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 three obstacles :

Options

A. `(143)/(216)`
B. `(100)/(243)`
C. `(216)/(243)`
D. `(100)/(216)`

Answer

To solve the problem of finding the probability that a person crosses the first three obstacles in the obstacle course, we will follow these steps:

### Step 1: Calculate Probability for the First Obstacle (n = 1)
At the first obstacle, the person rolls a die once (n = 1). The condition to cross the obstacle is that the sum of the points must be greater than \(2^1 = 2\).

- **Total Outcomes**: The possible outcomes when rolling a die once are {1, 2, 3, 4, 5, 6}. Thus, there are 6 total outcomes.
- **Favorable Outcomes**: The outcomes that are greater than 2 are {3, 4, 5, 6}. Hence, there are 4 favorable outcomes.

**Probability (P1)**:
\[ P_1 = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3} \]

### Step 2: Calculate Probability for the Second Obstacle (n = 2)
At the second obstacle, the person rolls the die twice (n = 2). The condition to cross this obstacle is that the sum of the points must be greater than \(2^2 = 4\).

- **Total Outcomes**: The total outcomes when rolling a die twice are \(6 \times 6 = 36\).
- **Favorable Outcomes**: We need to find pairs (x, y) such that \(x + y > 4\). The pairs that satisfy this condition are:
- For (1, _): (1, 4), (1, 5), (1, 6) → 3 outcomes
- For (2, _): (2, 3), (2, 4), (2, 5), (2, 6) → 4 outcomes
- For (3, _): (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) → 5 outcomes
- For (4, _): (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) → 6 outcomes
- For (5, _): (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) → 6 outcomes
- For (6, _): (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) → 6 outcomes

Counting these gives us:
- 3 + 4 + 5 + 6 + 6 + 6 = 30 favorable outcomes.

**Probability (P2)**:
\[ P_2 = \frac{30}{36} = \frac{5}{6} \]

### Step 3: Calculate Probability for the Third Obstacle (n = 3)
At the third obstacle, the person rolls the die three times (n = 3). The condition to cross this obstacle is that the sum of the points must be greater than \(2^3 = 8\).

- **Total Outcomes**: The total outcomes when rolling a die three times are \(6^3 = 216\).
- **Favorable Outcomes**: We need to find combinations (x, y, z) such that \(x + y + z > 8\). The total combinations can be calculated by finding the total combinations that yield sums of 3 to 8 and subtracting them from 216.

Calculating the sums:
- The minimum sum (1+1+1) = 3.
- The maximum sum (6+6+6) = 18.

The combinations yielding sums ≤ 8 can be calculated systematically or through generating functions, but for simplicity, we can use the complementary counting approach.

After calculating or using a generating function, we find that the number of outcomes where \(x + y + z \leq 8\) is 56.

Thus, the favorable outcomes for crossing the third obstacle are:
\[ 216 - 56 = 160 \]

**Probability (P3)**:
\[ P_3 = \frac{160}{216} = \frac{20}{27} \]

### Step 4: Calculate the Overall Probability
The overall probability that the person crosses all three obstacles is the product of the individual probabilities:
\[ P = P_1 \times P_2 \times P_3 = \frac{2}{3} \times \frac{5}{6} \times \frac{20}{27} \]

Calculating this gives:
\[ P = \frac{2 \times 5 \times 20}{3 \times 6 \times 27} = \frac{200}{486} = \frac{100}{243} \]

### Final Answer
The probability that a person crosses the first three obstacles is \(\frac{100}{243}\).

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