Question

What is the probability that, for a random viewing of circuit boards 1 and 2, that the total number of defects observed will be at least 6

Answer 1

0.3125

Explanation:

From the full question in your book, the probability can be computed as:

`P(X+Y=6) = P \{X=1,Y=5\} \cup \{X=2,Y=4\} \cup \{X=3,Y=3\} \cup \{X=4,Y=2\} \cup \{X=5,Y=1\}`

`= P(X=1,Y=5) + P(X=2,Y=4)+P(X=3,Y=3)+P(X=4,Y=2)+P(X=5,Y=1)`

since all events are enclosed in brackets, they are mutually exclusives.

Then:

`= P(X=1)P(Y=5)+P(X=2)P(Y=4)+P(X=3)P(Y=3)+P(X=4)P(Y=2)+P(X=5)P(Y=1)`

as X and Y are independent;

=
`(0.25*0.25 + 0.25*0.25 + 0.25*0.25+0.25*0.25 + 0.25*0.25)`

= 0.3125

Therefore;

P(X+Y= 6) = 0.3125