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?
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.
To find the probability of getting a red ball on the first draw and a white ball on the second draw, we need to break the problem down into two steps.
Step 1: Probability of drawing a red ball on the first draw (R):
There are a total of 16 balls in the box, and 3 of them are red. So, the probability of drawing a red ball on the first draw is:
P(R) = Number of red balls / Total number of balls
= 3 / 16
= 3/16
Step 2: Probability of drawing a white ball on the second draw (W):
Since the first ball is replaced before the second draw, the total number of balls remains the same. Now, there are 4 white balls in the box, so the probability of drawing a white ball on the second draw is:
P(W) = Number of white balls / Total number of balls
= 4 / 16
= 1/4
Step 3: Determining the combined probability:
Since the two draws are independent events, we can find the combined probability by multiplying the individual probabilities together:
P(R and W) = P(R) * P(W)
= (3/16) * (1/4)
= 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.