Answer to Question #313859 in Statistics and Probability for TIN

Question

In an automobile plant two tasks are performed by robots. The first entails welding two joints; the second, tightening three bolts. Let X denote the number of defective welds and Y the number of improperly tightened bolts produce per car.

Table 1:

x/y 0 1 2 3

0 .840 .030 .020 .010

1 .060 .010 .008 .002

2 .010 .005 .004 .001

Use table 1 to find each of these probabilities,

a. The probability that exactly two defective welds and one improperly tightened bolt will be produced by the robots.

b. The probability that at least one defective weld and at least one improperly tightened bolt will be produced.

c. The probability that at most one defective weld will be produced.

Answer 1

Solution

(a) $P(X=2,Y=1)$

$=f_{xy}(2,1)=0.005$

(b) $P(X\ge1~~ and~~Y\ge1)$

$=[P(1,1)+P(1,2)+P(1,3)+P(2,1)+P(2,2)+P(2,3)]$

$=0.01+0.008+0.002+0.005+0.004+0.001$

$=0.030$

(c) $P(X\le1 ~~and~~Y)$

$=[P(0,0)+P(0,1)+P(0,2)+P(0,3)+P(1,0)+P(1,1)+P(1,2)+P(1,3)]$

$=0.840+0.030+0.020+0.010+0.060+0.010+0.008+0.002$

$=0.980$