Six ordinary dice are rolled. The probability that at least half of them will show at least 3 is

Options

A. `41xx (2^4)/(3^6)`
B. `(2^4)/(3^6)`
C. `20xx(2^4)/(3^6)`
D. none of these

Answer

To find the probability that at least half of the six ordinary dice rolled will show at least a 3, we can follow these steps:

### Step 1: Determine the Probability of One Die Showing at Least 3
When rolling a single die, the outcomes that show at least a 3 are 3, 4, 5, and 6. The total number of favorable outcomes is 4 (3, 4, 5, 6) out of 6 possible outcomes.

\[ P(\text{at least 3}) = \frac{4}{6} = \frac{2}{3} \]

### Step 2: Determine the Probability of One Die Not Showing at Least 3
The outcomes that do not show at least a 3 are 1 and 2. The total number of unfavorable outcomes is 2.

\[ P(\text{not at least 3}) = \frac{2}{6} = \frac{1}{3} \]

### Step 3: Define the Random Variable
Let \( X \) be the random variable representing the number of dice that show at least a 3 when 6 dice are rolled. We want to find the probability that at least half of the dice (i.e., at least 3) show at least a 3.

### Step 4: Use the Binomial Probability Formula
The number of successes (dice showing at least a 3) follows a binomial distribution with parameters \( n = 6 \) (the number of trials) and \( p = \frac{2}{3} \) (the probability of success). The probability mass function is given by:

\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

where \( q = 1 - p = \frac{1}{3} \).

### Step 5: Calculate the Required Probability
We need to calculate \( P(X \geq 3) \), which is the sum of the probabilities from \( X = 3 \) to \( X = 6 \):

\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]

Calculating each term:

1. **For \( X = 3 \)**:
\[ P(X = 3) = \binom{6}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^{3} = 20 \cdot \frac{8}{27} \cdot \frac{1}{27} = \frac{160}{729} \]

2. **For \( X = 4 \)**:
\[ P(X = 4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^{2} = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729} \]

3. **For \( X = 5 \)**:
\[ P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{1} = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{64}{729} \]

4. **For \( X = 6 \)**:
\[ P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{0} = 1 \cdot \frac{64}{729} = \frac{64}{729} \]

### Step 6: Sum the Probabilities
Now, we sum the probabilities:

\[ P(X \geq 3) = \frac{160}{729} + \frac{240}{729} + \frac{64}{729} + \frac{1}{729} = \frac{465}{729} \]

### Final Answer
Thus, the probability that at least half of the dice will show at least a 3 is:

\[ \frac{465}{729} \]