Two dies are thrown n times in succession. The probability of obtaining double - six atleast once is

Options

A. A) `1/36`
B. B) `1-(35/36)^(n)`
C. C) `(1/36)^(n)`
D. D) `(35/36)^(n)`

Answer

To solve the problem of finding the probability of obtaining double six at least once when two dice are thrown \( n \) times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Probability of Double Six:**
When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). The only outcome that results in double six is (6, 6). Thus, the probability of getting double six in a single throw of two dice is:
\[ P(\text{double six}) = \frac{1}{36} \]

**Hint:** Remember that the total outcomes when rolling two dice is the product of the number of faces on each die.

2. **Calculate the Probability of Not Getting Double Six:**
The probability of not getting double six in a single throw is:
\[ P(\text{not double six}) = 1 - P(\text{double six}) = 1 - \frac{1}{36} = \frac{35}{36} \]

**Hint:** Use the complement rule: \( P(A') = 1 - P(A) \).

3. **Extend to \( n \) Trials:**
If the dice are thrown \( n \) times, the probability of not getting double six in all \( n \) trials is:
\[ P(\text{not double six in } n \text{ trials}) = \left(\frac{35}{36}\right)^n \]

**Hint:** When independent events occur multiple times, multiply their probabilities.

4. **Calculate the Probability of Getting Double Six at Least Once:**
The probability of getting double six at least once in \( n \) trials is the complement of not getting double six in all trials:
\[ P(\text{at least one double six}) = 1 - P(\text{not double six in } n \text{ trials}) = 1 - \left(\frac{35}{36}\right)^n \]

**Hint:** Again, use the complement rule to find the probability of at least one occurrence.

### Final Answer:
Thus, the probability of obtaining double six at least once when two dice are thrown \( n \) times is:
\[ P(\text{at least one double six}) = 1 - \left(\frac{35}{36}\right)^n \]