math 302 solutions - win.tue.nlthulshof/302/solutions2-302.pdf · homework 2 math 302 2013-14...
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MATH 302 SOLUTIONS
Q [ points]. In this problem, explain all factors arising in your solution.(a) Compute the probability that a poker hand contains:
(a) One pair (aabcd with a,b,c,d distinct card ranks). e answer is ..(b) Two pairs (aabbc with a,b,c distinct card ranks). e answer is ..
(b) Poker dice is played by simultaneously rolling dice. Compute the probabilities ofthe following outcomes:(a) All ve dice have distinct numbers. e answer is ..(b) One pair (aabcd with a,b,c,d distinct numbers). e answer is ..(c) Two pairs (aabbc with a,b,c distinct numbers). e answer is ..
Solution . (a) ( points) e size of the sample space is the number of poker hands,which is
(525
).
(a) ere are(13
1
)ways to choose the face value of a and for each of these
(123
)ways
to choose the face values of b,c,d . ere are(4
2
)choices of suits for the cards
with value a, and(4
1
)3 such choices for the other cards. erefore
P (pair) =(13
1
)(123
)(42
)(41
)3(525
) = 0.4226.
(b) ere are(13
2
)ways to choose the face values of a,b and for each of these
(111
)ways to choose the face value of c . ere are
(42
)2 choices of suits for the cardswith value a,b, and
(41
)such choices for the other card. erefore
P (two pair) =(13
2
)(111
)(42
)2(41
)(525
) = 0.04754.
(b) ( points) ink of the dice as having different colors, so that there are 65 possibleoutcomes.(a) ere are ways to assign a value to the rst die, ways to assign a value to
the second die, ..., so the probability is 6×5×4×3×2/65 = 0.09259.(b) ere are
(61
)(53
)ways to choose the values of a and b,c,d , and there are 5!/2!
ways to assign these values to the dice, so the probability is 6−5(6
1
)(53
)5!/2! =
0.4630.(c) ere are
(62
)(41
)ways to choose the values of a,b and c , and there are 5!/(2!2!)
ways to assign these values to the dice, so the probability is 6−5(6
2
)(41
)5!/(2!2!) =
0.2315.
Q [ points]. A coin is tossed until a head appears twice in a row. What is thesample space for this experiment? If the coin is fair, what is the probability that it will betossed exactly four times?
Solution ( points). e sample space is S = {2,3,4, . . .} where outcome j ∈ S is the out-come that the experiment terminates on the j th toss. Another way of describing the samplespace is S = {HH, THH, TTHH,HTHH, TTTHH, THTHH,HTTHH,…}. (Other answersare also possible.)( points) e outcome 4 ∈ S corresponds to the possibilities TTHH, HTHH, and thereare 24 = 16 possible results in a sequence of tosses, so the desired probability is 2
16 = 18 .
Q [ points]. Let E ,F,G be three events. Find expressions for the events thatof E ,F,G :
(a) only F occurs,(b) both E and F occur but not G ,(c) at least one event occurs,(d) at least two events occur,(e) all three events occur,(f) none occurs,(g) at most one occurs,(h) at most two occur.
Solution . .(a) F ∩E c ∩Gc ,(b) E ∩F ∩Gc ,(c) E ∪F ∪G ,(d) (E ∩F )∪ (E ∩G)∪ (F ∩G),(e) E ∩F ∩G ,(f) E c ∩F c ∩Gc ,(g) (E ∩F )c ∩ (E ∩G)c ∩ (F ∩G)c ,(h) (E ∩F ∩G)c .
Q [ points]. Show that the probability that exactly one of the events E or Foccurs (not both) is equal to P (E)+P (F )−2P (E ∩F ).
Solution . e event A that exactly one of the events E or F occurs (not both) is A =(E ∩F c )∪ (F ∩E c ), and this union is disjoint. erefore P (A) = P (E ∩F c )+P (F ∩E c ).We also have the disjoint unions E = (E ∩ F )∪ (E ∩ F c ) and F = (F ∩ E)∪ (F ∩ E c ), soP (E ∩F ) = P (E)−P (E ∩F c ) and P (F ∩E) = P (F )−P (F ∩E c ). Inserting these last twoequations into the equation for P (A) gives the desired result.
Q [ points]. Show that P (E c ∩F c ) = 1−P (E)−P (F )+P (E ∩F ).
Solution . Since E c ∩F c = (E ∪F )c (by De Morgan’s law), we haveP (E c ∩F c ) = 1−P (E ∪F ) = 1− [
P (E)+P (F )−P (E ∩F )],
which gives the desired result.