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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,

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

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,

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

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

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

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

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

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

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