Question

4. in a bank there are 15 coin vending machines: out of 15 machines 5 are defective 3 machines are chosen at random in the bank find the probability that at least one of these is defective?

Answer 1

There are 15 coin vending machines. Out of 15, there are 5 defective machines. 3 machines out of 15 are selected. Let n be the number of machines selected.

n=3

Let p be the probability of defective machine.

p = Number of defective machines/ Total number of machines

p = 5/15 = 1/3

So we have n=3 and p=1/3. Also we know probability of getting defective machine is fixed. So x = Getting defective machine follows Binomial distribution with n=3 and p=1/3

The probability function of x is given as

P(X=x) = $nCx p^{x}(1-p)^{n-x}$

where n= sample size, p=probability of success

Here we have to find probability that atleast one of the 3 selected is defective

It means minimum 1 is defective so there can 1 or 2 or all 3 defectives

P(X >= 1) = P(X=1) + P(X=2) + P(X=3)

= 1 - P(X=0)

Now P(X=0) = $3C0 (1/3)^{0} (1 - 1/3)^{3-0}$

= 1 * 1* (2/3)^3

= 0.296

So P(X >=1) = 1- 0.296

= 0.704

So the probability that getting atleast one defective out of 3 selected is 0.704