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 of the 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 ( 52 5 ) . (a) ere are ( 13 1 ) ways to choose the face value of a and for each of these ( 12 3 ) 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 )( 12 3 )( 4 2 )( 4 1 ) 3 ( 52 5 ) = 0.4226. (b) ere are ( 13 2 ) ways to choose the face values of a , b and for each of these ( 11 1 ) ways to choose the face value of c . ere are ( 4 2 ) 2 choices of suits for the cards with value a , b , and ( 4 1 ) such choices for the other card. erefore P (two pair) = ( 13 2 )( 11 1 )( 4 2 ) 2 ( 4 1 ) ( 52 5 ) = 0.04754. (b) ( points) ink of the dice as having different colors, so that there are 6 5 possible outcomes. (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 × 2/6 5 = 0.09259. (b) ere are ( 6 1 )( 5 3 ) 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 )( 5 3 ) 5!/2! = 0.4630. (c) ere are ( 6 2 )( 4 1 ) 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 )( 4 1 ) 5!/(2!2!) = 0.2315.

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Page 1: MATH 302 SOLUTIONS - win.tue.nlthulshof/302/Solutions2-302.pdf · Homework 2 Math 302 2013-14 Author: Tim Hulshof Created Date: 9/21/2013 8:17:01 PM

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.

Page 2: MATH 302 SOLUTIONS - win.tue.nlthulshof/302/Solutions2-302.pdf · Homework 2 Math 302 2013-14 Author: Tim Hulshof Created Date: 9/21/2013 8:17:01 PM

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.

Page 3: MATH 302 SOLUTIONS - win.tue.nlthulshof/302/Solutions2-302.pdf · Homework 2 Math 302 2013-14 Author: Tim Hulshof Created Date: 9/21/2013 8:17:01 PM

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.