When two dice are thrown n number of times, the probability of getting a doublet atleast once in greater than `80%` and the least value of n is `lambda`, then the value of `lambda` is equal to

Options

A. 62
B. 71
C. 80
D. 91

Answer

To solve the problem, we need to find the least value of \( n \) such that the probability of getting at least one doublet when two dice are thrown \( n \) times is greater than 80%.

### Step-by-step Solution:

1. **Understanding the Problem**:
- A doublet occurs when both dice show the same number. The possible doublets when throwing two dice are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Thus, there are 6 doublets.
- The total number of outcomes when throwing two dice is \( 6 \times 6 = 36 \).

2. **Calculating the Probability of Not Getting a Doublet**:
- The probability of not getting a doublet in a single throw is:
\[ P(\text{not a doublet}) = 1 - P(\text{doublet}) = 1 - \frac{6}{36} = \frac{30}{36} = \frac{5}{6} \]

3. **Probability of Not Getting a Doublet in \( n \) Throws**:
- If two dice are thrown \( n \) times, the probability of not getting a doublet in any of the \( n \) throws is:
\[ P(\text{no doublet in } n \text{ throws}) = \left(\frac{5}{6}\right)^n \]

4. **Probability of Getting at Least One Doublet**:
- Therefore, the probability of getting at least one doublet in \( n \) throws is:
\[ P(\text{at least one doublet}) = 1 - P(\text{no doublet in } n \text{ throws}) = 1 - \left(\frac{5}{6}\right)^n \]

5. **Setting Up the Inequality**:
- We want this probability to be greater than 80%:
\[ 1 - \left(\frac{5}{6}\right)^n > 0.8 \]
- Rearranging gives:
\[ \left(\frac{5}{6}\right)^n < 0.2 \]

6. **Taking Logarithms**:
- Taking the logarithm of both sides:
\[ \log\left(\left(\frac{5}{6}\right)^n\right) < \log(0.2) \]
- This simplifies to:
\[ n \cdot \log\left(\frac{5}{6}\right) < \log(0.2) \]
- Since \( \log\left(\frac{5}{6}\right) \) is negative, we can reverse the inequality:
\[ n > \frac{\log(0.2)}{\log\left(\frac{5}{6}\right)} \]

7. **Calculating the Values**:
- Using logarithm values:
\[ \log(0.2) \approx -0.6990 \]
\[ \log\left(\frac{5}{6}\right) \approx \log(5) - \log(6) \approx 0.6990 - 0.7781 \approx -0.0791 \]
- Now substituting these values:
\[ n > \frac{-0.6990}{-0.0791} \approx 8.83 \]

8. **Finding the Least Integer Value**:
- Since \( n \) must be an integer, we round up to the next whole number:
\[ n \geq 9 \]

### Conclusion:
Thus, the least value of \( n \) such that the probability of getting at least one doublet is greater than 80% is \( \lambda = 9 \).