To solve the problem of finding the probability of getting 1, 3, or 4 at most \( n \) times when a die is thrown \( 2n + 1 \) times, we can follow these steps:
### Step 1: Define the probabilities
Let \( p \) be the probability of getting a 1, 3, or 4 on a single throw of the die. Since there are 3 favorable outcomes (1, 3, and 4) out of 6 possible outcomes on a die, we have:
\[
p = \frac{3}{6} = \frac{1}{2}
\]
Let \( q \) be the probability of not getting a 1, 3, or 4, which is:
\[
q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}
\]
### Step 2: Use the binomial distribution
The number of times we get 1, 3, or 4 in \( 2n + 1 \) throws follows a binomial distribution with parameters \( n = 2n + 1 \) and \( p = \frac{1}{2} \).
We want to find the probability of getting 1, 3, or 4 at most \( n \) times. This can be expressed using the binomial probability formula:
\[
P(X \leq n) = \sum_{k=0}^{n} \binom{2n+1}{k} p^k q^{(2n+1-k)}
\]
### Step 3: Substitute the values of \( p \) and \( q \)
Substituting \( p \) and \( q \) into the equation, we have:
\[
P(X \leq n) = \sum_{k=0}^{n} \binom{2n+1}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{(2n+1-k)}
\]
This simplifies to:
\[
P(X \leq n) = \sum_{k=0}^{n} \binom{2n+1}{k} \left(\frac{1}{2}\right)^{2n+1}
\]
### Step 4: Factor out the constant
Since \( \left(\frac{1}{2}\right)^{2n+1} \) is a common factor, we can factor it out:
\[
P(X \leq n) = \left(\frac{1}{2}\right)^{2n+1} \sum_{k=0}^{n} \binom{2n+1}{k}
\]
### Step 5: Use the symmetry of the binomial distribution
By the symmetry of the binomial distribution, we know that:
\[
\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}
\]
Thus, half of this sum (which corresponds to \( k \leq n \)) is:
\[
\sum_{k=0}^{n} \binom{2n+1}{k} = \frac{1}{2} \cdot 2^{2n+1} = 2^{2n}
\]
### Step 6: Substitute back into the equation
Now substituting back, we find:
\[
P(X \leq n) = \left(\frac{1}{2}\right)^{2n+1} \cdot 2^{2n} = \frac{1}{2}
\]
### Final Answer
Thus, the probability of getting 1, 3, or 4 at most \( n \) times when a die is thrown \( 2n + 1 \) times is:
\[
\frac{1}{2}
\]
