# Homework Solutions

7.

Q.E.D

8.

1.4.4 Solution: The final calculation of
refers not
to a
*
single
*
draw of one ball from an urn containing three,
but rather to a composite experiment comprising more than one
stage. While it is true that {two black, one white } is the only
fixed collection of balls for which a random choice is black with
probability
, the composition of the urn is
*
not
determined
*
prior to the final draw.

After all, if Carroll's argument were correct then it would apply
also in the situation when the urn originally contains just one
ball, either black or white. The final probability is now
, implying that the original ball was one half black
and one half white! Carroll was himself aware of the fallacy in
this argument.

1.4.5 Solution: (a) (i)

(ii).

(iii).

(b) Let , and suppose the presenter possesses a coin which falls with heads upward with probability . He flips the coin before the show, and adopts strategy (i) if and only if the coin shows heads, and otherwise strategy(iii). The probability in question is now

You never lose by swapping, but whether you gain depends on the
presenter's protocol.

(c) Let D denote the first door chosen, and consider the following protocols:
(iv) If D conceals a goat, open it. Otherwise open one of the
other two doors at random. In this case
.

(v) If D conceals a car, open it. Otherwise open the unique
remaining door which conceals a goat. In this case
.

1.5.1 Solution:

1.5.2 Solution: Suppose and . If , then and are determined by distinct independent rolls, and are therefore independent.

For the case
we have that

=P(
,
, and
rolls show
same number)

P(
and
rolls both show
shows
)

so we have pair-wise independence.

But

Therefore not independence.

1.5.7 Solution:

(a)

(b)

(c) Only in the trivial cases when children are either almost
surely boys or almost surely girls.

(d) No.

1.8.5 Solution:

1.8.6 Solution:

1.8.19 Solution:

(d)

(c)

(b)

(a)

1.8.20 Solution: We condition on the reuslt of the first toss. If this is a head, then we require an odd number of heads in the next tosses. Similarly, if the first toss is a tail, we require an even number of heads in the next tosses.

Hence
=Prob. of even hands after n tosses

=P(even number of n-1 tosses)
P(tails on nth)

+P(odd number of n-1 tosses)
P(heads on nth)

with

As an alternative to induction, we may seek a solution of the
form
Hence

1.8.30 In general , there are
different combinations.
ways of having different birthdays,
probability of being all different,
of two of them are the same. let