only need part a b c (4) Random draws. An urn contains tiles numbered 1 through
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only need part a b c
(4) Random draws. An urn contains tiles numbered 1 through 6. Draw a pair of tiles randomly (a) Let M denote the larger of the 2 numbers drawn. Give the possible values of M and find the probability distribution function of M (b) Suppose that M = k and draw another tile from the 4 remaining in the urn. Find the conditional probability that the number on the last tile drawn is greater than M given that Mk (c) Find the (unconditional) probability that the number on the last tile drawn is greater than M (d) Find the conditional probability that M 4 given that the number on the last tile drawn is 6 More random draws. Now suppose that the urn contains tiles numbered 1 through n Draw 2 tiles randomly and let M denote the larger of the 2 numbers drawn (e) Suppose that M -k and draw another tile from the n-2 tiles remaining in the urn Let N denote the number on the last tile drawn. Find the conditional probability that N is greater than M given that M-k (f) Give an expression for the (unconditional) probability that N is greater than MExplanation / Answer
Part (a)
The possible values M can take are: 2, 3, 4, 5, 6.
To evaluate the probabilities:
First, two tiles out of 6 can be drawn in 6C2 = 15 ways. Various possibilities (not considering the order) for M = 2, 3, 4, 5, 6 and the corresponding probabilities are given below:
M
2
3
4
5
6
Total
Possibilities
(1, 2)
(1, 3),
(2, 3)
(1, 4), (2, 4),
(3, 4)
(1, 5), (2, 5) (3, 5), (4, 5)
(1, 6), (2, 6) (3, 6), (4, 6), (5, 6)
-
Probability
1/15
2/15
3/15
4/15
5/15
15/15 = 1
ANSWER
Part (b)
Let X = number on the new tile drawn from the remaining 4 tiles.
If M = k, then the tile numbered k and another tile with a number less than k have already been drawn. So, of the remaining 4 tiles, (6 - k) tiles will have numbers greater than k and (k - 2) tiles will have numbers less than k. So, the conditional probability that the number on the last tile is greater than M given M = k = P(X > k/M = k) = (6 - k)/4 = 1.5 – (k/4) ANSWER
Part (c)
Unconditional probability that the number on the last tile is greater than M
= [k = 2 (1) 5][P(X > k/M = k) x P(M = k)]
= [P(X > 2/M = 2) x P(M = 2)] +
[P(X > 3/M = 3) x P(M = 3)] +
[P(X > 4/M = 4) x P(M = 4)] +
[P(X > 5/M = 5) x P(M = 5)]
= {1 x (1/15)} + {(3/4) x (2/15)} + {(2/4) x (3/15)} + {(1/4) x (4/15)}
= 20/60
= 1/3 ANSWER
M
2
3
4
5
6
Total
Possibilities
(1, 2)
(1, 3),
(2, 3)
(1, 4), (2, 4),
(3, 4)
(1, 5), (2, 5) (3, 5), (4, 5)
(1, 6), (2, 6) (3, 6), (4, 6), (5, 6)
-
Probability
1/15
2/15
3/15
4/15
5/15
15/15 = 1
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