Page 77 - Probability and Statistical Inference

54 1. Notions of Probability

at random. Given that the selected marble was green, find the probability that
(i) the urn #1 was selected;
(ii) the urn #2 was selected;
(iii) the urn #3 was selected.
{Hint: Try and use the Bayes Theorem.}
1.4.10 (Exercise 1.2.1 Continued) One has three fair dice which are red,
yellow and brown. The three dice are rolled on a table at the same time. Con-
sider the following events:
A : The total score from all three dice is 10
B : The total score from the red and brown dice exceeds 8
Are A, B independent events?
1.4.11 (Example 1.4.4 Continued) Show that P(A) = P(B) assuming that
there are respectively m, n green and blue marbles in the urn to begin with.
This shows that the result P(A) = P(B) we had in the Example 1.4.4 was not
just a coincidence after all.
1.4.12 Show that . {Hint: Can the binomial theorem, namely
(a+b) = , be used here?}
n
1.4.13 Suppose that we have twenty beads of which eight are red, seven are
green and five are blue. If we set them up in a row on a table, how many
different patterns are possible?
1.4.14 Suppose that four boys and four girls are waiting to occupy a chair
each, all placed in a row adjacent to each other. If they are seated randomly,
what is the probability that the boys and girls will alternate? {Hint: The total
possible arrangement is 8!. Note that each arrangement could start with a boy
or a girl.}
1.4.15 Suppose that we have n different letters for n individuals as well as n
envelopes correctly addressed to those n individuals. If these letters| are ran-
domly placed in these envelopes so that exactly one letter goes in an envelope,
then what is the probability that at least one letter will go in the correct enve-
lope? Obtain the expression of this probability when we let n go to ∞. {Hint:
th
Let A be the event that the i letter is stuffed in its correct envelope, i = 1, ..., n.
i
We are asked to evaluate P(A ∪ ... ∪ A ). Apply the result from the Exercise
n
1
1.3.6. Observe also that P(A ) = (n–1)!/n!, P(A ∩ A ) = (n–2)!/n!, P(A ∩ A j
i
i
j
i
∩ A ) = (n–3)!/n!, ..., P(A ∩ ... ∩ A ) = 1/n!, for i ≠ j ≠ k... . Now all there
n
k
1
is left is to count the number of terms in the single, double, triple sums and so
on. The answer for P(A ∪...∪ A ) should simplify to 1 – 1/2! + 1/3! – ... +
n
1
(–1) 1/n! which is approximately 1 – e for very large n. See (1.6.13).}
n–1
–1