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Answered Same Day May 14, 2021

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Suraj answered on May 14 2021

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1. Let X random variable denotes the possible profits for the player on each ticket.
A. probability distribution is given by
     X
     -2
     8
     78
     P(X)
     0.8
     0.1
     0.1
B. E(X) == = -2* 0.8 + 8* 0.1 + 78*0.1 = 7
C. Since the mean is 7$ so If the player play a long run he will eventually make money.
2. n = 40 (total items)
p = Probability of defective = 0.02
X = No. of defective items
Now, we will find λ which is the parameter of Poisson distribution.
λ = np = .02 x 40 = 0.8
Now as we know, Poisson distribution
A. P(X= 40) = /40! = 0.0
B. P(X ≥1) = 0.55
C. Probability of defective is = 0.02
Total items = 40
So 40* 2/100 = 4/5 items are defective out of 40.
3.
(A). Since we know that the of the probability distribution is 1.
Hence P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1
0.08 + 0.10 + 0.12 + 0.05 + 0.20 + P(5) + 0.12 =1
0.67 + P(5)= 1
P(5) = 0.33
(B).P(both day they rent 4 cars ) = 2* P(4) = 2* 0.20 =0.40
4.
     Gender
    Passed
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     Total
     Male...

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