Questions: Two balls are drawn in succession out of a box containing 3 red and 5 white balls. Find the probability that the second ball was red, given that the first ball was: (A) replaced before the second draw. (B) not replaced before the second draw. (A) Find the probability that the second ball drawn was red, given that the first ball drawn was replaced before the second draw. (Type a decimal or a fraction.)

Solution Steps

When the first ball drawn is replaced, the total number of balls remains the same for the second draw. The probability of drawing a red ball in any draw is given by:

\[ P(\text{Red}) = \frac{\text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{3}{8} \]

Thus, the probability that the second ball drawn is red, given that the first ball was replaced, is:

\[ P(\text{Second Red | First Replaced}) = \frac{3}{8} = 0.375 \]

When the first ball drawn is not replaced, we need to consider two scenarios: the first ball drawn is red or white.

1. Case 1: First ball is red

   • If the first ball is red, there are now 2 red and 5 white balls left. The probability of this scenario is:

   \[ P(\text{First Red}) = \frac{3}{8} \]

   • The probability of drawing a red ball second, given the first was red, is:

   \[ P(\text{Second Red | First Red}) = \frac{2}{7} \]

   • Therefore, the contribution to the total probability from this case is:

   \[ P(\text{First Red}) \cdot P(\text{Second Red | First Red}) = \frac{3}{8} \cdot \frac{2}{7} = \frac{6}{56} \]

2. Case 2: First ball is white

   • If the first ball is white, there are still 3 red and now 4 white balls left. The probability of this scenario is:

   \[ P(\text{First White}) = \frac{5}{8} \]

   • The probability of drawing a red ball second, given the first was white, is:

   \[ P(\text{Second Red | First White}) = \frac{3}{7} \]

   • Therefore, the contribution to the total probability from this case is:

   \[ P(\text{First White}) \cdot P(\text{Second Red | First White}) = \frac{5}{8} \cdot \frac{3}{7} = \frac{15}{56} \]

Now, we can combine both cases to find the total probability that the second ball drawn is red without replacement:

\[ P(\text{Second Red | First Not Replaced}) = \frac{6}{56} + \frac{15}{56} = \frac{21}{56} = \frac{3}{8} = 0.375 \]

Final Answer