Space shuttle disaster. On January 28, 1986, the space shuttle Challenger exploded, killing all seven astronauts aboard. An investigation concluded that the explosion was caused by the failure of the O ring seal in the joint between the two lower segments of the right solid rocket booster. In a report made 1 year prior to the catastrophe, the National Aeronautics and Space Administration (NASA) claimed that the probability of such a failure was about 1/ 60,000, or about once in every 60,000 flights. But a risk-assessment study conducted for the Air Force at about the same time assessed the probability to be 1/35, or about once in every 35 missions. (Note: The shuttle had flown 24 successful missions prior to the disaster.) Given the events of January 28, 1986, which risk assessment—NASA's or the Air Force's—appears more appropriate?

In a report one year before the catastrophe, NASA claimed that the probability of space shuttle challenger failure was $\left(1/60,000\right)$about one in every 60,000 flights. At the same time,the air force study assessed the probability of beinga failure $\left(1/35\right)$or about one in every 35 missions since the shuttle had followed 24 successful missions before the disaster.

We choose geometric distribution since it is observable from the provided information.

The probability function of the geometric is defined as,

$$\begin{gathered} P\left(X=x\right)=\left\{q^x·p x=0,1,2,...0⩽p⩽1 \\ 0otherwise \\ \right. \end{gathered}$$

It follows a lack of memory. It means the event has not occurred before k. Let

Thus, Y is the additional time needed for the event's occurrence.

Then,

$\begin{array}{rcl}P\left(Y=x|X⩾k\right)& =& P\left(X=k\right)\\ & =& q^x·p\\ & & \end{array}$

The probability of NASA's assessment of the failure of the space shuttle is,

$\begin{array}{rcl}P\left(X=24\right)& =& {\left(\frac{5999}{60,000}\right)}^{24}.\left(\frac{1}{60,000}\right)\\ & =& 1.667\times {10}^{-5}\\ & & \end{array}$

The probability of air force assessment of the failure of the space shuttle is,

$\begin{array}{rcl}P\left(X=24\right)& =& q^{24}·p\\ & =& {\left(\frac{34}{35}\right)}^{24}\left(\frac{1}{35}\right)\\ & =& 1.4249\times {10}^{-2}\\ & & \end{array}$

Since air force risk assessed probability $1.4249\times {10}^{-2}$is much higher than NASA risk assessed probability$1.667\times {10}^{-5}$

Thus, it can conclude that the air force risk assessment report is more appropriate than the NASA report.