ssrs export to pdf barcode font (g) The probability of neither four nor club can be denoted by P(4r > Cr). But 4r > Cr in Software

Encoding Quick Response Code in Software (g) The probability of neither four nor club can be denoted by P(4r > Cr). But 4r > Cr

(g) The probability of neither four nor club can be denoted by P(4r > Cr). But 4r > Cr
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CHAPTER 1 Basic Probability
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Therefore, P(4r > Cr) P[(4 < C)r] 1 1 1 P(4 < C) [P(4) B 1 13 P(C) 1 4 1 R 52 P(4 > C)] 9 13
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We could also get this by noting that the diagram favorable to this event is the complement of the event shown circled in Fig. 1-8. Since this complement has 52 16 36 sample points in it and each sample point is assigned probability 1 > 52, the required probability is 36 > 52 9 > 13.
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Fig. 1-8
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1.7. A ball is drawn at random from a box containing 6 red balls, 4 white balls, and 5 blue balls. Determine the probability that it is (a) red, (b) white, (c) blue, (d) not red, (e) red or white.
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(a) Method 1 Let R, W, and B denote the events of drawing a red ball, white ball, and blue ball, respectively. Then P(R) ways of choosing a red ball total ways of choosing a ball 6 4 6 15 2 5
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Method 2 Our sample space consists of 6 4 5 15 sample points. Then if we assign equal probabilities 1 > 15 to each sample point, we see that P(R) 6 > 15 2 > 5, since there are 6 sample points corresponding to red ball. 4 4 (b) P(W) 6 4 5 15 5 5 1 (c) P(B) 6 4 5 15 3 2 3 (d) P(not red) P(Rr) 1 P(R) 1 by part (a). 5 5 (e) Method 1 P(red or white) P(R < W ) ways of choosing a red or white ball total ways of choosing a ball 6 6 4 4 5 10 15 2 3
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This can also be worked using the sample space as in part (a). Method 2 P(R < W) P(Br) 1 P(B) 1 1 3 2 by part (c). 3
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Method 3 Since events R and W are mutually exclusive, it follows from (4), page 5, that P(R < W) P(R) P(W) 2 5 4 15 2 3
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CHAPTER 1 Basic Probability
Conditional probability and independent events 1.8. A fair die is tossed twice. Find the probability of getting a 4, 5, or 6 on the first toss and a 1, 2, 3, or 4 on the second toss.
Let A1 be the event 4, 5, or 6 on first toss, and A2 be the event 1, 2, 3, or 4 on second toss. Then we are looking for P(A1 > A2). Method 1 P(A1 > A2) P(A1) P(A2 u A1) P(A1) P(A2) 3 4 6 6 1 3
We have used here the fact that the result of the second toss is independent of the first so that P(A2 u A1) P(A2). Also we have used P(A1) 3 > 6 (since 4, 5, or 6 are 3 out of 6 equally likely possibilities) and P(A2) 4 > 6 (since 1, 2, 3, or 4 are 4 out of 6 equally likely possibilities). Method 2 Each of the 6 ways in which a die can fall on the first toss can be associated with each of the 6 ways in which it can fall on the second toss, a total of 6 6 36 ways, all equally likely. Each of the 3 ways in which A1 can occur can be associated with each of the 4 ways in which A2 can occur to give 3 4 12 ways in which both A1 and A2 can occur. Then P(A1 > A2) This shows directly that A1 and A2 are independent since P(A1 > A2) 1 3 12 36 1 3
1.9. Find the probability of not getting a 7 or 11 total on either of two tosses of a pair of fair dice.
The sample space for each toss of the dice is shown in Fig. 1-9. For example, (5, 2) means that 5 comes up on the first die and 2 on the second. Since the dice are fair and there are 36 sample points, we assign probability 1 > 36 to each.
3 4 6 6
P(A1) P(A2)
Fig. 1-9
If we let A be the event 7 or 11, then A is indicated by the circled portion in Fig. 1-9. Since 8 points are included, we have P(A) 8 > 36 2 > 9. It follows that the probability of no 7 or 11 is given by P(Ar) 1 P(A) 1 2 9 7 9
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