Answer to Question #166679 in Statistics and Probability for Bikram Chaudhary

Question

At the begining of each day, the three motors required for the operation of a machine are checked and tested. These motors then have probabilities of 7/8, 5/6 and 2/3 of continuously operating through out a working day. If the production can be maintained as long as two of three motors are operating, calculate the probability of the machine not failing on a particular day. Calculate also the probability of the machine failing for a particular day.

Answer 1

Solution

Let :

Motor not failing be denoted by F'

Motor failing be denoted by F

P(machine not failing)= P(at least 2 motors don't fail)

$=P(f'f'f') +P(f'f'f) +P(f'ff') +P (ff'f')$

$$=({7\over 8}*{5 \over 6} * {2 \over 3})+({7\over 8}*{5 \over 6} * {1 \over 3})+({7\over 8}*{1 \over 6} * {2 \over 3})$$

$+({1\over 8}*{5 \over 6} * {2 \over 3})$

$$={35 \over 72}+{35 \over 144} +{7 \over 72}+{5 \over 72}={43 \over 48}$$

$$\therefore P(machine \space running \space continuously) = {43 \over 48}$$

Probability of machine failing

$$P(machine \space fail) =1-P(machine \space not \space fail)$$

$$=1 - {43 \over 48}$$

$={5 \over 48}$