Question 1199668: Suppose that you have 10 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards with replacement. Round your answers to four decimal places.
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?
They are independent events
They are dependent events
Answer by math_tutor2020(3817)   (Show Source):
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Part (a)
G1 = first card is green
P(G1) = probability the first card is green
P(G1) = (10 green)/(10 green + 5 yellow)
P(G1) = (10 green)/(15 total)
P(G1) = 10/15
P(G1) = 2/3
P(G2 given G1) = probability event G2 happens given event G1 happened
P(G2 given G1) = (number of green left)/(number total)
P(G2 given G1) = (10 green)/(15 total)
P(G2 given G1) = 2/3
We arrive at the same fraction as before.
This is because the number of cards stays the same when replacement is done.
P(G1 and G2) = P(G1)*P(G2 given G1)
P(G1 and G2) = P(G1)*P(G2)
P(G1 and G2) = (2/3)*(2/3)
P(G1 and G2) = 4/9
P(G1 and G2) = 0.4444
Answer: 0.4444
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Part (b)
Y1 = the first card is yellow
P(Y1) = probability the first card is yellow
P(Y1) = 5/15 = 1/3
P(Y1 and Y2) = P(Y1)*P(Y2 given Y1)
P(Y1 and Y2) = (1/3)*(1/3)
P(Y1 and Y2) = 1/9
P(At least 1 green) = 1 - P(2 yellow)
P(At least 1 green) = 1 - P(Y1 and Y2)
P(At least 1 green) = 1 - 1/9
P(At least 1 green) = 9/9 - 1/9
P(At least 1 green) = 8/9
P(At least 1 green) = 0.8889
Answer: 0.8889
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Part (c)
As mentioned earlier in part (a),
P(G2 given G1) = 2/3 = 0.6667 approximately
Answer: 0.6667
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Part (d)
Answer: independent
Reason: Event G1 does not affect G2, or vice versa, because the card was put back (aka replacement).
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