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The concept of O-ring failure probability in this context is crucial for understanding the reliability of space shuttle launches. In the space shuttle Challenger launch, each of the six O-rings used in the shuttle's solid rocket boosters had a probability of failure. This probability was calculated based on the temperature at the time of launch using a specific formula: \[ p(t) = \frac{e^{a+b \cdot t}}{1+e^{a+b-t}} \] Where: - \( a = 5.085 \) and \( b = -0.1156 \) are constants, - \( t \) is the temperature in degrees Fahrenheit.For Challenger's launch at \( 31^\circ F \), the failure probability was computed as \( p(31) = 0.8178 \). This figure indicates a high likelihood that each O-ring might fail. It's important to recognize here that a higher probability of failure puts the shuttle at risk during launch.

The binomial probability model is perfect for scenarios where there are repeated trials with two possible outcomes in each trial. In this model, the outcomes are often referred to as "success" or "failure." In our context, the 'success' is actually the failure of an O-ring.For the Challenger launch case:- **Number of trials (n)**: Total number of O-rings on the shuttle, which is 6. - **Probability of "success" (failure) in one trial (p)**: Calculated as \( p(31) = 0.8178 \).Hence, the number of failing O-rings, represented by \( X \), follows a Binomial distribution \( B(n=6, p=0.8178) \). This model helps determine the probability of different numbers of O-ring failures, useful for calculating the risk and preparing safety measures accordingly.

Calculating the probability that at least one O-ring fails involves a few steps using binomial probability rules. 1. **Complement Rule**: We simplify calculations by using \[ P(X \geq 1) = 1 - P(X = 0) \]. This rule helps us avoid computing multiple scenarios and only focus on determining the likelihood of zero failures first.2. **Probabilities for Zero Failures**: - Formula for zero failures: \[ P(X = 0) = \binom{6}{0} \cdot (0.8178)^0 \cdot (1 - 0.8178)^6 = (0.1822)^6 \] - This is the chance all O-rings work perfectly.3. **Final Probability Calculation**: - Substitute into the complement rule: \[ P(X \geq 1) = 1 - (0.1822)^6 \] - Simplify, leading to \[ P(X \geq 1) \approx 0.999812 \]The final result here of 0.999812 suggests a very high likelihood, approximately 99.98%, that at least one O-ring will fail. This high risk underlines the need for strict safety and reliability protocols in the design and launch procedures of space shuttles.