2022 Spring. Solutions of M262 Midterm problems.
1a. 16 balls are randomly distributed to 7 empty boxes. Given that one of
the boxes contains 9 balls find the probability that there is no empty box.
Solution: Let A be the even that one of the boxes contains 9 balls and B be
the event that no box is empty. Then
16 7 7
9 1 6
P (A) =
716
(we choose a box containing 9 balls, choose 9 balls which go to this box and
distribute remaining 7 balls.
16 7
9 2 7!
P (B ∩ A) =
716
(we choose 9 balls which go to this box, out of reamining 7 balls choose 2 balls
which go to the same box). Therefore,
16 7
P (B ∩ A) 7! 70
P (B|A) = = 9 2 = .
16 7 7
P (A) 9 1 6
64
1b. There are three fair dice:
† one face of the first die is white and five faces are black,
† two faces of the second die are white and four faces are black,
† three faces of the third die are white and three faces are black.
A randomly selected die is rolled twice. Given that the first outcome is a black
face find the probability that the second outcome is also black.
Solution: Let A be the event that the selected die has one white face, B be
the event that the selected die has two white faces and C be the event that the
selected die has three white faces. Then by Bayes’ rule
1
3· 65 5
P (A) = 1 =
3 · 56 + · 64 + 31 ·
1
3
3
6
12
1
3· 46 4
P (B) = 1 =
3 · 56 + · 46 + 31 ·
1
3
3
6
12
1
3· 36 3
P (C) = 1 =
3 · 65 + · 46 + 31 ·
1
3
3
6
12
Therefore, the probability that the second outcome is also black is
1 09-25-2025 13:09:00 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/249824289/22262midterm-solpdf/
1a. 16 balls are randomly distributed to 7 empty boxes. Given that one of
the boxes contains 9 balls find the probability that there is no empty box.
Solution: Let A be the even that one of the boxes contains 9 balls and B be
the event that no box is empty. Then
16 7 7
9 1 6
P (A) =
716
(we choose a box containing 9 balls, choose 9 balls which go to this box and
distribute remaining 7 balls.
16 7
9 2 7!
P (B ∩ A) =
716
(we choose 9 balls which go to this box, out of reamining 7 balls choose 2 balls
which go to the same box). Therefore,
16 7
P (B ∩ A) 7! 70
P (B|A) = = 9 2 = .
16 7 7
P (A) 9 1 6
64
1b. There are three fair dice:
† one face of the first die is white and five faces are black,
† two faces of the second die are white and four faces are black,
† three faces of the third die are white and three faces are black.
A randomly selected die is rolled twice. Given that the first outcome is a black
face find the probability that the second outcome is also black.
Solution: Let A be the event that the selected die has one white face, B be
the event that the selected die has two white faces and C be the event that the
selected die has three white faces. Then by Bayes’ rule
1
3· 65 5
P (A) = 1 =
3 · 56 + · 64 + 31 ·
1
3
3
6
12
1
3· 46 4
P (B) = 1 =
3 · 56 + · 46 + 31 ·
1
3
3
6
12
1
3· 36 3
P (C) = 1 =
3 · 65 + · 46 + 31 ·
1
3
3
6
12
Therefore, the probability that the second outcome is also black is
1 09-25-2025 13:09:00 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/249824289/22262midterm-solpdf/
