SOLUTION: You choose a tile at random from a bag containing 2A's, 3B's, and 4C's. You replace the first tile in the bag and then choose again. Find each probability.
P(A and A)
P(A and B
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Probability-and-statistics
-> SOLUTION: You choose a tile at random from a bag containing 2A's, 3B's, and 4C's. You replace the first tile in the bag and then choose again. Find each probability.
P(A and A)
P(A and B
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Question 1141253: You choose a tile at random from a bag containing 2A's, 3B's, and 4C's. You replace the first tile in the bag and then choose again. Find each probability.
P(A and A)
P(A and B)
P(B and C) Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! There are 9 tiles in the bag
:
P(A and A) = (2/9) * (2/9) = 4/81
:
P(A and B) = (2/9) * (3/9) = 6/81 = 2/27
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P(B and C) = (3/9) * (4/9) = 12/81 = 4/27
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