Additional Problem Set 6 Solutions

1
.
​
1
.
1
.
The event would be of “all coin flips being the same” ​
2
8
=
1
4
1
.
2
.
The event would be “Having exactly one coin turn up on Tails”​
3
8
1
.
3
.
The event would be “Having the first coin turn up on Heads”​
4
8
=
1
2
2
.
P(A⋃B⋃C)=
|A⋃B⋃C|
|Ω|
by the inclusion exclusion principle we have​
|A⋃B⋃C|
|Ω|
=
|A|+|B|+|C|-|A⋂B|-|A⋂C|-|B⋂C|+|A⋂B⋂C|
Ω
=
|A|
|Ω|
+
|B|
|Ω|
+
|C|
|Ω|
-
|A⋂B|
|Ω|
-
|A⋂C|
|Ω|
-
|B⋂C|
|Ω|
+
|A⋂B⋂C|
|Ω|
=P(A)+P(B)+P(C)-P(A⋂B)-P(A⋂C)-P(B⋂C)+P(A⋂B⋂C)
3
.
Let T be the event of drawing the card numbered 2 exactly 3 times.
Let S be the event of the sum of the three draws being 12.
​
Let us compute the size of T.
Choose the places in which the 2’s appear C(3,2)
Choose the remaining number 4
So
12
​
​
Let us compute the size of S by computing it as a union of disjoint subsets.
​
S
1
="drawing two 2's and an 8"
​
​
S
2
="Drawing three 4's"
​
​
S
3
="Drawing a 2, a 4, and a 6"
​
​
​
|
S
1
|=3,|
S
2
|=1,|
S
3
|=6
, so
|S|=10
.
Note that
S
1
=S⋂T
so P(T|S) =
3/10
4
.
We will define our sample space Ω to be choosing a coin and then the result of tossing it 6 times.​Let H be the event “Choosing the coin with two heads”Let E be the event “Getting 6 heads after tossing the chosen coin 6 times”​P(E) =
1
65
1
6
2
64+
1
65
The first summand calculating the probability of choosing a regular coin and it obtaining heads, and the second summand the probability of choosing the two headed coinP(H) = P(H⋂E) =
1
65
since the coin with two heads always turns up heads.​So P(H|E) =
1
65
1
65
1
6
2
64+
1
65
=
1
1+1
=
1
2
​​
5
.
Let
w=thenumberofballsinUrn1,b=thenumberofblackballsinUrn1.
​​Thus P(“drawing a white ball”) =
1
2
w
w+b
+
1
2
20-w
40-(w+b)
with constraints
0<=w<=25
, and
w<=b<=50
.​
In[]:=
Reverse@SortByFlattenTablew,b,
1
2
w
Max[w+b,1]
+
1
2
(20-w)
Max[40-(w+b),1]
,{w,0,20},{b,0,20},1,Last[[1;;5]]
Out[]=
19,20,
29
39
,1,0,
29
39
,18,20,
14
19
,2,0,
14
19
,17,20,
27
37

Taking the first few elements we see that the highest probability is
29
39
=0.74359
so we want 19 white and 20 black in one urn and then 1 white in the other.