Can someone please post this up or show me how to do this (1 Viewer)

daryl-d · Sep 20, 2009

Can some please show me how to do q6 and 7 from the nepean trial.

thanks

ninetypercent · Sep 20, 2009

Question 6

a) $T_5 = \binom{15}{4}1^{11}(2x)^4$

$= 21840 x^4$

b)i P(red) = 3/8

and P(black) = 5/8

Quarter black means 3 black, 9 red

$\binom{12}{3}[\frac{3}{8}]^9[\frac{5}{8}]^3$

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

$\frac{\frac{12!}{(k+1)!(11-k)!}\cdot \frac{5}{8}}{\frac{12!}{(12-k)!k!}\cdot \frac{3}{8}} > 1$

$\frac{5}{k+1} \cdot \frac{12-k}{3} > 1$

$8k < 57$

$k < 7\tfrac{1}{8}$

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

$A_{1} = 1000(1+0.056)^1$

$A_{2} = 1000(1.056)^2 + 1.2\cdot 1000(1.056)^1$

Let M = 1.056

$A_{3} = 1000M^3 + 1.2 \cdot 1000M^2 + 11.2^2 \cdot 1000M$

$A_{3} = 1000M (1.2^2 + 1.2M + M^2)$

ii) $A_{n} = 1000M (M^{n-1} + 1.2M^{n-2}+1.2^2M^{n-3}..1.2^{n-1})$

$= 1000M \cdot \frac{a(r^n - 1)}{r - 1}$

$= 1000M \cdot 1.2^{n-1} \cdot \frac{\frac{M}{1.2})^n - 1}{\frac{M}{1.2}-1}$

$= 1000 \cdot 1.056 \cdot 1.2^{n-1} \cdot \frac{\frac{1.056}{1.2})^n - 1}{\frac{1.056}{1.2}-1}$

$= 1056 \cdot (1.2)^{n-1} \cdot \frac{(0.88)^n - 1}{-0.12}$

$= 8800 (1.2)^{n-1} [1 - (0.88)^n]$

iii) $A_{30} = 1703156.94$

Amount invested

$=1000 + 1000(1.2) + 1000(1.2)^2...1000(1.2)^{29}$

$= 1000 \cdot \frac{(1.2)^{30}-1}{1.2 - 1}$

$= 1181881.57$

interest = $521 275.37

percentage interest = (521275.37/1703 156.94) x 100

= 30.6%

Can someone please post this up or show me how to do this (1 Viewer)

daryl-d

Member

ninetypercent

ninety ninety ninety

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