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K n^ k n
n P k p p k
b). Binomial probability law with parameters n and ap :
1 1 1 1
k n^ k n
n P k ap ap k
c). Binomial probability law with parameters n and (1-a)p :
2 2 2 2
[(1 ) ] 1 (1 )
k n k n
n P k a p a p k
c). Multinomial probability law with parameters n and p 1 =p, p 2 =ap, and p 3 =(1-a)p:
1 2 1 2
1 2
1 2 1 2 1 2 1 2
1 1 1 2
, , ( ) [(1 ) ]
k k n k k n
k k n
n P k k n k k ap a p p k k n k k
n a a p k k n k k
−^ −
a)
P k [ = 0 ]= p
P k [ = (^1) ] = p (^) ( 1 − p )
[ ] ( )
2 P k = 2 = p 1 − p
Problem 1: [GAR]2.
Problem 2: [GAR]2.
[ ] [ ] [ ] [ ] ( )
3 P k = 3 = 1 − P k = 0 − P k = 1 − P k = 2 = 1 − p
b) In general:
( ) ( 1 )^0
k P k = − p p ≤ k < m
1 1 1
0 0 0
m m m m k k
k k k
p P m P k p p p p p p
− − −
= = =
= − = − − = − − = − ×
∑ ∑ ∑
m m m
p p p p p
a). For a car to pay $k, it needs to be in the lot for (k-1)/2 < t < k/
1 1 2 2 2 2
[ ]= [ ] [ ] [ ]
= ( 1) for 0,1,2,
k k k
k k k k P Pay k dollars P t P t P t
e e e e k
− − − −
− = − = L
P($k) =
1 1 1 1 1 1 2 2 2 2 2 2 1 2 (1 ) (1 ) ( ) (1 )
k k k e e e e e e e
− − − − − − (^) − − − = − = − k=1,2,3,…
It is a geometric probability law with p =
1 e^2
− .
b). Unconditional
1 1 5 2 4 2 2 2 P k [ 5] ( e ) (1 e ) e e
− − (^) − −
= = − = −. For all values of k namely ( 1 ≤ k ≤ 5 ) it
is similar to part a.
Conditional: Means we have
5 1
(1 5) [ 5 / max 5] [ ]
k P k is P t e e
−
− −^ −
For other values of k 1 ≤ k ≤ 4 it is the same as part a. You can easily check that sum of all
probabilities is equal to 1.
P [k tosses required until heads comes up twice]
= P [one of the first k-1 tosses is heads AND the k
th
toss is heads] = P A ( I B )
Problem 3: [GAR]2.
Problem 4: [GAR]2.
y ( ) y ( ) x y x Sy
F y P x ≤ ∈
4
4 4 4
y ( )
y
y
F y
−^ ≤^ <
+^ =
4 4 4 4
y
y
+^ +^ =^ ≤^ <
+^ +^ +^ =
4
2 4
y
y
≤^ <
Fy(y)
-4 -2 2 4 y
1 1!^1
K n^ k^ k n^ k n
n (^) n P k p p p p p k K n k p
+ = − = − ×
+^ +^ −^ −^ −
k^ n^ k n n
p n^ P n^ k n k p p P k P k k p k n k p k
= × × − × − ⇒ + =
a) Let L be the duration of the match. If Fischar wins a match of L games, then L −1 draws
must occur before he wins. Summing over all possible lengths:
10 10 10 1 10
1 1
Fischer wins 0.3 0.4 0.3 1 0. 0.3 0.3 1 0.3 7
l l
l l
P
−
= =
= = = × = −
− ^
∑ ∑
b)
If is the length of the match, then iff 1 draws occur, followed by a win by either
player. The probability of a win by either player is 0.7. So:
X X = L L −
1
9
0 otherwise
l
X
l
P l P X l l
− (^) = = = = (^) =
Note that l =10 iff 9 draws occur.
a). probability of having 2 = P ( y = 2 ) = P 2
probability of not having 2 = 1 − P 2 (^) = P 0 (^) + P 1
3 7 3 7 2 2 2 0 1
So: number of 2's in 10 symbol =3 1 3 3
P P P P P P
b) Using Multinomial experiment. If Ni shows number of symbol 's in 10 symbols then i
N (^) 2 = 3, N (^) 0 + N 1 (^) = 7 ⇒ N (^) 1 = 7 − N (^) 0. when N 0 goes from 0 to 7. Therefore:
Problem 7:
Problem 8:
Problem 9:
c)
PX ( x )
e 1
e
−
x
FX (^) ( x )
x
e 1
e
−
x < 1 1
d)
( 2 1 )
1 1
k Z k k
P Z odd P Z P Z P k e e
∞ ∞ − −
= =
= = = + = + ⋅ ⋅ ⋅ = (^) ∑ − = − ∑
(^2 ) 2 2 2 1
k
k
e e e (^) e e e e e e e e e e e e e
∞ − − − =
∑
