wel (1) Pr( first = black, second = white ) = x3 = 18 = 3 (2) Pr( both are black ) = Pr(1st = black, 2nd = black) = (3) Pr(B and D) = (B,D) total 0.2= 20% = 10 = 5/00 -8 x ²² = 30 = 1/2 8

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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ort Answer
(1) Pr( first = black, second = white ) = x3 = 18 = 2
(2) Pr( both are black ) = Pr(1st - black, 2nd = black) = x ² = 30 = 12
72
-
=
(B,D)
2 = 0.2 = 20%
(3) Pr(B and D) =
total
Pr(A and C)
4/10
4
(4) Pr(A given C)
5/10
5
Pr(C)
(5) 8P3 =
=
8!
(8-3)!
8!
5!
8.7.6.5.4.3.2.18x7x6 = 336
5x4×3× 2 × 1
8!
(6) 8C3 =
8.7.6.5.4.3.2.1
8.7.6
336
X
315! (3 × 2 × 1)(5 x 4× 3 × 2 × 1)
3.2.1
= 56
6
06\365(2)
(-)
$65
06 365(2)
(7) P = 100, APR = 6% =.06, n = 365, Y = 2 → A = 100(1+
365
ny
(8) If A = P(1 + APR)", then P = A/ (1 + APR) = 8000 / 2.5 = 3200
(9) b = 6,000,, increase = m = 10% of 6,000 = 600, x = 2 years
y = mx + b = 600(2) + 6,000 = 1,200 + 6,000 = 7,200
t
Q = Qox (1+r)' = 6,000(1+0.10)²
(10) Qo= 6,000, increase = r = 10% = 0.10, 1 = 2 years
-
=
=
X
Transcribed Image Text:ort Answer (1) Pr( first = black, second = white ) = x3 = 18 = 2 (2) Pr( both are black ) = Pr(1st - black, 2nd = black) = x ² = 30 = 12 72 - = (B,D) 2 = 0.2 = 20% (3) Pr(B and D) = total Pr(A and C) 4/10 4 (4) Pr(A given C) 5/10 5 Pr(C) (5) 8P3 = = 8! (8-3)! 8! 5! 8.7.6.5.4.3.2.18x7x6 = 336 5x4×3× 2 × 1 8! (6) 8C3 = 8.7.6.5.4.3.2.1 8.7.6 336 X 315! (3 × 2 × 1)(5 x 4× 3 × 2 × 1) 3.2.1 = 56 6 06\365(2) (-) $65 06 365(2) (7) P = 100, APR = 6% =.06, n = 365, Y = 2 → A = 100(1+ 365 ny (8) If A = P(1 + APR)", then P = A/ (1 + APR) = 8000 / 2.5 = 3200 (9) b = 6,000,, increase = m = 10% of 6,000 = 600, x = 2 years y = mx + b = 600(2) + 6,000 = 1,200 + 6,000 = 7,200 t Q = Qox (1+r)' = 6,000(1+0.10)² (10) Qo= 6,000, increase = r = 10% = 0.10, 1 = 2 years - = = X
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