Introduction

This document aims to calculate the probability of a safe landing for an airplane based on the functionality of its engines.

Problem Statement

Each of the four engines on an airplane functions correctly on a given flight with a probability of \(0.99\). The engines function independently of each other. The plane can make a safe landing if at least two of its engines are functioning correctly. We want to find the probability of a safe landing.

Solution

Let \(E\) be the event that an engine functions correctly, and let \(E'\) be the event that an engine malfunctions. The probabilities are:

\[P(E) = 0.99\] \[P(E') = 1 - P(E) = 0.01\]

Now, we’ll consider three cases:

  1. All four engines are functioning.
  2. Three engines are functioning.
  3. Two engines are functioning.

The probability of each case is calculated using the independence of the engine events.

Probability Calculations

All Four Engines Functioning

\[ P(\text{All Functioning}) = P(E \cap E \cap E \cap E) = 0.99^4 = 0.96059601 \]

Three Engines Functioning

\[ P(\text{Three Functioning}) = \binom{4}{3} \times P(E)^3 \times P(E') = \binom{4}{3} \times 0.99^3 \times 0.01 = 0.03940399 \]

Two Engines Functioning

\[ P(\text{Two Functioning}) = \binom{4}{2} \times P(E)^2 \times P(E')^2 = \binom{4}{2} \times 0.99^2 \times 0.01^2 = 0.00000016 \]

Total Probability of Safe Landing

Adding these probabilities gives the total probability of a safe landing:

\[ P(\text{Safe Landing}) = P(\text{All Functioning}) + P(\text{Three Functioning}) + P(\text{Two Functioning}) = 0.96059601 + 0.03940399 + 0.00000016 = 0.9996 \]

The probability of a safe landing for the airplane is \(P(\text{Safe Landing}) = 0.9996\).

Probability Calculations using R

p_fail <- 0.01
p_works <- 1 - p_fail
all_four_fail <- choose(4, 4) * p_fail^4
three_fail <- choose(4, 3) * p_fail^3 * p_works
p_safe_landing <- 1 - (all_four_fail + three_fail)
cat("Probability of a safe landing:", p_safe_landing)
## Probability of a safe landing: 0.999996