Final answer:
The probability of drawing the balls in the order B, R, B, R, and B is 4/35. There are 4 black balls and 3 red balls in the bag, and the balls are drawn without replacement.
Step-by-step explanation:
To find the probability of drawing the balls in the order B, R, B, R, and B, we need to consider the number of favorable outcomes and the total number of possible outcomes.
In this case, there are 4 black (B) balls and 3 red (R) balls in the bag. The probability of drawing a black ball first is 4/7. After drawing a black ball, there will be 3 black balls and 3 red balls left in the bag, so the probability of drawing a red ball next is 3/6. Following the same logic, the probability of drawing a black ball, then a red ball, then a black ball, and finally a red ball again is:
P(B, R, B, R) = (4/7) x (3/6) x (3/5) x (2/4) = 4/35
Since the balls are drawn without replacement, the probability of drawing a black ball for the last time is 3/3 = 1.
Therefore, the overall probability of drawing the balls in the given order is:
P(B, R, B, R, B) = (4/35) x (1) = 4/35