Probability

P(A[sub:rjef0grk]1[/sub:rjef0grk]) = .12
P(A[sub:rjef0grk]2[/sub:rjef0grk]) = .07
P(A[sub:rjef0grk]3[/sub:rjef0grk]) = .05

P(A[sub:rjef0grk]1[/sub:rjef0grk] U A[sub:rjef0grk]2[/sub:rjef0grk]) = .13
P(A[sub:rjef0grk]1[/sub:rjef0grk] U A[sub:rjef0grk]3[/sub:rjef0grk]) = .14
P(A[sub:rjef0grk]2[/sub:rjef0grk] U A[sub:rjef0grk]3[/sub:rjef0grk]) = .10

P(A[sub:rjef0grk]1[/sub:rjef0grk] AND A[sub:rjef0grk]2[/sub:rjef0grk] AND A[sub:rjef0grk]3[/sub:rjef0grk]) = .01 (Don't know how to type the and symbol, which is an upside down U)

(a) What is the probability that the system has both type 1 and type 2 defects?
P(A[sub:rjef0grk]1[/sub:rjef0grk] AND A[sub:rjef0grk]2[/sub:rjef0grk]) = P(A[sub:rjef0grk]1[/sub:rjef0grk]) + P(A[sub:rjef0grk]2[/sub:rjef0grk]) - P(A[sub:rjef0grk]1[/sub:rjef0grk] U A[sub:rjef0grk]2[/sub:rjef0grk]) = .12 + .07 - .13 = .06


(b) What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?
P(A[sub:rjef0grk]1[/sub:rjef0grk] AND A[sub:rjef0grk]2[/sub:rjef0grk] AND A'[sub:rjef0grk]3[/sub:rjef0grk]) = P(A[sub:rjef0grk]1[/sub:rjef0grk] AND A[sub:rjef0grk]2[/sub:rjef0grk]) - P(A[sub:rjef0grk]1[/sub:rjef0grk] AND A[sub:rjef0grk]2[/sub:rjef0grk] AND A[sub:rjef0grk]3[/sub:rjef0grk]) = .06 - .01 = .05

(c) What is the probability that the system has at most two of these defects?
???

Do the first two look right? Also, I'm stuck on part C, if someone could help me get back on track, I'd appreciate it.

Yes, you have made a good start.
Now you need probability of "not all three".

pka said:

Yes, you have made a good start.
Now you need probability of "not all three".

Click to expand...

(P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] AND P(A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh]) + P(A[sub:uw5oo0lh]3[/sub:uw5oo0lh] - (P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] U A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = .12 +.05 - .14 = .03
(P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh] AND P(A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh]) + P(A[sub:uw5oo0lh]3[/sub:uw5oo0lh] - (P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh] U A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = .07+ .05 - .10 = .02


P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] U A[sub:uw5oo0lh]2[/sub:uw5oo0lh] U A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh]) + P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh]) + P(A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) - (P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] AND A[sub:uw5oo0lh]2[/sub:uw5oo0lh]) - (P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] AND A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) - (P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh] AND A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) + (P(A[sub:uw5oo0lh]1[/sub:uw5oo0lh] AND P(A[sub:uw5oo0lh]2[/sub:uw5oo0lh] AND A[sub:uw5oo0lh]3[/sub:uw5oo0lh]) = .12 +.07 + .05 - .06 - .03 - .02 + .01 = .14

So the probability that the system has at most 2 defects is .14

Does this look right? If so, is there a quicker way to do it?

thanks

Hello, Hansel13!

Part (c) is easier than you think . . .

\(\displaystyle \begin{array}{c}P(A_1) \:=\: 0.12 \\ P(A_2) \:=\: 0.07 \\ P(A_3) \:=\: 0.05 \end{array} \qquad \begin{array}{c} P(A_1 \cup A_2) \:=\: 0.13 \\ P(A_1 \cup A_3) \:=\: 0.14 \\ P(A_2 \cup A_3) \:=\: 0.10 \end{array} \qquad \begin{array}{c} P(A_1 \cap A_2 \cap A_3) \:=\: 0.01 \end{array}\)

(c) What is the probability that the system has at most two of these defects?

Click to expand...

\(\displaystyle \text{The opposite of "at most two defects" is "all three defects."}\)


\(\displaystyle \text{We are told that: }\(A_1 \;\cap\; A_2\; \cap\; A_3) \:=\:0.01 \quad\hdots \;\;\text{the probability of all three defects.}\)


\(\displaystyle \text{Therefore: }\(\text{at most two}) \:=\(not\text{ all three}) \:=\:1 - 0.01 \:=\:0.99\)