SOLUTION: #1a A bag contains 4 yellow balls and 6 green balls. Two balls are drawn without replacement. What is the probability that the second ball is yellow given that the first ball is

Algebra ->  Probability-and-statistics -> SOLUTION: #1a A bag contains 4 yellow balls and 6 green balls. Two balls are drawn without replacement. What is the probability that the second ball is yellow given that the first ball is       Log On


   



Question 860250: #1a
A bag contains 4 yellow balls and 6 green balls. Two balls are drawn without replacement. What is the probability that the second ball is yellow given that the first ball is green? 8/10?
#2b
A bag contains 6 red jelly beans and 4 black jelly beans. Two jelly beans are drawn at random from the bag one after another.
(a)If the first jell bean is replaced in the bag before the second one is drawn, find the probability that both jelly beans are red. 10/8?
(b)If the first jelly is NOT replaced in the bag before the second one is drawn, find the probability that both jelly beans are red. 8/10?
#3c A mathematics class consists of 16 engineering majors, 12 science majors, and 4 liberal art majors.
(a) What is the probability that a student selected at random will be a science or engineering major? 30/32?
(b) Five of the engineering students, 6 of the science majors, and 2 of the liberal arts majors are female. What is the probability that a student selected at random is an engineering major or a female? 11/13?
#4d A sample of 5 fuses is drawn from a lot containing 12 good fuses and 3 defective fuses. Find the probability that:
(a) all of the fuses are good
(b) at least 1 fuse is defective
By thursday the 10th. Thank you

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
You do understand the concept of TOTAL, that's a start
1) 10, 2) 10, 3)32
Probability is Particular%2FTOTAL, P(1st event and 2nd event)= P%281st+event%29%2A+P%282nd+event%29
1)w/o replacement P(2nd yellow given 1st green) = %286%2F10%29%284%2F9%29
2) a)replacement %286%2F10%29%285%2F10%29, b) %286%2F10%29%285%2F9%29w/o replacement
3) a)0r12%2F32+%2B+16%2F32 b)++16%2F32+%2B+13%2F32+-+5%2F32P(A or B) = P(A) + P(B) - P(A and B)
4) p(def) = 3/12 = .25, draw 5 a) .75%5E5 b)+1+-+.75%5E5