Chapter 2. Random variables
Problem 42
1. Let D1 (D2) be the outcome on the first (second) die
𝑝𝐴 (1) = 𝑃(𝐷1 = 1 ∩ 𝐷2 = 1) = 1⁄36
𝑝𝐴 (2) = 𝑃[(𝐷1 = 1 ∩ 𝐷2 = 2) ∪ (𝐷1 = 2 ∩ 𝐷2 = 1) ∪ (𝐷1 = 2 ∩ 𝐷2 = 2)] = 3⁄36
…
2. 𝑃(𝐴 𝑤𝑖𝑛𝑠)
= 𝑝𝐴 (1) ∗ 0 + 𝑝𝐴 (2) ∗ 1⁄6 + 𝑝𝐴 (3) ∗ 2⁄6 + 𝑝𝐴 (4) ∗ 3⁄6 + 𝑝𝐴 (5) ∗ 4⁄6 + 𝑝𝐴 (6) ∗ 5⁄6
= 125⁄216 = 0.579
Problem 43
Let Gk denote game k, 𝑘 = 1, 2, 3
𝑝(1) = 𝑃(𝐴 𝑤𝑖𝑛𝑠 𝐺1) = 1⁄3
𝑝(2) = 𝑃[(𝐵 𝑤𝑖𝑛𝑠 𝐺1 ∩ 𝐴 𝑤𝑖𝑛𝑠 𝐺2) ∪ (𝐵 𝑤𝑖𝑛𝑠 𝐺1 ∩ 𝐵𝑤𝑖𝑛𝑠 𝐺2) ∪ (𝐶 𝑤𝑖𝑛𝑠 𝐺1 ∩ 𝐴 𝑤𝑖𝑛𝑠 𝐺2)
∪ (𝐶 𝑤𝑖𝑛𝑠 𝐺1 ∩ 𝐴 𝑤𝑖𝑛𝑠 𝐺2)] = 4 ∗ (1⁄3)2 = 4⁄9
𝑝(3) = 1 − 𝑝(1) − 𝑝(2) = 2⁄9
Problem 44
Let 𝑅𝑘 be the result of the first roll, 𝑘 = 1, 2 (not counting rolls where the outcome is smaller than
the previous outcome resulting in at most two rolls)
𝑝(1) = 𝑃(𝑅1 = 1 ∩ 𝑅2 = 1) = 1⁄6 ∗ 1⁄6 = 1⁄36
𝑝(2) = 𝑃[(𝑅2 = 2 ∩ 𝑅1 = 1) ∪ (𝑅2 = 2 ∩ 𝑅1 = 2)] = 1⁄6 ∗ 1⁄6 + 1⁄6 ∗ 1⁄5 = 11⁄180
𝑝(3) = 1⁄6 ∗ 1⁄6 + 1⁄6 ∗ 1⁄5 + 1⁄6 ∗ 1⁄4 = 37⁄360
Problem 45
𝑝(1) = 𝑃[(𝑎𝑙𝑙 𝑜𝑟𝑑𝑒𝑟𝑠 𝑡𝑜 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑟 1) ∪ (𝑎𝑙𝑙 𝑜𝑟𝑑𝑒𝑟𝑠 𝑡𝑜 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑟 2) ∪ … ] =
∑4𝑘=1 𝑃(𝑎𝑙𝑙 𝑜𝑟𝑑𝑒𝑟𝑠 𝑡𝑜 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑟 𝑘) = 4 ∗ (1⁄4)4 = (1⁄4)3 = 1⁄64
𝑝(2) =
4!
4!
4!
4
( )∗[
+
+
]
2 2!2!0!0! 3!1!0!0! 1!3!0!0!
4
4
=
6∗[6+8]
256
= 21⁄64 : two suppliers can be selected from four
4
suppliers in ( ) ways. For each combination of suppliers, say supplier 1 and supplier 2, the four
2
orders can be split 2-2, 1-3 or 3-1.
4!
4
( ) ∗ [3 ∗ 2! 1! 1! 0!]
𝑝(3) = 3
= 9⁄16
44
𝑝(4) =
4!
= 3⁄32
44
Problem 46
Let 𝑝1𝐴 (𝑘) be the probability that the first arrival of team A arrives in position k.
𝑝1𝐴 (1) = 3⁄6 = 0.5
𝑝1𝐴 (2) = 3⁄6 ∗ 3⁄5 = 0.3: a B-runner followed by a A-runner
𝑝1𝐴 (3) = 3⁄6 ∗ 2⁄5 ∗ 3⁄4 = 0.15: two B-runners followed by a A-runner
𝑝1𝐴 (4) = 3⁄6 ∗ 2⁄5 ∗ 1⁄4 ∗ 1 = 0.05: three B-runners followed by a A-runner
Let 𝑝2𝐴 (𝑘) be the probability that the second arrival of team A arrives in position k.
𝑝2𝐴 (2) = 3⁄6 ∗ 2⁄5 = 0.2: A-runner followed by A-runner
𝑝2𝐴 (3) = 3⁄6 ∗ 3⁄5 ∗ 2⁄4 + 3⁄6 ∗ 3⁄5 ∗ 2⁄4 = 0.3: A-B-A or B-A-A
𝑝2𝐴 (4) = 3⁄6 ∗ 3⁄5 ∗ 2⁄4 ∗ 2⁄3 + 3⁄6 ∗ 2⁄5 ∗ 3⁄4 ∗ 2⁄3 + 3⁄6 ∗ 3⁄5 ∗ 2⁄4 ∗ 2⁄3 = 0.3:
A-B-B-A or B-B-A-A or B-A-B-A
𝑝2𝐴 (5) = 1 − 𝑝2𝐴 (2) − 𝑝2𝐴 (3) − 𝑝2𝐴 (4) = 0.2
Problem 47
Let ⌊𝑞⌋ denote the largest integer smaller or equal to q. Let Dk be the event that R is divisible by k, let
𝐷𝑘\ℎ1 , ℎ2 , … , ℎ𝑛 be the event that R is divisible by k but not by ℎ1 , ℎ2 , … , ℎ𝑛 .
𝑃(𝑋 = 7) = 𝑃(𝐷7) = ⌊1000⁄7⌋⁄1000 = 0.142
𝑃(𝑋 = 5) = 𝑃(𝐷5) − 𝑃(𝐷5 ∩ 𝐷7) = ⌊1000⁄5⌋⁄1000 − ⌊1000⁄35⌋⁄1000 = 0.2 − 0.028 = 0.172
𝑃(𝑋 = 3) = 𝑃(𝐷3) − 𝑃(𝐷5 ∩ 𝐷3) − 𝑃(𝐷3 ∩ 7) + 𝑃(3 ∩ 5 ∩ 7)
= ⌊1000⁄3⌋⁄1000 − ⌊1000⁄15⌋⁄1000 − ⌊1000⁄21⌋⁄1000 + ⌊1000⁄105⌋⁄1000
= 0.229
𝑃(𝑋 = 2) = 𝑃(𝐷2) − 𝑃(𝐷2 ∩ 𝐷3) − 𝑃(𝐷2 ∩ 𝐷5) − 𝑃(2 ∩ 7) + 𝑃(𝐷2 ∩ 𝐷3 ∩ 𝐷5)
+ 𝑃(𝐷2 ∩ 𝐷3 ∩ 𝐷7) + 𝑃(𝐷2 ∩ 𝐷5 ∩ 𝐷7) − 𝑃(𝐷2 ∩ 𝐷3 ∩ 𝐷5 ∩ 𝐷7)
= ⌊1000⁄2⌋⁄1000 − ⌊1000⁄6⌋⁄1000 − ⌊1000⁄10⌋⁄1000 − ⌊1000⁄14⌋⁄1000
+ ⌊1000⁄30⌋⁄1000 + ⌊1000⁄42⌋⁄1000 + ⌊1000⁄70⌋⁄1000 − ⌊1000⁄210⌋⁄1000
= 0.229
𝑃(𝑋 = 0) = 1 − 𝑃(𝑋 = 2) − 𝑃(𝑋 = 3) − 𝑃(𝑋 = 5) − 𝑃(𝑋 = 7) = 0.228
Problem 48
𝑃(𝑋 = 6) = 1⁄6
𝑃(𝑋 = 5) = 5⁄6 ∗ 2⁄6 = 5⁄18
𝑃(𝑋 = 4) = 5⁄6 ∗ 4⁄6 ∗ 3⁄6 = 5⁄18
…
Problem 49
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
= −4) = (1⁄4)4
= −2) = 4 ∗ (1⁄4)3 ∗ 3⁄4
= 0) = 6 ∗ (1⁄4)2 ∗ (3⁄4)2
= +2) = 4 ∗ (3⁄4)3 ∗ 1⁄4
= +4) = (3⁄4)4
Problem 50
1. 𝑃(𝑑𝑒𝑚𝑎𝑛𝑑 ≤ 100) = 0.8
2. When demand is 70: 𝑋 = 70 ∗ 50 + 30 ∗ 20 = 4100
When demand is 80: 𝑋 = 80 ∗ 50 + 20 ∗ 20 = 4400
When demand is 90: 𝑋 = 90 ∗ 50 + 10 ∗ 20 = 4400
When demand is 100 or more: 𝑋 = 1000 ∗ 50 = 5000
Problem 51
1. 𝑃(𝐴 𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅1) = 𝑃((𝐴 𝑡ℎ𝑟𝑜𝑤𝑠 𝐻, 𝑜𝑡ℎ𝑒𝑟𝑠 𝑡ℎ𝑟𝑜𝑤 𝑇) ∪ (𝐴 𝑡ℎ𝑟𝑜𝑤𝑠 𝑇, 𝑜𝑡ℎ𝑒𝑟𝑠 𝑡ℎ𝑟𝑜𝑤 𝐻) )
= 1⁄2 ∗ (1⁄2)𝑁−1 + 1⁄2 ∗ (1⁄2)𝑁−1 = (1⁄2)𝑁−1
2. 𝑃(𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅1) = 𝑃((𝐴 𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅1) ∪ (𝐵 𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅1) ∪ … ) = 𝑁 ∗ (1⁄2)𝑁−1
3. 𝐿𝑒𝑡 𝑞 = 𝑁 ∗ (1⁄2)𝑁−1
𝑃(𝑋 = 𝑥) = (1 − 𝑞)𝑥−1 ∗ 𝑞
𝑃(𝑋 = 𝑟 + 1) = (1 − 𝑞)𝑟
, 𝑥 = 1, 2, … , 𝑟
𝑃(𝐴 𝑖𝑠 𝑣𝑖𝑐𝑡𝑖𝑚) = 𝑃((𝐴 𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅1) ∪ (𝐴 𝑣𝑖𝑐𝑡𝑖𝑚 𝑖𝑛 𝑅2) ∪ … )
= (1⁄2)𝑁−1 + (1 − 𝑞) ∗ (1⁄2)𝑁−1 + (1 − 𝑞)2 ∗ (1⁄2)𝑁−1 + ⋯ + (1 − 𝑞)𝑟−1 ∗ (1⁄2)𝑁−1
= (1⁄2)𝑁−1 ∗
1 − (1 − 𝑞)𝑟 1 − (1 − 𝑞)𝑟
=
𝑞
𝑁
Problem 52
𝑃(𝑋 = 1) = 𝑃(𝑅 = 1 ∩ (𝑋 = 1|𝑅 = 1)) = 0.1 ∗ 0.1 = 0.01
𝑃(𝑋 = 2) = 𝑃(𝑅 = 1 ∩ (𝑋 = 2|𝑅 = 1)) + 𝑃(𝑅 = 2 ∩ (𝑋 = 2|𝑅 = 2)) = 1⁄10 ∗ 1⁄10 + 1⁄10 ∗ 1⁄9
= 1⁄10 ∗ (1⁄10 + 1⁄9)
…
𝑃(𝑋 = 𝑘) = 1⁄10 ∗ (1⁄10 + 1⁄9 + ⋯ + 1⁄(10 − 𝑘 + 1))
...
𝑃(𝑋 = 10) = 1⁄10 ∗ (1⁄10 + 1⁄9 + ⋯ + 1⁄1)
Problem 53
Let 𝐶𝑘 denote the 𝑘 𝑡ℎ child.
𝑃(𝑋 = 2) = 𝑃[(𝐶1 = 𝐵 ∩ 𝐶2 = 𝐺) ∪ (𝐶1 = 𝐺 ∩ 𝐶2 = 𝐵)] = .5 ∗ .5 + .5 ∗ .5 = .5
𝑃(𝑋 = 3) = 𝑃[(𝐶1 = 𝐵 ∩ 𝐶2 = 𝐵 ∩ 𝐶3 = 𝐺) ∪ (𝐶1 = 𝐺 ∩ 𝐶2 = 𝐺 ∩ 𝐶3 = 𝐵)] =. 53 +. 53 = .25
…
Problem 54
Let 𝐶𝑘 be the outcome of the 𝑘 𝑡ℎ toss with the coin
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
= 2) = 𝑃(𝐶1 = "𝐻") = 1⁄2
= 4) = 𝑃(𝐶1 = "𝑇" ∩ 𝐶2 = "𝐻") = 1⁄2 ∗ 1⁄2 = 1⁄4
= 8) = 1⁄2 ∗ 1⁄2 ∗ 1⁄2 = 1⁄8
= 16) = 1⁄2 ∗ 1⁄2 ∗ 1⁄2 ∗ 1⁄2 = 1⁄16
= 32) = 1 − 𝑃(𝑋 = 2) − 𝑃(𝑋 = 4) − 𝑃(𝑋 = 8) − 𝑃(𝑋 = 16) = 1⁄16
Problem 55
𝑥
1. 𝐹(𝑥) = ∫1 1⁄8 ∗ (𝑢 − 1) ∗ 𝑑𝑢 = 1⁄16 ∗ (𝑥 − 1)2 , 1 ≤ 𝑥 ≤ 5
=0 , 𝑥<1
=1 , 𝑥>5
2. 𝐹(𝑥) = 1⁄16 ∗ (𝑥 − 1)2 = 𝑅
→
𝑥 = 1 + √16 ∗ 𝑅
3. 𝐹(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(𝑋 ≤ 1⁄2 ∗ (𝑥 + 4)) = 1⁄16 ∗ (𝑦⁄2 + 1)2 , −2 ≤ 𝑦 ≤ 6
4. 𝑓(𝑦) = 𝑑𝐹(𝑦)⁄𝑑𝑦 = 1⁄16 ∗ (𝑦⁄2 + 1) , −2 ≤ 𝑦 ≤ 6
Problem 56
1. Trivial
2. 𝑦 = 𝑥 1⁄2 → 𝑥 = 𝑦 2 → 𝐹(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑦 2 ) = 𝑦 2 , 0 ≤ 𝑦 ≤ 1
𝑓(𝑦) = 𝑑𝐹(𝑦)⁄𝑑𝑦 = 2 ∗ 𝑦 , 0 ≤ 𝑦 ≤ 1
3.
𝑣 = 𝑥 2 → 𝑥 = 𝑣 1⁄2 → 𝐹(𝑣) = 𝑃(𝑉 ≤ 𝑣) = 𝑃(𝑋 ≤ 𝑣 1⁄2 ) = 𝑣 1⁄2 , 0 ≤ 𝑣 ≤ 1
𝑓(𝑣) = 𝑑𝐹(𝑣)⁄𝑑𝑣 = 1⁄2 ∗ 𝑣 −1⁄2 , 0 ≤ 𝑣 ≤ 1
Problem 57
𝑥
1. 𝐹(𝑥) = ∫4 4⁄𝑢2 ∗ 𝑑𝑢 = 1 − 4⁄𝑥 , 𝑥 ≥ 4
2. 𝐹(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(√𝑋 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑦 2 ) = 1 − 4⁄𝑦 2 , 𝑦 ≥ 2
𝑓(𝑦) = 𝑑𝐹(𝑦)⁄𝑑𝑦 = 8⁄𝑦 3 , 𝑦 ≥ 2
3. 𝐹(𝑣) = 𝑃(𝑉 ≤ 𝑣) = 𝑃(1⁄𝑋 ≤ 𝑣) = 1 − 𝑃(𝑋 ≤ 1⁄𝑣 ) = 4 ∗ 𝑣 , 0 ≤ 𝑣 ≤ 1⁄4
𝑓(𝑣) = 𝑑𝐹(𝑣)⁄𝑑𝑣 = 4 , 0 ≤ 𝑣 ≤ 1⁄4
Problem 58
1. 𝑌 = 2 ∗ 𝜋 ∗ 𝑋
𝑥
𝐹(𝑥) = ∫ 𝑑𝑢 = 𝑥 − 1 , 1 ≤ 𝑥 ≤ 2
1
𝐹(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(2𝜋𝑋 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑦⁄2𝜋) = 𝑦⁄2𝜋 − 1 , 2𝜋 ≤ 𝑦 ≤ 4𝜋
𝑓(𝑦) = 𝑑𝐹(𝑦)⁄𝑑𝑦 = 1⁄2𝜋 , 2𝜋 ≤ 𝑦 ≤ 4𝜋
2. 𝑊 = 𝜋 ∗ 𝑋 2
𝐹(𝑤) = 𝑃(𝑊 ≤ 𝑤) = 𝑃(𝜋𝑋 2 ≤ 𝑤) = 𝑃 (𝑋 ≤ √𝑤⁄𝜋) = √𝑤⁄𝜋 − 1 , 𝜋 ≤ 𝑤 ≤ 4𝜋
𝑓(𝑤) = 𝑑𝐹(𝑤)⁄𝑑𝑤 = 1⁄2𝜋 ∗ (𝑤⁄𝜋)−1⁄2 ,
𝜋 ≤ 𝑤 ≤ 4𝜋
Problem 59
1
3. 𝐷 = 𝑌 − 𝑋 = 1 − 𝑋 − 𝑋 = 1 − 2𝑋 → 𝑋 = 2 ∗ (1 − 𝐷)
𝑓(𝑑) = 2 ∗ 1⁄2 = 1 , 0 ≤ 𝑑 ≤ 1
4. 𝑆 =
𝑋
1−𝑋
→ 𝑋=
𝑓(𝑠) = 2 ∗
5. 𝐿 =
1−𝑋
𝑋
𝑆
1+𝑆
→
1
= (1+𝑆)2
1
, 0≤𝑠≤1
(1 + 𝑠)2
1
→ 𝑋 = 1+𝐿 →
𝑓(𝑙) = 2 ∗
𝑑𝑋
𝑑𝑆
1
, 𝑙≥1
(1 + 𝑙)2
𝑑𝑋
𝑑𝐿
1
= (1+𝐿)2