In this problem, the player rolls a fair die 10 times. The probability of rolling a 4 on any single roll is $\frac{1}{6}$, and the probability of not rolling a 4 is $1 - \frac{1}{6} = \frac{5}{6}$.

Since the situation follows a binomial distribution, we can use the binomial probability formula to calculate the probabilities. The binomial distribution is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ Where:

• $n$ = number of trials (rolls) = 10,
• $k$ = number of successful outcomes (fours rolled),
• $p$ = probability of a success (probability of rolling a 4) = $\frac{1}{6}$,
• $\binom{n}{k}$ = binomial coefficient (number of ways to choose $k$ successes from $n$ trials).

Part (a): Probability of rolling exactly 3 fours in 10 rolls

We need to calculate the probability of rolling exactly 3 fours in 10 rolls, i.e., $P(X = 3)$.

$P(X = 3) = \binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{10-3}$

First, calculate the binomial coefficient:

$\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Now, substitute the values:

$P(X = 3) = 120 \times \left( \frac{1}{6} \right)^3 \times \left( \frac{5}{6} \right)^7$ $P(X = 3) = 120 \times \frac{1}{216} \times \left( \frac{5}{6} \right)^7$ $P(X = 3) \approx 120 \times \frac{1}{216} \times 0.2791$ $P(X = 3) \approx 120 \times 0.001292$ $P(X = 3) \approx 0.1549$

So, the probability of rolling exactly 3 fours is approximately 0.1549.

Part (b): Probability of rolling fewer than 2 fours in 10 rolls

This is asking for the probability of rolling 0 or 1 fours in 10 rolls, i.e., $P(X < 2) = P(X = 0) + P(X = 1)$.

For $P(X = 0)$:

$P(X = 0) = \binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} = 1 \times 1 \times \left( \frac{5}{6} \right)^{10}$ $P(X = 0) \approx 0.1615$

For $P(X = 1)$:

$P(X = 1) = \binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{9}$ $P(X = 1) = 10 \times \frac{1}{6} \times \left( \frac{5}{6} \right)^9$ $P(X = 1) \approx 10 \times 0.1667 \times 0.1938$ $P(X = 1) \approx 0.323$

Now, sum the probabilities:

$P(X < 2) = P(X = 0) + P(X = 1)$ $P(X < 2) \approx 0.1615 + 0.323$ $P(X < 2) \approx 0.4845$

So, the probability of rolling fewer than 2 fours is approximately 0.4845.

---

Summary of Results:

• (a) The probability of rolling exactly 3 fours is approximately 0.1549.
• (b) The probability of rolling fewer than 2 fours is approximately 0.4845.

Would you like more detailed explanations for any part of this solution, or do you have other questions?

Related Questions:

1. How does changing the number of rolls (e.g., 5 or 20) affect the probability of rolling exactly 3 fours?
2. What is the probability of rolling at least 1 four in 10 rolls?
3. How can you compute the expected number of fours rolled in 10 rolls?
4. What is the probability of rolling exactly 5 fours in 10 rolls?
5. How does the binomial distribution apply to other real-world situations, such as flipping a coin multiple times?

Tip: Always remember that when using the binomial distribution, the number of trials $n$, the probability of success $p$, and the number of successes $k$ are key parameters in determining the probabilities.