Question: Question 24
A test has 10 problems. Each problem has 4 choices (A, B, C, D), only one of which is correct. For a statistical experiment, Joe decides to roll a fair tetrahedral dice to pick his response for each problem. The tetrahedral dice has four letters (A, B, C, D) on its four faces. Joe picks the letter facing down. Let $p$ be the probability that Joe will get exactly 8 problems correct. What is the value of $4^{10}p$?
1. **State the problem:**
Joe answers 10 multiple-choice questions by randomly selecting one of 4 choices (A, B, C, D) with equal probability. We want to find the value of $4^{10}p$ where $p$ is the probability that Joe gets exactly 8 correct answers.
2. **Identify the distribution and formula:**
This is a binomial probability problem where:
- Number of trials $n = 10$
- Number of successes $k = 8$
- Probability of success on each trial $q = \frac{1}{4}$ (since only one correct choice out of 4)
The binomial probability formula is:
$$p = \binom{n}{k} q^k (1-q)^{n-k}$$
3. **Calculate the binomial coefficient:**
$$\binom{10}{8} = \frac{10!}{8!2!} = \frac{10 \times 9}{2 \times 1} = 45$$
4. **Calculate the probability $p$:**
$$p = 45 \times \left(\frac{1}{4}\right)^8 \times \left(1 - \frac{1}{4}\right)^2 = 45 \times \left(\frac{1}{4}\right)^8 \times \left(\frac{3}{4}\right)^2$$
5. **Calculate $4^{10}p$:**
$$4^{10}p = 4^{10} \times 45 \times \left(\frac{1}{4}\right)^8 \times \left(\frac{3}{4}\right)^2$$
Rewrite powers of 4:
$$4^{10}p = 45 \times 4^{10} \times \frac{1}{4^8} \times \frac{9}{16} = 45 \times 4^{10-8} \times \frac{9}{16} = 45 \times 4^2 \times \frac{9}{16}$$
Since $4^2 = 16$:
$$4^{10}p = 45 \times 16 \times \frac{9}{16}$$
6. **Simplify by canceling 16:**
$$4^{10}p = 45 \times \cancel{16} \times \frac{9}{\cancel{16}} = 45 \times 9 = 405$$
**Final answer:**
$$\boxed{405}$$