ssrs 2008 r2 barcode font Special Probability Distributions in Software

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CHAPTER 4 Special Probability Distributions
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Method 2 (using formula) (a) P(3 heads) 3 1 3 1 0 a ba b a b 2 3 2 1 8 3 8
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(b) P(2 tails and 1 head) (c) P(at least 1 head)
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3 1 2 1 1 a ba b a b 2 2 2 P(1, 2, or 3 heads) P(1 head)
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P(2 heads)
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P(3 heads) 3 1 3 1 0 a ba b a b 2 3 2 7 8 ,
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3 1 1 1 2 a ba b a b 2 1 2 Alternatively, P(at least 1 head)
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3 1 2 1 1 a ba b a b 2 2 2
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P(no head) 3 1 0 1 3 a ba b a b 2 0 2 P(0 tails or 1 tail) P(0 tails) P(1 tail) 3 1 2 1 a ba b a b 2 2 2 1 2 7 8
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(d) P(not more than 1 tail)
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3 1 3 1 0 a ba b a b 2 3 2
It should be mentioned that the notation of random variables can also be used. For example, if we let X be the random variable denoting the number of heads in 3 tosses, (c) can be written P(at least 1 head) P(X 1) P(X 1) P(X 2) P(X 3) 7 8
We shall use both approaches interchangeably.
4.2. Find the probability that in five tosses of a fair die, a 3 will appear (a) twice, (b) at most once, (c) at least two times.
Let the random variable X be the number of times a 3 appears in five tosses of a fair die. We have Probability of 3 in a single toss Probability of no 3 in a single toss (a) P(3 occurs twice) P(X 2) P(X 5 1 2 5 3 a ba b a b 6 2 6 1) P(X 0) 625 3888 P(X 1) p q 1 6 1 p 5 6
(b) P(3 occurs at most once)
5 1 0 5 5 a ba b a b 6 0 6 3125 7776 (c) P(3 occurs at least 2 times) P(X P(X 2) 2) P(X 3) P(X 4) 3125 7776
5 1 1 5 4 a ba b a b 6 1 6 3125 3888
5) 5 1 5 5 0 a ba b a b 6 5 6
5 1 2 5 3 a ba b a b 6 2 6 625 3888 125 3888
5 1 3 5 2 a ba b a b 6 3 6 25 7776 1 7776
5 1 4 5 1 a ba b a b 6 4 6 763 3888
CHAPTER 4 Special Probability Distributions
4.3. Find the probability that in a family of 4 children there will be (a) at least 1 boy, (b) at least 1 boy and at least 1 girl. Assume that the probability of a male birth is 1> 2.
(a) P(1 boy) P(3 boys) Then P(at least 1 boy) P(1 boy) 1 4 Another method P(at least 1 boy) (b) P(at least 1 boy and at least 1 girl) 1 1 1 P(no boy) P(no boy) 1 16 1 16 1 1 4 a b 2 1 1 16 15 16 3 8 1 4 P(2 boys) 1 16 15 16 P(3 boys) P(4 boys) 4 1 1 1 3 a ba b a b 2 1 2 4 1 3 1 1 a ba b a b 2 3 2 1 , 4 1 , 4 P(2 boys) P(4 boys) 4 1 2 1 2 a ba b a b 2 2 2 4 1 4 1 0 a ba b a b 2 4 2 3 8 1 16
P(no girl) 7 8
We could also have solved this problem by letting X be a random variable denoting the number of boys in families with 4 children. Then, for example, (a) becomes P(X 1) P(X 1) P(X 2) P(X 3) P(X 4) 15 16
4.4. Out of 2000 families with 4 children each, how many would you expect to have (a) at least 1 boy, (b) 2 boys, (c) 1 or 2 girls, (d) no girls
Referring to Problem 4.3, we see that (a) Expected number of families with at least 1 boy (b) Expected number of families with 2 boys (c) P(1 or 2 girls) P(1 girl) P(1 boy) P(2 girls) P(2 boys) 1 4 3 8 5 8 1250 125 2000a 15 b 16 1875 3 2000a b 8 750
2000 P(2 boys)
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