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Llana [10]
3 years ago
10

People please help me in this math homework

Mathematics
2 answers:
viktelen [127]3 years ago
8 0
For the last 2 questions,

1. 9/10
2. 7/12
natita [175]3 years ago
4 0
At the top 3/6 and 1/2 are equal to 0.5
exercise 1 is 9/12
exercise 2 is 2/5
 and at the bottom it is 9/10 for the first one and for the second on it is 7/12
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Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the
kodGreya [7K]

Answer:

  • p=0.7103 (4-game series)
  • p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters  n=4, p=0.6:

P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\  i=4,5,6,7

#We find probabilities for the different values of i:

P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659

Hence, probability of the stronger team winning overall is:

=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103

#Define Y as the random variable for winning 2/3 games.:

P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

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