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Can someone please post this up or show me how to do this (1 Viewer)
- Thread starter daryl-d
- Start date
ninetypercent
ninety ninety ninety
Question 6
a)^4)
b)i P(red) = 3/8
and P(black) = 5/8
Quarter black means 3 black, 9 red
bii) The most likely number -> the term with the largest probability
C(k+1) > C(k)
12C(k+1) (3/8)^11-k (5/8)^k+1
------------------------------------------ > 1
12C(k) (3/8)^12-k (5/8)^k
!(11-k)!}\cdot \frac{5}{8}}{\frac{12!}{(12-k)!k!}\cdot \frac{3}{8}} > 1)
Need k = 7 - a term with 12C8
t9 = 12C8 (3/8)^4 (5/8)^8
the most likely combination is 4 red and 8 black
ci) Let An = value of investment at the end of n years
^1)
^2 + 1.2\cdot 1000(1.056)^1)
Let M = 1.056
)
ii))
}{r - 1})
^n - 1}{\frac{M}{1.2}-1})
^n - 1}{\frac{1.056}{1.2}-1})
^{n-1} \cdot \frac{(0.88)^n - 1}{-0.12})
^{n-1} [1 - (0.88)^n])
iii)
Amount invested
 + 1000(1.2)^2...1000(1.2)^{29})
^{30}-1}{1.2 - 1})
interest = $521 275.37
percentage interest = (521275.37/1703 156.94) x 100
= 30.6%
a)
b)i P(red) = 3/8
and P(black) = 5/8
Quarter black means 3 black, 9 red
bii) The most likely number -> the term with the largest probability
C(k+1) > C(k)
12C(k+1) (3/8)^11-k (5/8)^k+1
------------------------------------------ > 1
12C(k) (3/8)^12-k (5/8)^k
Need k = 7 - a term with 12C8
t9 = 12C8 (3/8)^4 (5/8)^8
the most likely combination is 4 red and 8 black
ci) Let An = value of investment at the end of n years
Let M = 1.056
ii)
iii)
Amount invested
interest = $521 275.37
percentage interest = (521275.37/1703 156.94) x 100
= 30.6%