A pair of dice is thrown 6 times. If getting a total of 7 is considered a success, what is the probability of at least 5 successes .

Answer

To solve the problem, we need to find the probability of getting a total of 7 when a pair of dice is thrown 6 times, and we want to calculate the probability of having at least 5 successes (i.e., getting a total of 7 at least 5 times).

### Step 1: Determine the probability of success (P)
When a pair of dice is thrown, the possible outcomes that give a sum of 7 are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)

There are 6 successful outcomes. The total number of outcomes when throwing two dice is 36 (6 faces on the first die multiplied by 6 faces on the second die).

Thus, the probability of getting a total of 7 (success) is:
\[ P = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6} \]

### Step 2: Determine the probability of failure (Q)
The probability of not getting a total of 7 (failure) is:
\[ Q = 1 - P = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Set up the binomial probability formula
We are interested in the probability of getting at least 5 successes in 6 trials. This can be calculated using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} P^k Q^{n-k} \]
where:
- \(n = 6\) (number of trials)
- \(k\) is the number of successes (5 or 6 in our case)
- \(P = \frac{1}{6}\)
- \(Q = \frac{5}{6}\)

### Step 4: Calculate the probability of exactly 5 successes (P5)
\[ P(X = 5) = \binom{6}{5} \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^{6-5} \]
Calculating \( \binom{6}{5} = 6 \):
\[ P(X = 5) = 6 \left(\frac{1}{6}\right)^5 \left(\frac{5}{6}\right)^1 = 6 \cdot \frac{1}{7776} \cdot \frac{5}{6} = \frac{30}{7776} \]

### Step 5: Calculate the probability of exactly 6 successes (P6)
\[ P(X = 6) = \binom{6}{6} \left(\frac{1}{6}\right)^6 \left(\frac{5}{6}\right)^{6-6} \]
Calculating \( \binom{6}{6} = 1 \):
\[ P(X = 6) = 1 \cdot \left(\frac{1}{6}\right)^6 \cdot 1 = \frac{1}{46656} \]

### Step 6: Combine the probabilities for at least 5 successes
Now, we need to sum the probabilities of getting exactly 5 successes and exactly 6 successes:
\[ P(X \geq 5) = P(X = 5) + P(X = 6) = \frac{30}{7776} + \frac{1}{46656} \]
To combine these fractions, we need a common denominator. The least common multiple of 7776 and 46656 is 46656.
\[ \frac{30}{7776} = \frac{30 \times 6}{7776 \times 6} = \frac{180}{46656} \]
Now, we can add:
\[ P(X \geq 5) = \frac{180}{46656} + \frac{1}{46656} = \frac{181}{46656} \]

### Final Answer
The probability of getting at least 5 successes when a pair of dice is thrown 6 times is:
\[ \frac{181}{46656} \]