QuestionIn a college, 70% of classes are taught by adjuncts. What's the chance a student taking 5 classes has no adjunct professors?

Studdy Solution

STEP 1

Assumptions1. The probability of a class being taught by an adjunct is70% or0.7. The student is taking5 classes3. Each class is independent of the others

STEP 2

We are looking for the probability that none of the5 classes are taught by adjuncts. This is a binomial probability problem, where we want0 "successes" (classes taught by adjuncts) in5 trials (classes). The formula for binomial probability is $(k; n, p) = C(n, k) \cdot (p^k) \cdot ((1-p)^{n-k})$ where- $(k; n, p)$ is the probability of $k$ successes in $n$ trials for an event with probability $p$ , - $C(n, k)$ is the combination of $n$ items taken $k$ at a time, - $p^k$ is the probability of $k$ successes, and- $(1-p)^{n-k}$ is the probability of $n-k$ failures.

STEP 3

In this case, we want0 successes in5 trials, and the probability of success is0.7. So we plug in these values into the formula $(0;5,0.7) = C(5,0) \cdot (0.7^0) \cdot ((1-0.7)^{5-0})$

STEP 4

Calculate the combination $C(,0)$ , which is1 because there is only one way to choose0 items from.
$C(,0) =1$

STEP 5

Calculate $0.7^0$ , which is1 because any number to the power of0 is1.
$0.7^0 =1$

STEP 6

Calculate $(1-0.)^{5-0}$ , which is $(0.3)^5$ .
$(1-0.)^{5-0} = (0.3)^5$

STEP 7

Calculate $(0.3)^5$ .
$(0.3)^5 =0.00243$

STEP 8

Now, multiply all the calculated values together to get the final probability.
$(0;5,0.7) =1 \cdot1 \cdot0.00243 =0.00243$ The probability that a given student who takes5 classes will have no professors who are adjuncts is0.00243.

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