All that is left in a packet of candy are 7 reds, 4 greens, and 2 blues, What is the probability that a random drawing yields a
green followed by a red assuming that the first candy drawn is put back into the packet?
2 answers:
Multiply 7(reds) by 4 (greens) to get 2/13
Probability of drawing a green = 4/13
Probability of drawing a red = 2/13
probability of drawing a green and then a red is 4/13 * 2/13 = 6/169
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2x+3=5x
3=3x
x=1
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