Hi,
Consider this problem involving Binomial distribution and
probability:
The probability that a particular make of light bulb is
faulty is 0.2. The light bulbs are packed in boxes of 12. A buyer
accepts a consignment of 50 boxes, if, when he chooses two boxes at
random, he finds that they contain no more than one faulty light
bulb altogether. Find the probability that he will accept the
consignment.
The usual approach is given below:
1. Let X be the number of faulty bulbs in a box of 12.
Then X ~ B(12, 0.2).
2. For the two boxes picked, there are three possible cases:
(i) both boxes 1 & 2 come with zero
faulty bulbs,
(ii) box 1 has zero faulty bulbs and box 2
has 1 faulty bulb.
(iii) box 1 has 1 faulty bulb and box 2
has zero faulty bulb.
Use multiplication and addition principles
of probability.
Another shorter approach we may consider is:
1. Let X be the number of faulty bulbs in two boxes.
Then X ~ B(24, 0.2).
2. The required probability is simply P(X <= 1).
The shorter approach comes from the application of The
Discrete Convolution Formula Theorem. Using this theorem, we have
the following result which is not covered in H2 Statistics:
If X_i ~ B(n, p) and X_1, X_2, ..., X_k are independent
random variables, then
X_1 + X_2 + ... + X_k ~ B(kn, p).
In particular, we have used the theorem on the Poisson
(where the sum of Poisson variables is Poisson) and Normal
(where the sum of Normal variables is Normal) distributions.
Thanks.
Cheers,
Wen Shih