3.21 b)

[Andropb]

Hi,
I think I might have a possible solution for the problem 3.21 b).

Let "a" be the number of breakdowns of A and "b" the number of breakdowns of B. Now, let's define the event:

R:{the pairs (a,b) | a + b < 4}

1.- Using intuition:
The pairs (a,b) that meet this condition are: R={(0,0);(0,1);(0,2);(0,3);(1,1);(1,2);(2,1);(1,0);(2,0);(3,0)}.
Then:
P(R)= P(a=0 ∩ b=0) + P(a=0 ∩ b=1) + ... + P(a=3 ∩ b=0)
Using that A and B are independent events:
P(R)= P(a=0)*P(b=0) + P(a=1)*P(b=0) + ... + P(a=3)*P(b=0)

2.- Using the Theorem of total probability:
Let's first notice that we can rewrite the condition:
a + b < 4 ➡ b < 4 - a
So we can rewrite the definition of the event R as follows: R:{(a,b) | b< 4 - a}
Then:
P(R) = P(b <4 - a)= P(b <4 | a=0)P(a=0) + P(b < 3 | a=1)P(a=1) + P(b <2 | a=0)P(a=2)P(a=2) + P(b < 1 | a=0)P(a=3)
Notice that:
° P(b <4 | a=0) = P( (b=0)∪(b=1)∪(b=2)∪(b=3) | (a=0)) = P((b=0) | (a=0)) + P((b=1) | (a=0)) + P((b=2) | (a=0)) +
P((b=3) | (a=0)) =
P(b=0) + P(b=1) + P(b=2) + P(b=3) ; (Given that A and B are independent events)

Similarly with the other terms. Then:
P(b <4 - a)= [P(b=0) + P(b=1) + P(b=2) + P(b=3)]P(a=0) + [P(b=0) + P(b=1) + P(b=2)]P(a=1) + [P(b=0) + P(b=1)]P(a=2) +
[P(b=0)]P(a=3)

The same idea applies for the part b).
I also think that the use of conditions can be helpful to solve the rest of the items, i.e. c). condition: a>b; d) condition: b=2a ; etc.
I hope this results useful for you.