Probability Problems---Check My Work and Correct Me?
A survey showed that 1 in 3 people entering the post office are wearing glasses. At noon, 4 people enter the post office.
A. What is the probability that one of those people is wearing glasses?
My Answer: 100%
B. What is the probability that one of those people is not wearing glasses?
My answer: 75%
C. What is the probability that at least 2 of the 4 people are wearing glasses?
My answer: 25%
D. The probability that at least 3 of the people are wearing glasses is given by P( is greater than or equal to 3)=P(4). Find the probability.
My Answer: How do I solve this type of problem?
A survey showed that 1 in 3 people entering the post office are wearing glasses. At noon, 4 people enter the post office.
A. What is the probability that one of those people is wearing glasses?
My Answer: 100%
B. What is the probability that one of those people is not wearing glasses?
My answer: 75%
C. What is the probability that at least 2 of the 4 people are wearing glasses?
My answer: 25%
D. The probability that at least 3 of the people are wearing glasses is given by P( is greater than or equal to 3)=P(4). Find the probability.
My Answer: How do I solve this type of problem?
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you have a 1/3 probability of someone wearing glasses
you have a 2/3 probability of someone not wearing glasses
let x = number of people wearing glasses
P(x = 1) = 4C1 * 1/3 * (2/3)^3 =
0.39506172839506172839506172839506
39.51% probability of one person wearing glasses
let y = number of people not wearing glasses
P(y = 1) = 4C1 * 2/3 * (1/3)^3 =
0.09876543209876543209876543209877
9.88% probability of one person not wearing glasses
let x = probability of number of people wearing glasses
P(x >= 2) = P(2 + 3 + 4)
P(x = 2) = 4C2 * (1/3)^2 * (2/3)^2
P(x = 3) = 4C3 * (1/3)^3 * 2/3
P(x = 4) = (1/3)^4
P(x >= 2) = 0.40740740740740740740740740740741
40.74% probability of at least 2 people are wearing glasses
P(x >= 3) = P(3 + 4)
0.11111111111111111111111111111111
11.11% probability of at least 3 people wearing glasses
you have a 2/3 probability of someone not wearing glasses
let x = number of people wearing glasses
P(x = 1) = 4C1 * 1/3 * (2/3)^3 =
0.39506172839506172839506172839506
39.51% probability of one person wearing glasses
let y = number of people not wearing glasses
P(y = 1) = 4C1 * 2/3 * (1/3)^3 =
0.09876543209876543209876543209877
9.88% probability of one person not wearing glasses
let x = probability of number of people wearing glasses
P(x >= 2) = P(2 + 3 + 4)
P(x = 2) = 4C2 * (1/3)^2 * (2/3)^2
P(x = 3) = 4C3 * (1/3)^3 * 2/3
P(x = 4) = (1/3)^4
P(x >= 2) = 0.40740740740740740740740740740741
40.74% probability of at least 2 people are wearing glasses
P(x >= 3) = P(3 + 4)
0.11111111111111111111111111111111
11.11% probability of at least 3 people wearing glasses