1
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2
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In this problem, we're looking
at a two stage process in
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which the first stage, we roll
a fair die which has four
4
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faces to obtain a number N,
where N belongs to the set 0,
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1, 2, and 3 with equal
probability.
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Now, given the result of the die
roll, N will toss a fair
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coin N times in getting K heads
from the coin tosses.
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9
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For instance, if from the first
die roll, we get N equal
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to 3, then we'll toss
a coin 3 times.
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Let's say the outcome is heads,
heads, and tails.
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And that will give
us K equal to 2.
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For part A, we're asked to
compute the PMF for N, which
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is a result of the
first die roll.
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Now, since we had assumed the
die roll was uniformly
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distributed in the set in the
set 0, 1, 2, and 3, we have
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that the chance of N being equal
to any little n is equal
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to 1/4 if n is in the set 0,
1, 2, 3, and 0 otherwise.
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If we were to plot this in
a figure, we'll have the
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following plot.
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For part B, things are getting
a little more complicated.
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This time, we want to compute
the joint PMF between N and K
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for N equal to little n and
K equal to little k.
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What we'll do first is to use
the law of conditional
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probability to break the joint
probability into the product
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of probability of K is equal to
little k conditional on N
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is equal to little n, multiply
by the probability that N is
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equal to little n.
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Now, the second term right here
is simply the PMF of N,
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which will be computed
earlier.
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00:02:09,270 --> 00:02:13,930
So this gives us 1/4 times
probability K equal to little
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00:02:13,930 --> 00:02:20,250
k, N equal to little n,
for all N in the set
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0, 1, 2, and 3.
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Now, clearly if N is not one
of those four values, this
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whole event couldn't have
happened in the first place,
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and hence will have P
and K equal to 0.
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We'll now go over all the cases
for little n in this
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expression right here.
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The first case is
the simplest.
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00:02:42,050 --> 00:02:45,830
If we assume that little n is
equal to 0, that means the die
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roll was 0, and hence we're
not tossing any coins
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00:02:48,300 --> 00:02:49,380
afterwards.
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00:02:49,380 --> 00:02:54,760
And this implies that we must
have K is equal to 0, which,
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mathematically speaking,
is equivalent to saying
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probability of K equal
to 0 conditional on N
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equal to 0 is 1.
48
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And K being any other value
conditional N equal to 0 is 0.
49
00:03:09,080 --> 00:03:13,100
So we're done with the case that
little n is equal to 0.
50
00:03:13,100 --> 00:03:19,850
Now, let's say little n is
in the set 1, 2, and 3.
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00:03:19,850 --> 00:03:23,150
In this case, we want to notice
that after having
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observed the value of N, all
the coin tosses for N times
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are conditionally independent
from each other.
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00:03:30,670 --> 00:03:34,610
What this means is now the total
number of heads in the
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subsequent coin toss is equal in
distribution to a binomial
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random variable with parameter
n and 1/2.
57
00:03:45,650 --> 00:03:50,180
And here says the number of
trials is n, and 1/2 is
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because the coin is fair.
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00:03:52,060 --> 00:03:54,320
And the reason it is a binomial
random variable,
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again, is because the coin
tosses are independent
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conditional on the outcome
of the die roll.
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00:03:59,720 --> 00:04:01,970
And now we're done, since we
know what the binomial
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distribution looks like given
parameter n and 1/2.
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00:04:06,020 --> 00:04:11,760
And we'll simply substitute
based on the case of n the
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00:04:11,760 --> 00:04:16,410
conditional distribution of K
back into the product we had
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00:04:16,410 --> 00:04:20,100
earlier, which in turn will
give us the joint PMF.
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00:04:20,100 --> 00:04:23,230
68
00:04:23,230 --> 00:04:26,450
This table summarizes the PMF
we were computing earlier.
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00:04:26,450 --> 00:04:30,670
P of N, K, little
n, and little k.
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00:04:30,670 --> 00:04:34,210
Now, as we saw before, if
n equal to 0, the only
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00:04:34,210 --> 00:04:36,920
possibility for k
is equal to 0.
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00:04:36,920 --> 00:04:40,330
And this is scaled by the
probability of n equal to 0,
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which is 1/4.
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00:04:41,810 --> 00:04:45,590
For any other values of n, we
see that the distribution of
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k, conditional n, is the same as
a binomial random variable
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00:04:50,750 --> 00:04:52,650
with n trials.
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And again, every entry here
is scaled by 1/4.
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00:04:56,210 --> 00:04:58,000
And this completes part B.
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00:04:58,000 --> 00:05:02,490
In part C, we're asked for
the conditional PMF of K
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conditioning on the value
of N being equal to 2.
81
00:05:06,320 --> 00:05:10,170
Now, as we discussed in part B,
when N is equal to 2, we're
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00:05:10,170 --> 00:05:13,530
essentially flipping a fair coin
twice, and this should
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00:05:13,530 --> 00:05:18,070
give us the same distribution as
a binomial random variable
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00:05:18,070 --> 00:05:21,060
with parameter 2 and 1/2.
85
00:05:21,060 --> 00:05:24,630
Now, 2 is the number of flips,
and 1/2 is the chance of
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00:05:24,630 --> 00:05:26,780
seeing a head in each flip.
87
00:05:26,780 --> 00:05:28,410
And that gives us the following
distribution.
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00:05:28,410 --> 00:05:31,720
89
00:05:31,720 --> 00:05:33,330
But there's another
way to see this.
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It's to write P K given N,
little k, and you go to 2 by
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00:05:39,330 --> 00:05:44,430
using the law of conditional
probability as P K, N, the
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00:05:44,430 --> 00:05:51,030
joint PMF, k n2, divided by
the probability that N is
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equal to 2.
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00:05:52,220 --> 00:05:56,210
Now, we know that probability n
equal to 2 is simply 1/4, so
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this gives us 4 times the joint
density K, N, k, 2.
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00:06:03,610 --> 00:06:07,690
In other words, in order to
arrive at the distribution
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right here, [INAUDIBLE]
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to go back to the table we had
earlier and look at the role
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where n is equal to 2 and
multiply each number by 4.
100
00:06:17,380 --> 00:06:22,310
101
00:06:22,310 --> 00:06:24,640
Finally, in part D, we're asked
for the conditional
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00:06:24,640 --> 00:06:31,450
distribution of N, write as
P N, given K of N equal to
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little n conditional
on K is equal to 2.
104
00:06:38,500 --> 00:06:41,420
Again, we'll apply the formula
for conditional probability.
105
00:06:41,420 --> 00:06:49,970
This is equal to the joint PMF
evaluated at n and 2 divided
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by the probability of
K being equal to 2.
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00:06:54,050 --> 00:06:57,790
Since we have computed the
entire table of the joint PMF,
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00:06:57,790 --> 00:06:59,110
this shouldn't be
too difficult.
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00:06:59,110 --> 00:07:04,055
In particular, for the
denominator, the probability
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that k is ever equal to
2, we just look at the
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column right here.
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00:07:10,150 --> 00:07:13,560
So the entries in this column
shows all the cases where k
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can be equal to 2.
114
00:07:15,250 --> 00:07:19,640
And in fact, we can see that k
can be equal to 2 only if n is
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00:07:19,640 --> 00:07:21,280
equal to 2 or 3.
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00:07:21,280 --> 00:07:24,740
Clearly, if you toss the coin
fewer than 2 times, there's no
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00:07:24,740 --> 00:07:26,760
chance that we'll get 2 heads.
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00:07:26,760 --> 00:07:30,340
So to get this probability right
here, we'll add up the
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number in these two cells.
120
00:07:34,310 --> 00:07:39,980
So we get P N, K, little
n, and 2 divided
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by 1/16 plus 3/32.
122
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123
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Now, the numerator, again, can
be read off from the table
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right here.
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00:07:51,080 --> 00:07:54,170
In particular, this tells us
that there are only two
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possibilities.
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Either n is equal to
2 or n equal to 3.
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00:07:59,140 --> 00:08:03,370
When n is equal to 2, we know
this quantity gives us 1/16
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reading off this cell divided
by 1/16 plus 3/32
130
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for n equal to 2.
131
00:08:14,590 --> 00:08:18,780
And the remaining probability
goes to the case where n is
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equal to 3.
133
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So this is 3 divided by 32,
1/16 plus 3/32, which
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simplifies to 2/5 and 3/5.
135
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136
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And this distribution gives
us the following plot.
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And this completes our problem.
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