The rule of an "obstacle course" specifies that at the `n^(th)` obstacle a person has to tos a fair 6 sided die n times. If the sum of points in these n tosses is bigger than `2^(n)`, the person is said to have crossed the obstacle.
Q. The probability that a person crosses the first two obstacles but fails to cross the third obstacle.

Options

A. `(36)/(243)`
B. `(116)/(216)`
C. `(35)/(243)`
D. `(143)/(243)`

Answer

To solve the problem, we need to find the probability that a person crosses the first two obstacles but fails to cross the third obstacle in an obstacle course where the rules involve rolling a fair six-sided die.

### Step-by-Step Solution:

1. **Understanding the First Obstacle:**
- At the first obstacle, the person rolls the die once (n=1).
- The sum of the points must be greater than \(2^1 = 2\).
- The possible outcomes when rolling a die are {1, 2, 3, 4, 5, 6}.
- The successful outcomes (greater than 2) are {3, 4, 5, 6}.
- Therefore, the probability of crossing the first obstacle is:
\[ P(\text{cross 1st}) = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3} \]

2. **Understanding the Second Obstacle:**
- At the second obstacle, the person rolls the die twice (n=2).
- The sum of the points must be greater than \(2^2 = 4\).
- The total possible outcomes when rolling the die twice is \(6 \times 6 = 36\).
- We need to count the outcomes where the sum is less than or equal to 4:
- (1,1) → 2
- (1,2), (2,1) → 3
- (1,3), (3,1), (2,2) → 4
- The total unsuccessful outcomes are 6. Therefore, the successful outcomes are:
\[ \text{Successful outcomes} = 36 - 6 = 30 \]
- The probability of crossing the second obstacle is:
\[ P(\text{cross 2nd}) = \frac{30}{36} = \frac{5}{6} \]

3. **Understanding the Third Obstacle:**
- At the third obstacle, the person rolls the die three times (n=3).
- The sum of the points must be greater than \(2^3 = 8\).
- The total possible outcomes when rolling the die three times is \(6^3 = 216\).
- We need to count the outcomes where the sum is less than or equal to 8.
- The combinations that yield sums less than or equal to 8 can be calculated systematically or through combinatorial counting:
- After counting, we find that there are 56 outcomes where the sum is less than or equal to 8.
- Therefore, the successful outcomes are:
\[ \text{Successful outcomes} = 216 - 56 = 160 \]
- The probability of failing to cross the third obstacle is:
\[ P(\text{fail 3rd}) = \frac{56}{216} = \frac{7}{27} \]

4. **Calculating the Overall Probability:**
- The overall probability that the person crosses the first two obstacles and fails the third obstacle is given by the product of the individual probabilities:
\[ P(\text{cross 1st and 2nd, fail 3rd}) = P(\text{cross 1st}) \times P(\text{cross 2nd}) \times P(\text{fail 3rd}) \]
\[ = \left(\frac{2}{3}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{56}{216}\right) \]
- Simplifying this:
\[ = \frac{2 \times 5 \times 56}{3 \times 6 \times 216} = \frac{560}{3888} \]
- Reducing this fraction:
\[ = \frac{35}{243} \]

### Final Answer:
The probability that a person crosses the first two obstacles but fails to cross the third obstacle is:
\[ \frac{35}{243} \]