SOLUTION: Suppose that you have 7 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement. G1 = the first card drawn is green G2 = t

Question 1202922: Suppose that you have 7 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement.
G1 = the first card drawn is green
G2 = the second card drawn is green
a. P(G1 and G2) =
b. P(At least 1 green) =
c. P(G2|G1) =

d. Are G1 and G2 independent
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
p(g1) = 7/12
p(g2) = 6/11

p(g1 and g2) = p(g1) * p(g2) = 7/12 * 6/11 = 42/132 = 7/22.

p(at least 1 greem) = 1 minus p(0 green) = 5/12 * 4/11 = 20/132 = 5/33.

p(g2 giiven g1) = p(g1 and g2) / p(g1) = (7/22) / (7/12) = 7/22 * 12/7 = 12/22 = 6/11.

g1 and g2 are not independent because p(g2) is dependent on p(g1).
in this particular case it's because g1 is picked without replacement which changes the probability for g2.

if g1 were picked with replacement, then the probability for g2 would be the same as the probability for g1, i.e. 7/12.

note that p(g1) is the probability for getting green on the first draw and p(g2) is the probbility for geting green on the second draw.

the fact that any color is drawn on the first draw changes the probability for any color on the second draw because the total number of possibilities is now one less.
if you drew green on the first draw, then p(g2) becomes 6/11 becasuse there is one less total colors and 1 less green in the pot.
if you drew yellow on the first draw, then p(g2) becomes 7/11 because there is one less total colors even though the number of green in the pot remains the same.
when you draw with replacement, the original probgabilities remain intact and then p(g1 and g2) becomes 7/12 * 7/12.
in this case, the events are independent of each other.

if the events are independent of each other than p(g2 given g1) would be equal to p(g1)

example:
indepdnednt events, then p(g2) = p(g1) = 7/12 because replacement resets the pot to what was there in the beginning.
in that case, p(g2 given g1) would be equal to p(g1 and g2) / p(g1) which would be equal to (7/12 * 7/12) / 7/12) = 7/12 which is the same as p(g1).

here's a reference:

https://www.statisticshowto.com/probability-and-statistics/dependent-events-independent/

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