Problem Sheet I solutions
Probability and Stochastic Processes
1. (a) The variable Y is called a Bernoulli or an indicator random variable. Note that, for
a Bernoulli random variable.
E[Y ] = p.1 + (1 − p).0 = p = P r(Y = 1) = 1/2.
(b) The probability of having j successes when the coin is flipped n times or the probability of j successes in any n independent experiments is defined
by the following
probability distribution on j = 0, 1, 2, ..., n. P r(X = j) = nj pj (1 − p)( n − j)
2. (a) There are 36 equally outcomes, just 10 of them contain exactly one six. The answer
5
is therefore 10
36 = 18
(b) A die shows an odd number with probability
1 1
1
2.2 = 4.
1
2;
by independence, P(both odd)=
(c) Consider S as the sum, and {i, j} as the event in which the first is i and the second
3
is j. Then P(S = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 36
.
(d) Similarly
P(S divisible by 3) = P(S = 3) + P(S = 9) + P(S = 12)
= {P(1, 2) + P(2, 1)}
+ {P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1)}
+ {P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) + P(6, 6)}
12
= 13 .
= 36
3. (a) By independence, P(n − 1 tails, followed by a head) = 2−n .
(b) If n is odd, P(#heads
cardinality of the set A. If n is
n = #tails) = 0; #A denotes the
1
even, there are n/2 sequences of outcomes with 2 n heads and 12 n tails. Any given
of heads and tails has probability 2−n ; therefore P(#heads = #tails) = 2−n
sequence
n
n/2 .
(c) There are n2 sequences containing 2 heads and n − 2 tails. Each sequence has
probability 2−n , and therefore P(exactly two heads) = n2 2−n .
(d) Clearly
P(at least 2 heads)=1-P(no heads)- (eaxactly one head)= 1 − 2−n − n1 2−n .
4. The sample space is S = {0, 1, ..., 9, so the sets of outcomes corresponding to the above
events are
A = {1, 3, 5, 7, 9}, B = {3, 6, 9}, C = {0, 1, 2, 3, 4}
.
If we assume that the outcomes are equally likely, then
P [A] = P [{1}] + P [{3}] + P [{5}] + P [{7}] + P [{9}] =
P [B] = P [{3}] + P [{6}] + P [{9}] =
1
3
10
5
10
Problem Sheet I solutions
Probability and Stochastic Processes
P [A] = P [{0}] + P [{1}] + P [{2}] + P [{3}] + P [{4}] =
5
10
Using the hint,
5
3
2
6
+
−
= .
10 10 10
10
where we used the fact that A ∩ B = {3, 9}, so P [A ∩ B] = 2/10.
Similarly,
P [A ∪ B] = P [A] + P [B] − P [A ∩ B] =
P [A ∪ B ∪ C] = P [A] + P [B] + P [C] − P [A ∩ B] − P [A ∩ C] − P [B ∩ C] + P [A ∩ B ∩ C]
=
3
5
2
2
1
1
5
+
+
−
−
−
+
10 10 10 10 10 10 10
=
9
10
5. The receiver can correct a single error, but it will make the wrong decision if the channel
introduces two or more signals. If, we view each transmission as a Bernoulli trial in which
a ”success” corresponds to the introduction of an
the probability
of two or more
3error, then
3
2
errors in three Bernoulli trials is P [K 2] = 2 (0.001) (.999)+ 3 (.001)3 3(10)−6
6. Choosing x and y between zero and one, the sample space would be unit square. If we
draw the areas corresponding to events A, B, C they all would have regions with area
equal to 1/2, therefore:
1
1
1
P [A] = , P [B] = , P [C] = .
2
2
2
Considering the area of a line corresponding to x = 0.5 which is zero, P [D] would be zero
as well.
2
Exercise 8:
Exercise 9: