Question

A shuttle launch depends on three key devices that may fail independently of each other with probabilities 0.01, 0.02, and 0.02, respectively. If any of the key devices fails, the launch will be postponed. Compute the probability for the shuttle to be launched on time, according to its schedule.

Answer 1

Answer:

$0.950796$

Step-by-step explanation:

Given that a shuttle launch depends on three key devices that may fail independently of each other with probabilities 0.01, 0.02, and 0.02, respectively.

Required probability =  the probability for the shuttle to be launched on time

= Probability that all three do not fail

Since each key device is independent of the other

we have

prob that all three do not fail = $(1-0.01)(1-0.02)(1-0.02)\\=0.99*0.98*0.98\\=0.950796$