a company places a rush order for wire of two thicknesses. consignments of each thickness are to be sent immediately when they are available. previous experience suggests that the probability is 0.8 that at least one of these consignments will arrive within 1 week. it is also estimated that , if the thinner wire arrives within 1 week, the probability is 0.4 the thicker wire will also arrive within a week. further, it is estimated that, if the thicker wire arrives within a week, the probability is 0.6 that the thinner wire will also arrive within 1 week.
Q: what is the probability that the thicker wire will arrive within 1 week?
Originally posted by Pen sean:a company places a rush order for wire of two thicknesses. consignments of each thickness are to be sent immediately when they are available. previous experience suggests that the probability is 0.8 that at least one of these consignments will arrive within 1 week. it is also estimated that , if the thinner wire arrives within 1 week, the probability is 0.4 the thicker wire will also arrive within a week. further, it is estimated that, if the thicker wire arrives within a week, the probability is 0.6 that the thinner wire will also arrive within 1 week.
Q: what is the probability that the thicker wire will arrive within 1 week?
Let A = thick arrives within 1 week
Let B = thin arrives within 1 week
Question asks for P(A).
P(A|B) = 0.4
P(A^B) / P(B) = 0.4
0.4P(B) = P(A^B) -(2)
P(B|A) = 0.6
P(B^A) / P(A) = 0.6
0.6P(A) = P(B^A) -(1)
(1)&(2) --- 0.4P(B) = 0.6P(A) -(3)
at least one arrives within 1 week =
A arrives not B, B arrives not A, A & B both arrives = 0.8 = P(AuB)
drawing venn diagram, both doesn't arrive = 0.2 = P(A'^B')
now find P(A).
P(B) = P(AuB) - P(A) + P(A^B)
P(B) = 0.8 - (2/3)P(B) + 0.4P(B)
P(B) = 12/19
P(A) = (0.4/0.6) P(B) = 8/19
dunno if correct. what is the answer.
yeah, i believe your answer is perfectly correct, from my answer key, it says it 0.6136, thank you, man...
P(A) is not enuff
P(A) + P(A +B) is the full answer.
P no wire come = 0.2
P thick wire alone = A x 0.4
P thin wire alone = B x0.6
P thick wire come, thin wire than come = A x 0.6
P thin wire, thick wire than come = B x 0.4
Originally posted by Pen sean:yeah, i believe your answer is perfectly correct, from my answer key, it says it 0.6136, thank you, man...
my ans is 8/19 = 0.421, how can be correct
Originally posted by Craappiboy:my ans is 8/19 = 0.421, how can be correct
Sorry about that, I mean P(B) =12/19 is 0.6136, so P(A) =8/19 is certainly right, the answer key also says it is .4211.
btw, man,do u have a method without using substitution? or this is the simplest way?
thanx alot :)
this should be the standard way of solving such qn. if without substiition, what method you gg to use?
Originally posted by skythewood:P no wire come = 0.2
P thick wire alone = A x 0.4
P thin wire alone = B x0.6
P thick wire come, thin wire than come = A x 0.6
P thin wire, thick wire than come = B x 0.4
by drawing venn diagram, P(A+B) is inside P(A)...