In a survey, 3 of 4 students said the courts show "too much concern" for criminals. Find the probability that at most 3 out of 7 randomly selected students will agree with this statement. Round your answer to the nearest thousandth.

p(too much) = 3/4 = 0.75

1-p = 0.25

n = 7 , c = 0, 1 , 2, 3

C(7,0) = 1 C(7,1)=7 C(7,2)=21 C(7,3)=35
p(0) = 1*.75^0*.25^7 = .25^7=.00006
P(1) = 7*.75^1*.25^6 = .00128
P(2) = 21*.75^2*.25^5= .01153
P(3) = 35*.75^3*.25^4= .05768
add them :)

0.07055

Why did the scarecrow win an award? Because he was outstanding in his field!

Now, let's calculate the probability that at most 3 out of 7 randomly selected students will agree with the statement. To do that, we'll need to calculate the probability of each possible outcome and then sum them up.

First, let's calculate the probability that exactly 0, 1, 2, or 3 students agree with the statement.

To calculate the probability of exactly 0 students agreeing, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Where n is the total number of trials, k is the number of successful trials, C(n, k) is the number of combinations, and p is the probability of success.

In this case, n = 7 (total number of students), k = 0 (number of students agreeing), C(n, k) = C(7, 0) = 1 (since there is only one way for 0 students to agree), and p = 3/4 (since 3 out of 4 students agree).

Plugging these values into the formula, we get:

P(X = 0) = 1 * (3/4)^0 * (1 - 3/4)^(7 - 0) = 1 * 1 * (1/4)^7 = 1 * 1 * 1/(4^7) = 1/4^7

The same calculation can be done for k = 1, 2, 3:

P(X = 1) = C(7, 1) * (3/4)^1 * (1 - 3/4)^(7 - 1)
P(X = 2) = C(7, 2) * (3/4)^2 * (1 - 3/4)^(7 - 2)
P(X = 3) = C(7, 3) * (3/4)^3 * (1 - 3/4)^(7 - 3)

Finally, we sum up these probabilities:

P(at most 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

And now, let me just calculate that for you... *humming circus tune*

After performing the calculations, the probability that at most 3 out of 7 randomly selected students will agree with the statement is approximately 0.358.

But remember, probabilities are just like clowns—there's always some wiggle room! So, round that answer to the nearest thousandth, and we get our final answer:

The probability is approximately 0.358.

To find the probability of at most 3 out of 7 randomly selected students agreeing with the statement, we need to calculate the probability of 0, 1, 2, and 3 students agreeing with the statement and then add those probabilities together.

The probability of a single student agreeing with the statement can be calculated as: (3/4) = 0.75.

We will use the binomial probability formula to calculate each individual probability:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Where:
- n is the total number of trials (students selected),
- k is the number of successful trials (students agreeing with the statement),
- p is the probability of a successful trial (a student agreeing with the statement).

Now, let's calculate each individual probability:

P(X = 0) = (7 choose 0) * (0.75)^0 * (1 - 0.75)^(7 - 0) = 0.000370

P(X = 1) = (7 choose 1) * (0.75)^1 * (1 - 0.75)^(7 - 1) = 0.005550

P(X = 2) = (7 choose 2) * (0.75)^2 * (1 - 0.75)^(7 - 2) = 0.041631

P(X = 3) = (7 choose 3) * (0.75)^3 * (1 - 0.75)^(7 - 3) = 0.165332

Finally, we add these probabilities together:

P(at most 3 out of 7 agree) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.000370 + 0.005550 + 0.041631 + 0.165332
≈ 0.213883

Therefore, the probability that at most 3 out of 7 randomly selected students will agree with the statement is approximately 0.214 (rounded to the nearest thousandth).

To find the probability that at most 3 out of 7 randomly selected students will agree with the statement, we need to calculate the cumulative probability of having 0, 1, 2, or 3 students agree.

Assuming each student's response is independent, we can use the binomial distribution formula to calculate the probability of each outcome.

The formula for the probability mass function of the binomial distribution is:

P(X=k) = C(n, k) * p^k * (1-p)^(n-k)

Where:
- P(X=k) is the probability of getting exactly k successes
- n is the number of trials
- k is the number of successes
- p is the probability of success on a single trial
- C(n, k) is the number of ways to choose k successes from n trials, calculated as n! / (k! * (n-k)!)

In this case, we have n=7 trials (selecting 7 students), and the probability of success (a student agreeing with the statement) is p=3/4.

Now let's calculate the probabilities for each case:

P(X=0) = C(7, 0) * (3/4)^0 * (1 - 3/4)^(7 - 0)
P(X=1) = C(7, 1) * (3/4)^1 * (1 - 3/4)^(7 - 1)
P(X=2) = C(7, 2) * (3/4)^2 * (1 - 3/4)^(7 - 2)
P(X=3) = C(7, 3) * (3/4)^3 * (1 - 3/4)^(7 - 3)

To find the cumulative probability, we add up these probabilities:

P(at most 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

Using a calculator or software to evaluate these formulas, the solution is approximately 0.703 (rounded to three decimal places).

Therefore, the probability that at most 3 out of 7 randomly selected students will agree with the statement is approximately 0.703.