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Random experiments, sample spaces, and events 1.1. A card is drawn at random from an ordinary deck of 52 playing cards. Describe the sample space if consideration of suits (a) is not, (b) is, taken into account.
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(a) If we do not take into account the suits, the sample space consists of ace, two, . . . , ten, jack, queen, king, and it can be indicated as {1, 2, . . . , 13}. (b) If we do take into account the suits, the sample space consists of ace of hearts, spades, diamonds, and clubs; . . . ; king of hearts, spades, diamonds, and clubs. Denoting hearts, spades, diamonds, and clubs, respectively, by 1, 2, 3, 4, for example, we can indicate a jack of spades by (11, 2). The sample space then consists of the 52 points shown in Fig. 1-5.
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Fig. 1-5
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CHAPTER 1 Basic Probability
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1.2. Referring to the experiment of Problem 1.1, let A be the event {king is drawn} or simply {king} and B the event {club is drawn} or simply {club}. Describe the events (a) A < B, (b) A > B, (c) A < Br, (d) Ar < Br, (e) A B, (f ) Ar Br, (g) (A > B) < (A > Br).
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(a) A < B (b) A > B (c) Since B {either king or club (or both, i.e., king of clubs)}. {both king and club} {club}, Br {king of clubs}. {heart, diamond, spade}.
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{not club}
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Then A < Br (d ) Ar < Br
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{king or heart or diamond or spade}. {not king of clubs} {any card but king of clubs}.
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{not king or not club}
This can also be seen by noting that Ar < Br (e) A B {king but not club}.
(A > B)r and using (b).
This is the same as A > Br (f) Ar Br
{king and not club}. {not king and club} Br Ar > (Br )r {any club except king}.
{not king and not not club }
This can also be seen by noting that Ar (g) (A > B) < (A > Br )
Ar > B. {king}.
{(king and club) or (king and not club)} A.
This can also be seen by noting that (A > B) < (A > Br )
1.3. Use Fig. 1-5 to describe the events (a) A < B, (b) Ar > Br.
The required events are indicated in Fig. 1-6. In a similar manner, all the events of Problem 1.2 can also be indicated by such diagrams. It should be observed from Fig. 1-6 that Ar > Br is the complement of A < B.
Fig. 1-6
Theorems on probability 1.4. Prove (a) Theorem 1-1, (b) Theorem 1-2, (c) Theorem 1-3, page 5.
(a) We have A2 A1 < (A2 A1) where A1 and A2 P(A2) so that Since P(A2 A1) P(A2 P(A1) A1) A1 are mutually exclusive. Then by Axiom 3, page 5: P(A2 P(A2) A1) P(A1) P(A1).
0 by Axiom 1, page 5, it also follows that P(A2)
(b) We already know that P(A) 0 by Axiom 1. To prove that P(A) by Theorem 1-1 [part (a)] and Axiom 2, P(A) (c) We have S S < \. Since S > \ P(S) P(S) 1
1, we first note that A ( S. Therefore,
\, it follows from Axiom 3 that P(S) P(\) or P(\) 0
CHAPTER 1 Basic Probability
1.5. Prove (a) Theorem 1-4, (b) Theorem 1-6. (a) We have A < Ar S. Then since A > Ar P(A < Ar) i.e., (b) We have from the Venn diagram of Fig. 1-7, (1) Then since the sets A and B A <B A < [B (A > B)] \, we have P(S) or P(Ar) 1 P(A) P(A) P(Ar) 1
(A > B) are mutually exclusive, we have, using Axiom 3 and Theorem 1-1, P(A < B) P(A) P(A) P[B P(B) (A > B)] P(A > B)
Fig. 1-7
Calculation of probabilities 1.6. A card is drawn at random from an ordinary deck of 52 playing cards. Find the probability that it is (a) an ace, (b) a jack of hearts, (c) a three of clubs or a six of diamonds, (d) a heart, (e) any suit except hearts, (f) a ten or a spade, (g) neither a four nor a club.
Let us use for brevity H, S, D, C to indicate heart, spade, diamond, club, respectively, and 1, 2 , c, 13 for ace, two, c , king. Then 3 > H means three of hearts, while 3 < H means three or heart. Let us use the sample space of Problem 1.1(b), assigning equal probabilities of 1 > 52 to each sample point. For example, P(6 > C) 1 > 52. (a) P(1) P(1 > H or 1 > S or 1 > D or 1 > C ) P(1 > H) P(1 > S) P(1 > D) P(1 > C )
1 1 1 1 1 52 52 52 52 13 This could also have been achieved from the sample space of Problem 1.1(a) where each sample point, in particular ace, has probability 1 > 13. It could also have been arrived at by simply reasoning that there are 13 numbers and so each has probability 1 > 13 of being drawn. 1 (b) P(11 > H) 52 1 1 1 (c) P(3 > C or 6 > D) P(3 > C ) P(6 > D) 52 52 26 1 1 1 13 1 c (d) P(H) P(1 > H or 2 > H or c13 > H) 52 52 52 52 4 This could also have been arrived at by noting that there are four suits and each has equal probability1>2 of being drawn. 1 3 (e) P(Hr) 1 P(H) 1 using part (d) and Theorem 1-4, page 6. 4 4 (f) Since 10 and S are not mutually exclusive, we have, from Theorem 1-6, P(10 < S) P(10) P(S) P(10 > S) 1 13 1 4 1 52 4 13 (4 < C)r.
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