In a certain state, it has been shown that only 56% of the high school graduates who are capable of college work actually enroll in colleges.Find the probability that, among the 11 capable high school graduates in this state, each number will enroll in college.
a. Exactly 7
b.All 11
c.At least 4
d. Exactly 2
e. Less than 3
f None
I wonder how anyone can measure who is capable of college work? Exactly what is "college work" anyway? From my observation of all these online students enrolled in "colleges", being told they are doing "college work", and getting "degrees", one wonders.
I have seen a number of online colleges (want me to list them?) whose students are performing at less than high school level, yet getting "college credit" for it. An it amazes me how these "colleges" backscratch each other in giving each other "accrediation" to one another.
Enough, but it is a worthy question.
Here Pr(each)=.56, so if each kid makes independent decisions..., and the probability for a kid not going to college is .47
Pr(7)=.53^7*.47^4 * ways to do this, or
ways to do this (selecting 7 out of 11)
= 11!/(7!4!)
SO the final answer to your question..
Pr(7 of 11)=.53^7*.47^4*11!/(4!*7!)= nice math problem. Enter this into your google search window:
.53^7*.47^4*11!/(4!*7!)=
check my logic..
I did the exactly 7, follow my steps to get the others. On the less than three, do zero first, then exactly1, and exactly 2, then add those to get less than three.
Good logic, but I wonder about the .47 in light of P(college) = .56
To find the probability for each scenario, we need to use the concept of probability distribution.
First, we need the probability of a high school graduate enrolling in college, which is given as 56% or 0.56.
a) To find the probability that exactly 7 capable high school graduates will enroll in college, we need to use the binomial probability formula:
P(7 out of 11) = C(11, 7) * (0.56)^7 * (1 - 0.56)^(11 - 7)
Here, C(11, 7) represents the number of combinations of choosing 7 out of 11.
b) To find the probability that all 11 capable high school graduates will enroll in college, we simply multiply the individual probabilities:
P(all 11) = (0.56)^11
c) To find the probability of at least 4 capable high school graduates enrolling in college, we need to find the probabilities for 4, 5, 6, 7, 8, 9, 10, and 11 and sum them up:
P(at least 4) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11)
d) To find the probability that exactly 2 capable high school graduates will enroll in college, we again use the binomial probability formula:
P(2 out of 11) = C(11, 2) * (0.56)^2 * (1 - 0.56)^(11 - 2)
e) To find the probability of less than 3 capable high school graduates enrolling in college, we need to find the probabilities for 0, 1, and 2, and sum them up:
P(less than 3) = P(0) + P(1) + P(2)
f) To find the probability that none of the 11 capable high school graduates will enroll in college, we can use the formula:
P(none) = (1 - 0.56)^11
By calculating the above probabilities using the given formulas, you would get the desired probabilities for each scenario.