Answer:
a) 131/450
b) 1233/1276
Step-by-step explanation:
P(bad) = P(1st batch)*P(bad 1st batch ) + P(2nd batch )*P(bad 2nd batch) + P(3rd batch )*P(bad 3rd batch)
p(bad) =(60/360)*(1/3) + (120/360)*(1/4 ) + (180/360)*(1/5)
= 43/180
And that of P(good )
= 1 - 43/180
= 137/180
a)
P(defective) = P(bad)*P(defective /bad) + P(good)*P(defective /good)
= (43/180)*(9/10) + (137/180)*(1/10)
= 131/450
b)
P(Bc I Dc ) = P(good)*P(not defective |good) / P(not defective)
= (137/180)*(1 - 1/10) / (1 - 131/450)
= 1233/1276
Answer:
it can be. It just depends on what you are interested in. If you are interested in a scientific field then it is an amazing way to use your time. But if you are bot interested in science, it is a waste personally.
Answer:
The probability that a ship that is declared defecive is sound is 0.375
Step-by-step explanation:
Let P(A|B) denote the conditional probability of A given B. We will make use of the equation
P(A|B) = P(A) × P(B|A) / P(B)
We have the probabilities:
- P(Declared Defective (detected) | Defective) = 0.95
- P(not Detected | Defective) = 1-0.95=0.05
- P(Declared Sound | Sound) = 0.97
- P(Declared Defective |Sound) = 1-0.97=0.03
We can calculate:
P(Declared Defective)= P(Detected | Defective)×P(Defective) + P(Declared Defective |Sound) ×P(Sound) = 0.95×0.05 + 0.03×0.95=0.076
P(S | Declared Defective) =
(P(Sound) × P(Declared Defective | Sound)) / P(Declared Defective)
=0.95×0.03 /0.076 =0.375
40% of 180 is 72
1/4 of 320 is 80
therefore 1/4 of 320 is bigger
The second choice should be correct
