Answer:
This is a problem of a binomial distribution, where the probability of success (in this case, the probability of a smoker starting before 21 years of age) is given as 80% or 0.8, and the number of trials is 9.
For a binomial distribution, the probability of observing exactly k successes (k smokers who started before 21) in n trials (9 smokers selected) is given by:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where:
- C(n, k) is the binomial coefficient, which is the number of ways to choose k successes from n trials.
- p is the probability of success (0.8 in this case).
- (1-p) is the probability of failure.
Let's calculate the probabilities for the three situations you asked about:
1. The probability that at least 3 of them started smoking before 21 years of age:
This is equal to 1 minus the probability that fewer than 3 started smoking before 21 (i.e., 0, 1, or 2 started smoking before 21).
2. The probability that at most 2 of them started smoking before 21 years of age:
This is the sum of the probabilities that exactly 0, 1, or 2 started smoking before 21.
3. The probability that exactly 6 of them started smoking before 21 years of age:
This is a straightforward calculation using the binomial formula.
Now, let's do the calculations:
1. Probability that at least 3 of them started smoking before 21 years of age:
P(X>=3) = 1 - (P(X=0) + P(X=1) + P(X=2))
Where P(X=k) can be calculated using the binomial distribution formula.
2. Probability that at most 2 of them started smoking before 21 years of age:
P(X<=2) = P(X=0) + P(X=1) + P(X=2)
3. Probability that exactly 6 of them started smoking before 21 years of age:
P(X=6) can be calculated directly using the binomial distribution formula.
Remember that the binomial coefficient C(n, k) can be calculated as n!/(k!(n-k)!), where "!" denotes factorial. Factorial of a number n, denoted by n!, is the product of all positive integers less than or equal to n.
Let's calculate these probabilities now:
1. P(X>=3) = 1 - [C(9, 0)*(0.8^0)*(0.2^9) + C(9, 1)*(0.8^1)*(0.2^8) + C(9, 2)*(0.8^2)*(0.2^7)]
2. P(X<=2) = C(9, 0)*(0.8^0)*(0.2^9) + C(9, 1)*(0.8^1)*(0.2^8) + C(9, 2)*(0.8^2)*(0.2^7)
3. P(X=6) = C(9, 6)*(0.8^6)*(0.2^3)
Now, let's calculate these binomial coefficients and probabilities:
1. P(X>=3) = 1 - [1*(0.8^0)*(0.2^9) + 9*(0.8^1)*(0.2^8) + 36*(0.8^2)*(0.2^7)] ≈ 0.9984
2. P(X<=2) = 1*(0.8^0)*(0.2^9) + 9*(0.8^1)*(0.2^8) + 36*(0.8^2)*(0.2^7) ≈ 0.0016
3. P(X=6) = 84*(0.8^6)*(0.2^3) ≈ 0.2785
Remember, these values are approximate, and rounding to four decimal places was done as per your request.
So, to answer your questions:
1. The probability that at least 3 of them started smoking before 21 years of age is approximately 0.9984.
2. The probability that at most 2 of them started smoking before 21 years of age is approximately 0.0016.
3. The probability that exactly 6 of them started smoking before 21 years of age is approximately 0.2785.