Mrs. Gomes found that 40% of students at her high school take chemistry. She randomly surveys 12 students. What is the probability that at most 4 students have taken chemistry? Round the answer to the nearest thousandth.

A)0.008
B)0.438
C)0.562
D)0.665

Use the binomial distribution,

p=P(chemistry)=0.4
n=12 (sample size)
r=number taking chemistry

P(n,r)=nCr*0.4^r*(1-0.4)^(n-r)
where nCr=n!/((n-r)!r!) is binomial coefficient.

For probability of getting AT MOST 4 taking chemistry, add up all possible cases, i.e. 0,1,2,3,4, or
P(12,0)+P(12,1)+P(12,2)+P(12,3)+P(12,4)
the answer should match one of the choices.

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To solve this problem, we need to use the binomial probability formula. The binomial probability formula is given by:

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

Where:
P(X = k) is the probability of having exactly k successes
C(n, k) is the number of combinations of n items taken k at a time
p is the probability of success (in this case, the probability that a randomly selected student has taken chemistry)
q is the probability of failure (1 - p)
n is the number of trials (in this case, the number of students surveyed)

In this problem, we are given that p = 0.4 (40% of students take chemistry), n = 12 (12 students surveyed), and we need to find the probability that at most 4 students have taken chemistry (k ≤ 4).

Now, let's calculate the probability for each value of k (0 to 4) and sum them up to get the final answer.

P(X = 0) = C(12, 0) * 0.4^0 * (1-0.4)^(12-0)
P(X = 1) = C(12, 1) * 0.4^1 * (1-0.4)^(12-1)
P(X = 2) = C(12, 2) * 0.4^2 * (1-0.4)^(12-2)
P(X = 3) = C(12, 3) * 0.4^3 * (1-0.4)^(12-3)
P(X = 4) = C(12, 4) * 0.4^4 * (1-0.4)^(12-4)

To calculate the combinations, you can use the formula:
C(n, k) = n! / (k!(n-k)!), where n! is the factorial of n.

After calculating the individual probabilities for each value of k, you need to sum them up:

P(at most 4 students have taken chemistry) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Finally, round the answer to the nearest thousandth and select the corresponding option from the given choices.

0.004

I'm not even in this class. Lol.