The value of probability P(A) and P(B) and P(C) is 2/5 and 9/10 and 1/6.
According to the statement
we have given that the an urn contains five balls labeled 1 through 5. And Three balls are removed from the urn one by one randomly.
And we have to find that the probability of the given conditions.
So, For this purpose, we know that the
On the first given condition
A. Since there is no information on the first draw, this probability simply equals:
P(A)=2/5
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate P(A) as:
P(A) = 2/5*1/4+3/5*2/4
P(A) = 8/20
P(A) = 2/5.
And on the second given condition:
B. The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
P(B) = 1−3/5*2/4*1/3
P(B) = 1−1/10
P(B) = 9/10
And the third given condition is :
C. There are (5/3)=10 ways to choose the three numbers, and only one correct way to draw them.
We thus find:
P(C) = (5/3)/5*4*3
P(C) = 10/60
P(C) = 1/6
So, The value of probability P(A) and P(B) and P(C) is 2/5 and 9/10 and 1/6.
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Question:
An urn contains five balls labeled 1 through 5. Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let A be the event that the second ball drawn is ball 4 or 5. Find P(A).
(ii) Let B be the event that the sample of three balls contains ball 1 or ball 2 or both. Find P(B).
(iii) Let C be the event that the three recorded numbers come in increasing order. Find P(C).
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