The probability of drawing a red ball on the first draw is the number of red balls (3) divided by the total number of balls (3+4+9=16). This is equal to 3/16.
Since the ball is replaced before the second draw, the probability of drawing a white ball on the second draw is also 4/16.
To find the probability of both events occurring, we multiply the probabilities. So the probability of drawing a red ball on the first draw and a white ball on the second draw is: (3/16) * (4/16) = 12/256.
Simplifying the fraction, the probability is 3/64.
A box contains 3 red balls, 4 white balls, and 9 black balls. Two balls are drawn at random from the box with replacement of the first before the second is drawn. What is the probability of getting a red (R) ball on the first draw and a white (W) ball on the second?
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