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a. A contest entrant has a 0.002 probability of winning $12,165. If this is the only prize and the fee is $35, then find the expected value of winning the contest. b. The probability of winning a lottery is 0.125. What is the probability of winning AT LEAST ONCE in twelve trials?

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Answer:

The right answer is:

(a) -10.67

(b) 0.7986

Step-by-step explanation:

(a)

According to the question,

X                          P(X)                  X.P(X)                 X2.P(X)

12130                  0.002               24.26                 294274

-35                      0.998               -34.93                1222.55      

Now,

[tex]\Sigma x.P(x) = -10.67[/tex]

or,

[tex]\Sigma x^2.P(x) = 295496.35[/tex]

hence,

The mean will be:

[tex]\Sigma x.P(x) = -10.67[/tex]

(b)

According to the question,

n = 12

p = 0.125

q = 1 - p

  = 0.875

Now,

⇒ [tex]P(X=x) = \binom{n}{x} p^x q^{n-x}[/tex]

By substituting the values, we get

⇒ [tex]P(X \geq 1)=1-(\binom{12}{0} 0.125^0. 0.875^{12-0})[/tex]

⇒                  [tex]=1-(1 (0.125^0) (0.875^{12}))[/tex]

⇒                  [tex]=1-(1(1.0)(0.2014))[/tex]

⇒                  [tex]=1-(0.2014)[/tex]

⇒                  [tex]=0.7986[/tex]    

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