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3 years ago

Suppose we want to choose 4 colors, without replacement, from 18 distinct colors.

Mathematics

1 answer:

amm18123 years ago

8 0

Answers:

(a) 73440
(b) 3060

=====================================================

Explanation for part (a)

Consider four slots labeled A,B,C,D.

We have

18 choices for slot A
17 choices for slot B
16 choices for slot C
15 choices for slot D

We started at 18 and counted down until we filled the four slots. The countdown is because we cannot repeat a color. Multiply out those values to get the answer: 18*17*16*15 = 73440

You can use the nPr permutation formula as an alternative method

$_nP_r = \frac{n!}{(n-r)!}$

in this case n = 18 and r = 4. The exclamation marks indicate factorial.

-----------------------------------

Explanation for part (b)

Let's say we had color labels A,B,C,... all the way up to R which is the 18th letter in the English alphabet. From those 18 letters, we can only pick four. Let's say we pick D,E,F,G.

Focusing solely on the set {D,E,F,G}, we only have one set and the order doesn't matter. So {D,E,F,G} is the same as {D,E,G,F}.

There are 4*3*2*1 = 24 ways to arrange those four items meaning that we'll need to divide the result of part (a) by 24 to get the result of part (b)

73440/24 = 3060

You could use the nCr combination formula to get the same result

$_nC_r = \frac{n!}{r!(n-r)!}$

where n = 18 and r = 4 are the same from last time.

The connection between nPr and nCr is this

$_nC_r = \frac{_nP_r}{r!}$

which can be rearranged into

$_nP_r = r!*\left( _nC_r \right)$

these formulas are handy to help us go back and forth between nCr or nPr.

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$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

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d. We have

$\displaystyle P\left(X$

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Compute the marginal density of $Y$ , then directly compute the expected value.

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Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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