continued: what is the probability that both are defective? (The first one is not replaced before the second one is selected.)
a. 1/145
b. 2/145
c. 1/78
d. 2/87
e. 1/260
f. 1/130
g. none of these
2007-03-04 16:06:55 · 5 answers · asked by justanobservation-notajudgement 3 in Science & Mathematics ➔ Mathematics
(5/30) * (4/29) = 2/87
Answer is D.
2007-03-04 16:09:23 · answer #1 · answered by ........ 5 · 1⤊ 0⤋
P=Pdef1 AND Pdef2
AND -> *
OR -> +
P=(5/30)*(4/29)=2/87
First draw has a chance of getting 1 on the total of 5 defective (5/30).
At second draw the total set is now only 29 bec 1 is already widrawn. Assume that this 1 set is defective bec your equation states that 1st draw is defective. So at 2nd draw you only have 4 def (4/29). If equation involves OR meaning any one can happen then use + (addition).
2007-03-04 16:25:47 · answer #2 · answered by Smoochum 2 · 0⤊ 0⤋
evaluate those questions: a million. what's the possibility that a randomly picked television is fallacious? 2. If the 1st television is fallacious and you pick a 2nd television, what's the possibility that that television is fallacious? Now basically multiply those mutually.
2016-12-14 11:00:36 · answer #3 · answered by ? 4 · 0⤊ 0⤋
g
5/30= 1/6
2/30 =1/15
so multiply the two and you get
1/90
2007-03-04 16:16:45 · answer #4 · answered by loktao s 2 · 0⤊ 0⤋
owwww. it hurts
2007-03-04 16:09:22 · answer #5 · answered by kristina43 5 · 0⤊ 0⤋