having a boy in each birth). We need to find the probability of having at least 1 boy, which is [tex]\( 1 - P(X = 0) \)[/tex], where [tex]\( X \)[/tex] is the number of boys.
[tex]\[ P(\text{at least 1 boy}) = 1 - P(X = 0) \][/tex]
The probability of having 0 boys (all girls) can be calculated using the binomial pmf:
[tex]\[ P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = (1) (0.5)^4 = 0.0625 \][/tex]
Thus, the probability of having at least 1 boy is:
[tex]\[ P(\text{at least 1 boy}) = 1 - 0.0625 = 0.9375 \][/tex]
### Summary of Results:
(a) Probability that at most two people reacted to the vaccine: [tex]\( \approx 0.423190 \)[/tex]
(b) Probability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin: [tex]\( \approx 0.964800 \)[/tex]
(c) Probability that in a family of 4 children, there will be at least 1 boy: [tex]\( \approx 0.9375 \)[/tex]
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