To calculate the total probability of successfully clearing the layout, we need to analyze the combinations of paths that allow the player to succeed.
Since the player must successfully navigate at least two paths, success can happen in one of two ways: either they succeed on exactly two paths, or they succeed on all three paths.
Here is the step-by-step mathematical breakdown and the exact LaTeX syntax for your analysis.
1. Identify the Complementary Probabilities (Failure Rates)
First, we find the probability of failing each independent path, denoted as $P(X')$ or $1 - P(X)$:
$P(A') = 1 - 0.75 = 0.25$
$P(B') = 1 - 0.54 = 0.46$
$P(C') = 1 - 0.60 = 0.40$
2. The Success Scenarios
There are 4 mutually exclusive scenarios where the player successfully navigates at least two paths:
Succeeds on A and B, fails on C: $P(A) \times P(B) \times P(C')$
Succeeds on A and C, fails on B: $P(A) \times P(B') \times P(C)$
Succeeds on B and C, fails on A: $P(A') \times P(B) \times P(C)$
Succeeds on all three (A, B, and C): $P(A) \times P(B) \times P(C)$
3. The LaTeX Formula for Total Probability
To find the total probability of success, $P(\text{Success})$, you sum these four scenarios together. Here is the exact LaTeX syntax you can copy and use:
Code snippet
P(\text{Success}) = [P(A) \cdot P(B) \cdot (1 - P(C))] + [P(A) \cdot (1 - P(B)) \cdot P(C)] + [(1 - P(A)) \cdot P(B) \cdot P(C)] + [P(A) \cdot P(B) \cdot P(C)]
Rendered, it looks like this:
$$P(\text{Success}) = [P(A) \cdot P(B) \cdot (1 - P(C))] + [P(A) \cdot (1 - P(B)) \cdot P(C)] + [(1 - P(A)) \cdot P(B) \cdot P(C)] + [P(A) \cdot P(B) \cdot P(C)]$$
Alternatively, a slightly shorter algebraic shortcut for "at least 2 out of 3" simplifies the equation to:
Code snippet
P(\text{Success}) = P(A)P(B) + P(A)P(C) + P(B)P(C) - 2P(A)P(B)P(C)
$$P(\text{Success}) = P(A)P(B) + P(A)P(C) + P(B)P(C) - 2P(A)P(B)P(C)$$
4. Step-by-Step Calculation
Plugging your specific values ($P(A)=0.75$, $P(B)=0.54$, $P(C)=0.60$) into the formula:
A and B only: $0.75 \times 0.54 \times 0.40 = 0.1620$
A and C only: $0.75 \times 0.46 \times 0.60 = 0.2070$
B and C only: $0.25 \times 0.54 \times 0.60 = 0.0810$
All three: $0.75 \times 0.54 \times 0.60 = 0.2430$
$$\text{Total Success Probability} = 0.1620 + 0.2070 + 0.0810 + 0.2430 = \mathbf{0.693}$$
The exact mathematical probability that a player clears this specific layout is $69.3\%$.
