Download MTH 202 : Probability and Statistics

MTH 202 : Probability and Statistics
Homework 2
6th January, 2017
(1) A man has n keys exactly one of which fits the lock. He tries
the keys one at a time, at each trial choosing at random from
the keys that were not tried earlier. Find the probability that
the r-th key tried is the correct key. [Ref : Exercise-3, Hoel,
Port, Stone, Page-46]
Solution : The total number of keys tried up to the r-th attempt (without replacement) is
n(n − 1) . . . (n − r + 2)(n − r + 1)
Now if the r-th key tried is the correct one, all the attempts
made up to (r − 1)-th attempt has to be a failure. Thus, keeping the correct key aside, the number of attempts (out of the
remaining (n − 1) keys) made up to the (r − 1)-th attempt is
(n − 1) . . . (n − r + 2)(n − r + 1)(1)
with only one choice left for the r-th attempt. Hence the required probability is
1
(n − 1) . . . (n − r + 2)(n − r + 1)
= .
n(n − 1) . . . (n − r + 2)(n − r + 1)
n
(2) If Sam and Peter are among n men who are arranged at random
in a line, what is the probability that exactly k men stand
between them? [Ref : Exercise-7, Hoel, Port, Stone, Page-46]
Solution : Keeping Sam and Peter and k men m1 , m2 , . . . , mk
out of the remaining n − 2 we treat the block :
∆ := Sam, m1 , m2 , . . . , mk , Peter
as a single component and distribute the remaining n − k − 2
men in left and right segment. Keeping the block ∆ fixed and
treating it as a object there are (n − k − 1)! ways to permute
these.
Now notice that :
(i) there are n−2
ways of choosing these k men,
k
(ii) there are 2! ways one can keep Sam and Peter at the ends
1
2
of these blocks (either left or right end),
(iii) once k men are chosen, there are k! ways of permuting them
within a block.
Hence out of the total n! ways to permute all people there are
2 n−2
k!(n − k − 1)! many correct alignments. Hence the rek
quired probability is
2 n−2
k!(n − k − 1)!
2(n − k − 1)
k
=
n!
n(n − 1)
(3) Suppose n balls are distributed in n boxes.
(i) What is the probability that exactly one box is empty?
(ii) Given that box 1 is empty, what is the probability that only
one box is empty?
(iii) Given that only one box is empty, what is the probability
that box 1 is empty?
[Ref : Exercise-10, Hoel, Port, Stone, Page-47]
Proof : Note that here both the balls and the boxes are distinguishable. For (i) let Ai denote the event that exactly box
number i is empty. Then as computed in Example 9 (where we
computed the probability of exactly box number 1 being empty;
the computation would be same for any particular box), we have
that
n
(n − 1)!
P (Ai ) = 2
(1 ≤ i ≤ n)
nn
Now the required probability is
n
n
n
n
[
X
(n − 1)!
n!
2
2
P ( Ai ) =
P (Ai ) = n ×
=
nn
nn
i=1
i=1
since the events A1 , . . . , An are mutually disjoint.
(ii) Let B1 denote the event that the box number 1 is empty.
While B1 happen the remaining boxes numbered 2, 3, . . . , n
could be arbitrarily filled. Hence we see here that A1 ⊆ B1
while Ai ∩ B1 = ∅ for any i 6= 1. Also while box number 1 is
empty, there are n balls each have a possible (n − 1) destinations, which mean there are (n − 1)n possible ways. Hence
P (B1 ) =
(n − 1)n
nn
3
Now the required conditional probability is
S
T
n
n
[
(n − 1)!
P (( ni=1 Ai ) B1 )
P (A1 )
2
P ( Ai |B1 ) =
=
=
P (B1 )
P (B1 )
(n − 1)n
i=1
(iii) Here we compute the conditional probability
TS
n
[
P (B1 ( ni=1 Ai ))
P (A1 )
1
Sn
Sn
=
=
P (B1 | Ai ) =
P ( i=1 Ai )
P ( i=1 Ai )
n
i=1
Sn
as we have noticed in part (i) that P ( i=1 Ai ) = nP (A1 ).
(4) Compute the probability of each of the following poker hands
of five cards :
(i) Straight flush (five cards of the same suit in a sequence),
(ii) Straight (five cards in a sequence, regardless of the suit),
(iii) One pair (face values of the form (w, w, x, y, z) where w, x, y
and z are distinct).
[Ref : Exercise-14, Hoel, Port, Stone, Page-47]
Solution : (i) Straight flush : five cards of the same suit in a
sequence (see Fig 1)
Figure 1
choices for a poker
Recall that there are always a total of 52
5
hand of five cards. Now note that there are 10 possible sequences that can be made out of a fixed suit :
[A, 2, 3, 4, 5], [2, 3, 4, 5, 6], . . . , [9, 10, J, Q, K], [10, J, Q, K, A]
which is possible because an ace ”A” can be treated as both
the least and the highest value in a sequence. Since there are
4
4 suits, the total number of such sequence is 4 × 10 and hence
the required probability is 40/ 52
5
(ii) Straight : five cards in a sequence, regardless of the suit
(see Fig 2)
Figure 2
Keeping the same symbols A, 2, 3, . . . , 10, J, Q, K as before there
are again 10 possible sequences out of these symbols. Now each
of these symbols can be chosen arbitrarily from the four possible suits. This give a required count 45 × 10. The required
probability is thus
45 × 10
52
5
(iii) One pair : face values of the form (w, w, x, y, z) where
w, x, y and z are distinct.
Recall that a face value mean the symbols A, 2, 3, . . . , 10, J, Q, K.
Now we can choose one face value among the 13 for the pair
and choose any two of the
four suits with the fixed face value.
4
These
would
give
13
×
choices. Note here that the number
2
4
comes because keeping the face value fixed the suits can be
2
permuted within them, e.g. in Fig. 3 (King of hearts, King of
spades) and (King of spades, King of hearts) would represent
the same hand. Now out of the remaining 12 face values 3 are
to be chosen and in each of the coordinates (x, y, z) a arbitrary
suit can occur. Since the face values are all different here, these
can be chosen in 12
× 43 ways. Hence the required probability
3
5
Figure 3
is
13 ×
4
2
×
12
3
× 43
52
5
(5) Cards are dealt from an ordinary deck of playing cards one at a
time until the first king appears. Find the probability that this
occurs with the n-th card dealt. [Ref : Exercise-16, Hoel, Port,
Stone, Page-48]
Solution : Since there are 52 cards are they are dealt without
replacement, the total number of choices up to the n-th stage
are
(52)n = 52(52 − 1) . . . (52 − n + 1)
(see the notation in Page-29, Hoel, Port, Stone). Now since
the event require the King to appear at n-th stage and there
are a total of 4 kings in the deck, the first (n − 1) times the
cards has to come from the remaining 52 − 4 = 48 choices, and
the last choice could be any of the 4 kings. Hence the required
probability is
(48)n−1 × 4
(52)n
(6) Find the probability that a poker hand of 5 cards will contain
no card smaller than 7, given that at least 1 card over 10, where
aces are treated as high cards. [Ref : Exercise-24, Hoel, Port,
Stone, Page-48]
Solution : The cards which at least 7 are the eight face values
given as : 7, 8, 9, 10, J, Q, K, A. Considering the four suits, there
6
are total 8 × 4 = 32 such cards. Similarly the cards over 10 are
J, Q, K, A and there are 4 × 4 = 16 such cards. To estimate
the probability we set C to be the event that no card is smaller
than 7 and D represent the event that at least one card is over
10. We compute
P (C|D) =
P (C ∩ D)
P (D)
To compute P (D) notice that out of the total 52
choices, there
5
36
would be 5 hands representing all five cards come from the
face values 2, 3, . . . , 10 (9 cards, 4 suits and hence 4 × 9 = 36
such choices). By removing these choices from the total we
compute P (D) as
52
− 36
5
5
P (D) =
52
5
Similarly while
computing P (C ∩ D) we notice that out of
the total 32
choices representing all cards have face values at
5
least 7, there are 16
choices representing no card are over 10,
5
i.e. these are among 7, 8, 9, 10 (4 cards, 4 suits and hence total
4 × 4 = 16 choices). By removing these from the total we have
P (C ∩ D) as
32
16
−
P (C ∩ D) = 5 52 5
5
Hence the required probability is
32
−
5
P (C|D) = 52
−
5
16
5
36
5
(7) If you hold three tickets to a lottery for which n tickets were
sold and 5 prizes are to be given, what is the probability that
you will win at least 1 prize? [Ref : Exercise-25, Hoel, Port,
Stone, Page-48]
Solution : We can compute the probability of no prizes won.
Since there are a total of n tickets sold, the
total number of
ways to purchase 3 tickets out of them is n3 . Now as there are
5 prizes among these (n ≥ 5), the number of ways of getting no
7
prize is n−5
(assuming the prizes to ticket numbers are already
3
decided). Hence the probability of winning at least one prize is
n−5
= 1 − P (winning no prize) = 1 − n3 3
Note that
n−5
3
n
3
n−3
5
n
5
=
which means in an
alternate way we can solve the same : there
are a total of n5 ways to assign the prizes to the tickets and
ensuring that no prizes to be won, there are n−3
ways to assign
5
the prizes to the tickets which are not purchased.
(8) A box of 10 washers contain 5 defective ones. What is the probability that two washers selected at random (without replacement) from the box are both good? [Similar to : Exercise-26,
Hoel, Port, Stone, Page-48]
Solution : The total number
of ways to select two washers
(without replacement) is 10
and the number of ways to choose
2 5
two goodones out of 5 are 2 . Hence the required probability
is 52 / 10
.
2