0:00
Hello everyone.
00:02
So it is given that failure succour for a mechanical process according to poisson process.
00:08
So first let us define the function of poison process.
00:14
It is equal to probability of x equals to x is e raised to the power minus lambda, lambda raised to the power x divided by x factorial where lambda is our parameter.
00:31
Right so for part a kind of probability that two failures occur in one hour so probability of two failure in hour is equal to probability of one probability of one let us denote capital m by major and small m by minor so it is given that a failure can either be major or be minor.
01:16
Probability of two failure in an hour is probability of one major and one minor plus probability of two major and zero minor plus probability of zero major and two minor.
01:48
So these are the only possible cases.
01:51
So this will be equals to e raised to the power minus okay so lambda for major is equals to 1 .5 and lambda for minor is equal to 3 so using this e raised to the power minus 1 .5 1 .5 raised to the power 1 divided by 1 factorial plus multiplied by e raised to the power minus 3 3 raised to the power 1 divided by 1 factorial.
02:31
Plus e -raise to the power minus 1 .5, 1 .5 raised to the power 2 divided by 2 factorial multiplied by e -raise to the power minus 3, 3 raised to the power 0, divided by 0 factorial.
02:55
Plus, e -raise to the power minus 1 .5, 1 .5 raised to the power 0 divided by 0 factorial, multiplied by e raised to the power minus 3, 3 raised to the power 2 whole divided by 2 factorial.
03:12
So in solving this we get this is equals to 0 .1125.
03:18
Now let us come to part b of the question.
03:21
In part b, you have to find the probability that in half an hour one major failure occurs.
03:29
So we have probably first we will find the expected number of major failures in half an hour...