Past experience in the production of a certain component has shown that the prop
ID: 3246544 • Letter: P
Question
Past experience in the production of a certain component has shown that the proportion of defectives is 0.03. Components leave the factory in boxes of 500. What is the probability that (1) a box contain 3 or more defectives. (2) two successive boxes contain 6 or more defectives between them? Past experience in the production of a certain component has shown that the proportion of defectives is 0.03. Components leave the factory in boxes of 500. What is the probability that (1) a box contain 3 or more defectives. (2) two successive boxes contain 6 or more defectives between them? Past experience in the production of a certain component has shown that the proportion of defectives is 0.03. Components leave the factory in boxes of 500. What is the probability that (1) a box contain 3 or more defectives. (2) two successive boxes contain 6 or more defectives between them?Explanation / Answer
Solution
Let X = number of defective components in a box of 500.
Then, X ~ B(500, p), where p = probability of finding a defective component in a box of 500.
Past experience in the production of a certain component has shown that the proportion of defectives is 0.03 =>
P = 0.03.
Back-up Theory
If X ~ B(n, p). p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n …………..(1)
For various combinations of x, n and p, these probabilities are readily given by Excel Function, which is what will be used here.
Part (1)
Probability that a box contains 3 or more defectives = P(X 3) = 1 – P(X 2)
= 1 – 3.30171E-05 = 0.99997 ANSWER
Part (2)
Probability that two successive boxes contain 6 or more defectives between them
= 1 - Probability that two successive boxes contain 5 or less defectives between them.
Now, if Y = number of defective components in the first box of 500 and Z = number of defective components in the second box of 500, then, what we want is = 1 – P{(Y + Z) 5} = 1 – [P(Y = y) x P{Z (5 – y)]
Various possibilities and the corresponding probabilities are tabulated below:
y
5 - y
P(y)
P{Z (5 – y)}
P(y) x P{Z (5 – y)}
0
5
2.4315E-07
0.002510579
6.10437E-10
1
4
3.76E-06
0.000754389
2.8365E-09
2
3
2.9014E-05
0.000181975
5.27982E-09
3
2
0.00014896
3.30171E-05
4.91816E-09
4
1
0.00057241
4.00314E-06
2.29145E-09
5
0
0.00175619
2.43146E-07
4.27011E-10
Total
1.63634E-08
1 - Total
0.999999984
So, required probability = 0.999999984 ANSWER
NOTE: All these probabilities can also be approximated by Poisson or Normal distributions.
y
5 - y
P(y)
P{Z (5 – y)}
P(y) x P{Z (5 – y)}
0
5
2.4315E-07
0.002510579
6.10437E-10
1
4
3.76E-06
0.000754389
2.8365E-09
2
3
2.9014E-05
0.000181975
5.27982E-09
3
2
0.00014896
3.30171E-05
4.91816E-09
4
1
0.00057241
4.00314E-06
2.29145E-09
5
0
0.00175619
2.43146E-07
4.27011E-10
Total
1.63634E-08
1 - Total
0.999999984