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There are 6 cards, 2 with stars on them and 4 with circles on them. Two cards are chosen at random.

All of the following probabilities represent dependent events in this scenario except:

A) picking a star, keeping it, and then picking another star.
B) picking a star, replacing it, and then picking a circle.
C) picking a circle, not replacing it, and then picking a star.
D) picking a circle, keeping it, and then picking another circle.


PLEASE CAN U HLP ME IN THIS ONE REALLY FAST??
Guest Mar 4, 2012
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TIARRA:

There are 6 cards, 2 with stars on them and 4 with circles on them. Two cards are chosen at random.

All of the following probabilities represent dependent events in this scenario except:

A) picking a star, keeping it, and then picking another star.
B) picking a star, replacing it, and then picking a circle.
C) picking a circle, not replacing it, and then picking a star.
D) picking a circle, keeping it, and then picking another circle.


PLEASE CAN U HLP ME IN THIS ONE REALLY FAST??



N=6
Ps=2/6 (2 star cards, total 6 cards)
Pc=4/6 (4 circle cards, total 6)

A)
probability for picking a star: Ps
now there is only 1 star card left in total of 5cards: Pss = 1/5 (depends on the first pick)
probability for picking another star: Pss
Total probability: Ps * Pss = 2/6 * 1/5 = 2/30 = 1/15

B)
probability for picking a star: Ps
probability for picking a circle: Pc (depends not on the first pick)
Total probability: Ps * Pc = 2/6 * 4/6 = 8/36 = 2/9

C)
probability for picking a circle: Pc
probability for picking a star: Pcs = 2 / 5
Total probability: Pc * Pcs = 4/6 * 2/5 = 8/30 = 4/15

D)
probability for picking a circle: Pc
probability for picking a circle: Pcc = 3/5 (depends on first pick)
Total probability: Pc * Pcc = 4/6 * 3/5 = 12/30 = 6 / 15 = 2/5
admin  Mar 4, 2012

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