Math 411 - Fall 2014 - 3rd Homework.
1. Decide if the following statements are true or false: (If true, provide a proof, if not
provide a counterexample)
– a) If E is independent of F and E is independent of of G, then E is independent
of F ∪ G.
– b) If E is independent of F and E is independent of G, and F G = ∅, then E
is independent of F ∪ G.
– c) If E is independent of F and E is independent of G, and E is independent
of F G, then G is independent of EF .
2. Suppose that there are n possible
Pn outcomes of a trial, with outcome i resulting with
probability pi , 1 ≤ i ≤ n,
i=1 pi = 1. If two independent trials are observed
compute
1. what is the probability that the results of the trials are the same.
2. what is the probability that the result of the second one is larger than the first
one.
3. You ask your neighbor to water a sickly plant while you are ine vacation. Without
water, it will die with probability 0.8; with water, it will die with probability 0.15.
You are 90% certain that your neighbor will remember water the plant. If the plant
is dead upon your return, what is the probability that your neighbor forgot to water
it?
4. A team plays in a tournament of two rounds. They estimate that the probability
to win the first opponent is 0.8 and to win the second game is 0.5. What is the
probability that the team wins the tournament?
5. What is more likely to occur?
1. A fair coin lands on heads.
2. Three independent trials, each of which is a success with probability 0.8, all
result in success.
Due to Wednesday October 1st.
Hints-Solutions
1. Check page 474 (3.22).
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2. Let E:=“the results are equal“, L:=the first result is larger, S:= the first result is
smaller. Let Ai,j := the i-th result in the j trial. 1 ≤ i ≤ n and 1 ≤ j ≤ 2. Then,
since Ai,1 , Ai,2 are independent and the events Ai,1 Ai,2 , 1 ≤ i ≤ n are mutually
disjoint,
n
X
n
P(E) = P(∪i=1 Ai,1 Ai,2 ) =
P(Ai,1 Ai,2 ) =
i=1
n
X
P(Ai,1 )P(Ai,2 ) =
i=1
n
X
p2i .
i=1
By symmetry we have that P (L) = P (S) and P (L) + P (S) + P (E) = 1.
3. Let D=“the plant is dead” and W:=“it was watered”. The
P(W c |D) =
P (D|W c )P (W c )
P(DW c )
=
P(D)
P (D|W c )P (W c ) + P (D|W )P (W )
4. 0.5 × 0.8.
5. P(SSS) = (P(S))3 = (0.8)3 > 21 .
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