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Answer:

your friend's contention is incorrect. Instead, a more laborious process of going through potential situations needs to be done.

Explanation:

First thing - we're not talking probability but instead the number of ways something(s) can be arranged (if we were talking probabilities, we'd have a ratio - the number of ways to meet a certain set of conditions divided by the total number of ways the things could be arranged).

The key here is the number of duplicates. There are an unknown number of them - we're told there are "some" but we don't know how many. So let's work with what we do know and see if we can formulate an answer from there.

Your friend has already figured out that we are talking about a combination problem (the general formula of which is:

#C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#)

The reason it's a combination problem is that we don't care about the order in which the books are picked (much like with a poker hand - we don't care how we got the hand but only that we got it).

If we weren't dealing with duplicates at all, we'd have the two people picking 8 books in total from the 40 (non-duplicate) available, which would give us:

#C_(40,8)=(40!)/((8!)(32!))#

So far so good - your friend is on track. So now to deal with the duplicates.

First I'll walk through a simple explanation as to why we're going to do what we're going to do, and then go back to our problem.

When we work with factorials, permutations, and combinations, what we're really doing is starting with a factorial (note the #40!# in the numerator of our combination) and then dividing by all the limitations.

I'll interweave an example into this discussion. Let's work with a 4-letter word, say ABCD.

A factorial takes all the members of a group and arranges them in a unique order. We can make #4! = 24# different arrangements of the 4 letters in ABCD.

A permutation limits the number of members being arranged. If we only pick 2 letters at a time, we end up with #(4!)/((4-2!))=24/2=12# different arrangements of the letters.

A combination eliminates the differences in the different orders of things - and so AB is the same as BA (much like a poker hand), and so we end up with #(4!)/(2!(4-2!))=24/(2xx2)=6# different arrangements of the letters.

To eliminate duplicates, we divide by the internal ordering that occurs between said duplicates. So let's say that instead of ABCD we have instead AABC. How many ways can we arrange these letters?

We keep the numerator at #4!# and now we divide by the internal ordering of the duplicate A (which is #2!#), which gives:

#(4!)/(2!)=12# different ways to arrange AABC.

So what happens when we take a population, choose from it, and that population has duplicates. How do we handle it?

Let's take combinations where we pick 2 from AABC. We'll end up with

AA, AB, AC, BC = 4 ways. But if we simply divide by #2!#, the number of duplicates we could have, we get:

#C_(4,2)/(2!)=(4!)/((2!)(2!)(2!))=3#, which isn't correct.

We can expand this thinking to the question at hand, with 40 books in total, 16 potential duplicates, and the picking of 8 books from that population. In short, your friend's contention that

#C_(40,8)/(2!)=(40!)/((8!)(32!)(2!))#

will eliminate the duplicate and come up with an accurate tally is incorrect. Instead, a more laborious process of going through potential situations needs to be done.

Answer:

#P("Spade" uu "Queen") = 4/13#

Explanation:

Whenever you solve a probability question involving two conditions, and you are being asked to find the probability that either will occur for a given action, you are looking for what is known as a "union probability". Formally speaking, if we say #A# represents "the card is a Spade", and #B# represents "the card is a Queen", then we are looking for the probability of "the card is a Spade or a Queen", or symbolically:

#P(A uu B)#

The trick here is that these two possible events are not disjoint events; in other words, it can be possible to pull a single card and have it be a Spade and a Queen at the same time. The formula for determining #P(A uu B)# takes this into consideration:

#P(A uu B) = P(A) + P(B) - P(A nn B)#

(This is read as "the probability of A union B is equal to the probability of A plus the probability of B minus the probability of the intersection of A and B".)

If we consider #P(A)# (the probability the card is a Spade), in a standard deck of 52 cards there are exactly 13 cards which are Spades. Thus, #P(A) = 13/52 = 1/4#. (This is intuitive, because there are 4 suits of cards with the same values/ranks within them and we're only interested in one of those four suits.)

If we consider #P(B)# (the probability the card is a Queen), in a standard deck of 52 cards there are exactly 4 cards which are Queens (in suits of Hearts, Spades, Clubs, and Diamonds). Thus, #P(B) = 4/52 = 1/13#. (Again, this is intuitive, because there are 13 unique values of cards, of which there is only one Queen value.)

However, the probability #P(A nn B)# represents the probability the card is a Spade and a Queen at the same time. Of all 52 cards in the deck, there is only one Queen of Spades, thus #P(A nn B) = 1/52#.

Thus:

#P(A uu B) = P(A) + P(B) - P(A nn B)#

# = 13/52 + 4/52 - 1/52 = 16/52 = 4/13#

Answer:

If the first ball is replaced: #P("different") = 35/72#

If the first ball is not replaced: #P("different") = 35/66#

Explanation:

There are #3# outcomes that can occur:

  • both are white
  • both are black
  • one of each color

The probability of one of each color is #1- P("same")#

If the first ball is replaced:

#P(WW) = 5/12 xx5/12 = 25/144#
#P(BB) = 7/12 xx7/12 = 49/144#

#:. P("same) = 25/144+49/144 = 74/144#

#P("different") = 1-74/144 = 70/144 = 35/72#

However, if the first ball is NOT replaced, the probability changes for the second ball. There are fewer of one color and one ball less.

#P(WW) = 5/12 xx4/11 = 20/132#

#P(BB) = 7/12 xx6/11 = 42/132#

#:. P("same) = 20/132+42/132 = 62/132#

#P("different") = 1-62/132 = 70/132 = 35/66#

Answer:

Only the following two combinations are possible:

#1" " , 1" ", 4" ", 5" "9#

#1" " , 1" ", 4" ", 6" "8#

Explanation:

I will assume that both the median and the mean are #4#.

The median is the value in the middle of a set of data arranged in order. There are #5# numbers, so the third one has to be #4#

#--- , ---, 4 ---, ---" "larr# the median is #4#

The mode is the value that occurs the most often. The mode is #1#. There are only two possible spaces for #1#, so there must be two of them.

#1" " , 1" ", 4 ---, ---" "larr# the mode is #1#

If the mean is #4# and there are five numbers, it means that the total of the five numbers is #5 xx 4 =20#

We already have a total of #6#, so the sum of the remaining teo numbers must be #20-6 =14#

Any combination of numbers bigger than 4 will do, except they cannot both be equal to #7#, because then there would be two modes.

Only the following two combinations are possible:

#1" " , 1" ", 4" ", 5" "9#

#1" " , 1" ", 4" ", 6" "8#

The following are not possible:

#1" " , 1" ", 4" ", 4" "10" "larr# there are two modes

#1" " , 1" ", 4" ",7" "7" "larr# there are two modes

#1" " , 1" ", 3" ", 4" "11" "larr# the median is not #4#

#1" " , 1" ", 2" ", 4" "12" "larr# the median is not #4#

Answer:

#E(X) = H/T #

Explanation:

This has got to be one of my favourite problems in mathematics...

Let #H# be the probability of flipping heads

Let #T# be the probability of flipping tails

#H and T# is not nessesarily #= 1/2# as its unfair...

Now we must use our knowledge of expected outcomes...

#E(X) = sum x_i P(X=x_i) = sumx_i p_i #

Now we can devise a descrete random variable #X#:

#X# - Number of wins...

Getting 0 wins would have a probability of #T#
Getting 1 win is getting a head and then tails, #HT#
.
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enter image source here

Hence we must use our #E(X) # formula...

#E(X) = (0*T) + (1*HT) + (2*H^2T) + (3*H^3T) + ...#

#=> E(X) = HT + 2H^2 T + 3H^3 T + ... #

Now factoring out #HT#:

# => E(X)= HT ( 1 + 2H + 3H^2 + 4H^3 + ... )#

This stage is the most difficult, we must recognise that:

# 1 + 2H + 3H^2 + ... = d/(dH) ( H + H^2 + H^3 + ... ) #

#=> E(X) = HT * d/(dH) ( H + H^2 + H^3 + ... ) #

Now we know #H + H^2 + H^3 + ... # is a infinte geometric series...

We know: #sum_(n=1) ^oo a*r^(n-1) = a/(1-r) ,|r|<1 #

#=> H + H^2 + H^3 + ... = H/(1-H) #

We can do this as #0 < H < 1 #

#=> E(X) = HT* d/(dH) ( H/(1-H) ) #

We can find the derivative using the quotient rule:

#y = u/v -> y' = ( vdu - udv )/v^2 #

#d/(dH) ( H/(1-H) ) = 1/(1-H)^2 #

#=> E(X) = (HT)/(1-H)^2 #

We know the total probability is #=1#

#=> H + T =1 #

#=> E(X) = (HT)/(T^2) #

#=> E(X) = H/T #

enter image source here

Answer:

#y = 1/pi sechx " For larger " x#

#y = 1/sqrt(2pi) sechx " For smaller " x#

Explanation:

A few ways of approximating the bell curve...

The general Bell curve:

#color(red)(y = 1/sqrt(2pi) e^(-1/2 x^2 )#

desmos.com

There are many reasons to why you may want to approximate the bell curve, one being that it is particulaly difficult to integrate definitely...

So by approximating, and finding a function that is more easy to integrate definitely, and has an antiderivative, this is much eaiser.

The first thing I can think of is utilising a function that already has #e^x# within it, then we consider #color(blue)(coshx -= 1/2 ( e^x + e^-x ) #

But by sketching, this doesnt quite have the general shape, so the reciprical could be a good idea:

#color(orange)(sech x -=1/coshx -= 2/(e^x+e^(-x) ) #

enter image source here

Now this looks more like it...

One main property of the bell curve, used in statistics regulaly...

#color(blue)(int_-oo ^ oo 1/sqrt(2pi) e^(-1/2 x^2 ) # #color(blue)(dx=1 #

This can be proven using polar substitution using the jacobian, a higher level of mathematics....

So one thing we could do, is understand what the total area under #y = sechx # is?

Finding: #int_ -oo ^oo sechx # # dx #

We can use the substitution #u = tanhx #

#=> du = sech^2 x dx #

#=> (du)/(sechx) = sechx dx #

So #int sechx dx => int (du)/sechx #

Now use #color(red)(1-tanh^2 x -= sech^2 x #

#=> sechx = sqrt(1-tanh^2 x ) #

#=> sechx = sqrt(1-u^2 ) #

#int (du)/sechx = int (du)/sqrt(1-u^2) #

Now we can use another trig substitution:

#u = sintheta #

#=> du = costheta d theta #

# int (du)/sqrt(1-u^2)#

#=> int (costheta d theta )/ (sqrt(1-sin^2 theta) )#

#=> int (costheta d theta ) / costheta #

#=> int d theta #

# = theta +c #

Using our substitutions...

#= sin^(-1) u + c #

#color(blue)(= sin^(-1) (tanhx ) + c #

Integrating indefinitely...

#[sin^(-1) (tanhx) ]_ -oo ^ oo #

#=> sin^(-1) ( tanh (oo) ) - sin^(-1)( tanh(-oo) )#

#=> sin^(-1) (1) - sin^(-1) (-1) #

#=> pi/2 - (-pi/2) #

#=> color(red)(int_-oo ^oo sechx dx = pi #

#=> color(red)(int_-oo ^oo 1/pi sechx dx = 1 #

So #color(blue)(y = 1/pi sechx # is a half decent approximation for the bell curve...

enter image source here

But as from looking from the graph we see that for small values of #x# this may not be the most accurate approximation

So #color(blue)(y = 1/sqrt(2pi) sechx # is a good aprpimation for small #x#:

As it has same value for #x=0# And for #x# being small

enter image source here

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