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Knowing that 200 undersea cables break every year globally, estimate the probability that 3 cables break in the baltic sea on the same day.

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To estimate the probability that 3 undersea cables break in the Baltic Sea on the same day, we'll begin by using known global data and making reasonable approximations.

**1. Average Daily Cable Breaks Globally:**

- There are approximately 200 undersea cable breaks globally per year.
- Number of days in a year ≈ 365.
- Average breaks per day globally: \( \frac{200 \text{ breaks}}{365 \text{ days}} \approx 0.548 \text{ breaks/day} \).

**2. Estimating Cable Breaks in the Baltic Sea:**

- To find how many breaks occur in the Baltic Sea, we need to estimate the proportion of cables in the Baltic Sea compared to the global total.
- Let's assume the Baltic Sea contains about 1% of the world's undersea cables (this is a rough estimate for the purpose of calculation).
- Therefore, expected breaks per year in the Baltic Sea: \( 1\% \times 200 = 2 \text{ breaks/year} \).
- Average breaks per day in the Baltic Sea: \( \frac{2 \text{ breaks}}{365 \text{ days}} \approx 0.00548 \text{ breaks/day} \).

**3. Using the Poisson Distribution:**

Because cable breaks are rare and occur independently, we can model the number of cable breaks per day as a Poisson distribution with rate \( \lambda = 0.00548 \).

The probability of observing \( k \) events in a Poisson distribution is:
\[ P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!} \]

We need to find \( P(3; 0.00548) \).

**4. Calculating the Probability:**

- \( \lambda = 0.00548 \).
- \( k = 3 \).
- \( e^{-\lambda} \approx e^{-0.00548} \approx 1 - 0.00548 = 0.99452 \) (since \( \lambda \) is very small).
- \( \lambda^3 = (0.00548)^3 \approx 164.566 \times 10^{-9} \).
- \( 3! = 6 \).

Now compute:
\[ P(3; 0.00548) = \frac{0.99452 \times 164.566 \times 10^{-9}}{6} \]

Simplifying:
\[ P(3; 0.00548) = \frac{163.664178 \times 10^{-9}}{6} \approx 27.277363 \times 10^{-9} \]

\[ P(3; 0.00548) \approx 0.0000000273 \]

**5. Interpreting the Result:**

- The probability is approximately \( 2.73 \times 10^{-8} \).
- This means there's about a 1 in 36 million chance (since \( \frac{1}{2.73 \times 10^{-8}} \approx 36,630,037 \)).

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**Answer: Approximately a 1 in 36 million chance—a probability of about 2.7 × 10^(–8) for three breaks in one day.

Tue, 19 Nov 2024 00:38 GMT