There are 200 machines in a factory, and they work independently with probabilit

Question

There are 200 machines in a factory, and they work independently with probability 0.002 of being broken within any given day. Estimate the smallest natural number n such that the probability that at least n machines will break on a given day is less than 0: 01. Telegraph passes information by using dots and dashes. When sending a signal "dot", the probability of receiving a "dash is 2/5. When sending a "dash", the probability of receiving a "dot" is 1/3. Among all signals received, the ratio of dots and dashes is 5: 3. Find (a) The probability that the sent signal was a "dot" if a "dot" was received: (b) The probability that the sent signal was a "dash" if a "dash" was received.

Explanation / Answer

(1)

Modelling this as a binomial distribution with parameters:

n = 200, p = 0.002

Let X denote the RV for the number of machines breaking down on a particular day.

So,

P(X >= n) = 1 - P(X < n)

Now,

P(X < n) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + ... + P(X=n-1)

So,

P(X >= n) = 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) + ... + P(X=n-1) ]

Generally speaking,

P(X=r) = 200Cr*(0.002)r*(0.998)200-r

We need to solve the inequality:

P(X >= n) < 0.01

OR

1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) + ... + P(X=n-1) ] < 0.01

By hit and trial method, putting n = 2, we get:

P(X >= n) = 1-0.93 = 0.07, which is not smaller than 0.01

Next, putting n = 3, we get:

P(X >= n) = 1-0.992 = 0.008, which is less than 0.01

So the correct answer is n = 3

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