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Stan guessed on all 10 questions of a multiple-choice quiz. Each question has 4 answer choices. What is the probability that he got at least 2 questions correct? Round the answer to the nearest thousandth.

[tex]\[
\begin{array}{c}
P(k \text{ successes }) = { }_n C_k p^k (1-p)^{n-k} \\
{ }_n C_k = \frac{n!}{(n-k)! \cdot k!}
\end{array}
\][/tex]

A. 0.211
B. 0.244
C. 0.756
D. 0.944


Sagot :

To find the probability that Stan guessed correctly on at least 2 out of 10 questions in a multiple-choice quiz where each question has 4 answer choices, we need to use the binomial probability formula:
[tex]\[ P(k \text { successes }) = \binom{n}{k} p^k (1 - p)^{n - k} \][/tex]

Here:
- [tex]\( n = 10 \)[/tex] (total number of questions)
- [tex]\( k \)[/tex] is the number of correct guesses
- [tex]\( p = \frac{1}{4} = 0.25 \)[/tex] (probability of guessing a question correctly)
- [tex]\( 1 - p = 0.75 \)[/tex] (probability of guessing a question incorrectly)

The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] is calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k! (n - k)!} \][/tex]

We need to sum the probabilities of getting 2 or more questions correct. That means we will calculate the probability for [tex]\( k = 2, 3, \ldots, 10 \)[/tex] and add them together.

The probabilities sum we need is:
[tex]\[ P(k \geq 2) = P(2) + P(3) + \cdots + P(10) \][/tex]

Rather than calculating each term individually, we use the solution we have derived through calculation to find that the probability sum turns out to be approximately [tex]\( 0.756 \)[/tex].

Therefore, the probability that Stan got at least 2 questions correct, rounded to the nearest thousandth, is:
[tex]\[ \boxed{0.756} \][/tex]