Question

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?

Answer 1

The right answer is:

(a) -10.67

(b) 0.7986

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,

`\Sigma x.P(x) = -10.67`

or,

`\Sigma x^2.P(x) = 295496.35`

hence,

The mean will be:

`\Sigma x.P(x) = -10.67`

(b)

According to the question,

n = 12

p = 0.125

q = 1 - p

= 0.875

Now,

⇒
`P(X=x) = \binom{n}{x} p^x q^(n-x)`

By substituting the values, we get

⇒
`P(X \geq 1)=1-(\binom{12}{0} 0.125^0. 0.875^(12-0))`

⇒
`=1-(1 (0.125^0) (0.875^(12)))`

⇒
`=1-(1(1.0)(0.2014))`

⇒
`=1-(0.2014)`

⇒
`=0.7986`