Solving Probability Problems: Binomial & Poisson Distribution Formulas

[NickLippy]

I have these four questions I'm having trouble solving...
Could anyone write the solutions, but with steps on how they did it so I can possibly learn! Thanks a lot

I'm pretty sure they use the binomial or poisson distribution formulas... can't get proper answers


1. In a large shipment of chips, 5% are defective. What is the probability that exactly two out of a sample of ten are defective?

2. On average, a system breaks down every 50hrs. Find the probability of more than two break-downs in a 24hr period.

3. Show that there are more families of 6 children split 4-2 than those with 3 boys and 3 girls.

4. People arrive at a bank at the rate of 60 per hour. Find the chance of getting 0,1,2 or 3 in the next minute.

 

[HallsofIvy]

You learn by doing things yourself. (1) is a "binomial distribution" problem, (2) and (4) are Poisson distribution, and (3) is just a matter of calculating the number of combinations.

 

[tacman]

1. To solve this problem using the binomial distribution formula, we first need to identify the values for n, p, and x.
n = 10 (sample size)
p = 0.05 (probability of defective chips)
x = 2 (number of defective chips we want to find the probability for)

Next, we can plug these values into the formula:
P(x) = nCx * p^x * (1-p)^(n-x)

P(2) = 10C2 * 0.05^2 * (1-0.05)^(10-2)
= 45 * 0.0025 * 0.951
= 0.1074

Therefore, the probability of exactly two out of a sample of ten chips being defective is 0.1074 or 10.74%.

2. To solve this problem using the poisson distribution formula, we first need to identify the value for λ (mean number of breakdowns in a 24hr period).
λ = 24/50 = 0.48

Next, we can plug this value into the formula:
P(x > 2) = 1 - P(x ≤ 2) = 1 - (e^-λ * (λ^0/0!) + e^-λ * (λ^1/1!) + e^-λ * (λ^2/2!))
= 1 - (0.6094 + 0.2923 + 0.1404)
= 0.0579

Therefore, the probability of more than two breakdowns in a 24hr period is 0.0579 or 5.79%.

3. To show that there are more families of 6 children split 4-2 than those with 3 boys and 3 girls, we can use the binomial distribution formula.
For 4-2 split, we have n = 6 (number of children), p = 0.5 (probability of having a boy or a girl), and x = 4 (number of boys we want to find the probability for).
P(x) = 6C4 * 0.5^4 * (1-0.5)^(6-4)
= 15 * 0.0625 * 0.25
= 0.2344

For 3 boys and