Probability question: Bag A contains 6 white beads and 3 black beads.Bag B contains 6 white beads and 4 black?

Question

beads. One bead is chosen at random from each bag. Find the probability that:

-both beads are back

-at least one of the two beads is white

Then, the beads are not replaced. A second bead is chosen at random from each bag. find the probablity that

-all four beads are white
-the beads are not all the same color.


PLS HELP...have been cracking my brain doing this.

Thanx a lot!!!!

Answer 1

p( both black) = 3/9 * 4/10 = 12/90 = 4/30 = 2/15

p(at least 1 white) = 1 - p( none white) = 1 - 2/15 = 13/15
p(none white) = p( both black)

p( 4 white) = 6/9 * 5/8 * 6/10 * 5/9

p(beads not same color) = 1 - (p(all white) + p(all black) )
p(all black) = 3/9 * 2/8 * 4/10 * 3/9
multiply out p(4 white) and p( all black) to get your answer.

Answer 2

P(both black)=(3/9)*(4/10)=2/15
P(at least one is black)=1-P(both white)=1-(6/9)(6/10)=9/15
P(all are white)=(6/9)(5/8)(6/10)(5/9)
P(all not the same color)=1-P(all white)-P(all black)