1. **Problem:** A student is taking a multiple choice test with 5 questions, each having 4 possible answers. Find the following probabilities: a. Probability of getting only the first question correct. b. Probability of getting any one question correct. c. Probability of getting all questions correct. 2. **Formula and rules:** For binomial probability, the formula is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where: - $n$ = number of trials (questions), - $k$ = number of successes (correct answers), - $p$ = probability of success on a single trial, - $\binom{n}{k}$ = number of combinations of $n$ items taken $k$ at a time. 3. **Step-by-step solutions:** a. Probability of getting only the first question correct means: - First question correct: $p = \frac{1}{4}$ - Other 4 questions incorrect: $1-p = \frac{3}{4}$ Since only the first is correct and the rest are incorrect, the probability is: $$P = \frac{1}{4} \times \left(\frac{3}{4}\right)^4 = \frac{1}{4} \times \frac{81}{256} = \frac{81}{1024} \approx 0.0791$$ b. Probability of getting any one question correct means exactly one correct out of five questions, but the correct one can be any of the five questions. So $k=1$, $n=5$, $p=\frac{1}{4}$: $$P(X=1) = \binom{5}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^4 = 5 \times \frac{1}{4} \times \frac{81}{256} = \frac{405}{1024} \approx 0.3955$$ c. Probability of getting all questions correct means $k=5$: $$P(X=5) = \binom{5}{5} \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{1024} \times 1 = \frac{1}{1024} \approx 0.00098$$ 4. **Explanation:** - For part a, since only the first question is correct and the rest are wrong, we multiply the probability of first correct by the probability of the rest wrong. - For part b, since any one question can be correct, we use the binomial coefficient to count all possible positions. - For part c, all must be correct, so the probability is the product of $p$ for all 5 questions. **Final answers:** - a. $\approx 0.0791$ - b. $\approx 0.3955$ - c. $\approx 0.00098$