Final answer:
In this case, the probability of each event Ai is 1/24.
Step-by-step explanation:
To find the probability of a match occurring on the ith draw, we can use the concept of permutations. In this case, there are 4 balls numbered 1 through 4. We are selecting the balls one at a time without replacement.
Let's consider the event A1, which denotes a match on the first draw. In order for a match to occur on the first draw, we need to select the ball numbered 1 first. There is only one ball numbered 1, so the probability of a match on the first draw is 1/4.
Similarly, for the event A2, which denotes a match on the second draw, we need to select the ball numbered 2 second. After selecting the ball numbered 1 on the first draw, there are now 3 remaining balls to choose from. So the probability of a match on the second draw is 1/3.
Following the same logic, the probability of a match on the third draw (event A3) is 1/2, and the probability of a match on the fourth draw (event A4) is 1/1.
To find the overall probability of each event Ai, we multiply the probabilities of each individual draw. So for A1, A2, A3, and A4, the probabilities are:
- P(A1) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
- P(A2) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
- P(A3) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
- P(A4) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
Notice that the probabilities are the same for each event. We can simplify the expression by noting that 3! (3 factorial) is equal to 3 * 2 * 1. Similarly, 4! (4 factorial) is equal to 4 * 3 * 2 * 1.
So, P(Ai) = 3! / 4! = (3 * 2 * 1) / (4 * 3 * 2 * 1) = 1/24.
Therefore, the probability of each event Ai is 1/24.
Your question is incomplete, but most probably the full question was:
An urn contains four balls numbered 1 through 4.The balls are selected one at a time without replacement.A match occurs if the ball numberedm is the mth ballselected. Let the event Ai denote a match on the ith draw i = 1, 2, 3, 4.
Q: Show that P(Ai) = 3! /4! for each i.