5 numbers chosen randomly without replacement. "B" represents number of even numbers, this random variable has...
5 numbers chosen randomly without replacement. "B" represents number of even numbers, this random variable has this probability:
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| p(B=x) | 0.02693 | 0.15989 | .33858 | .31977 | .13464 | .02020 |
number of odd #s chosen would then be 5-x, if x is even #s chosen. "C" represents difference b/w # of even and # of odd chosen, --> C= 2B-5
a. probability that exactly 1 even # chosen?
b. probability at most 1 even # chosen?
c. prob. that more even than odd #s chosen?
d. show that mean of B is =2.4359 .
Homework Answers
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a. P(X=1) = .15989
b. P(X<=1) = P(X=0)+P(X=1) = .02693+.15989 = 0.18682
c. P(more even than odd) = P(X=3) + P(X=4) + P(X=5) = .31977+.13464+.02020 = 0.47461
d. The mean is x weighted by p(B=x) = 0*0.02693 + 1*0.15989 +2*.33858+3*.31977+4*.13464+5*.02020 = 2.4359 <Hence Proved>
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