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 ball on the first draw and a white ball on the second
The probability of drawing a red ball on the first draw is 3/16 (since there are 3 red balls out of a total of 16 balls).
Since the first ball is replaced before the second draw, the probability of drawing a white ball on the second draw is still 4/16.
To find the probability of both events happening, we multiply the probabilities together: (3/16) * (4/16) = 12/256.
Simplifying, we get 3/64.
Therefore, the probability of drawing a red ball on the first draw and a white ball on the second draw is 3/64.
To find the probability of getting a red ball on the first draw and a white ball on the second, we can use the concept of independent events, as the ball is replaced before the second draw.
Step 1: Determine the probability of drawing a red ball on the first draw.
Out of a total of 16 balls in the box, there are 3 red balls. Hence, the probability of drawing a red ball on the first draw is:
P(Red on first draw) = 3/16.
Step 2: Determine the probability of drawing a white ball on the second draw.
As the first ball is replaced before the second draw, the probability of drawing a white ball on the second draw remains the same.
Out of a total of 16 balls in the box, there are 4 white balls. Hence, the probability of drawing a white ball on the second draw is also:
P(White on second draw) = 4/16 = 1/4.
Step 3: Multiply the probabilities from steps 1 and 2.
To find the combined probability of both events occurring, we multiply the probabilities together:
P(Red on first draw and White on second draw) = P(Red on first draw) * P(White on second draw)
= (3/16) * (1/4)
= 3/64.
Therefore, the probability of getting a red ball on the first draw and a white ball on the second draw is 3/64.