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?

Part (a)

If you win $12165, then you really net 12165-35 = 12130 dollars when you consider the ticket fee. So this is the true amount of money you win, or take home at the end of the day. This is before taxes.

Multiply 0.002 with 12,130 to get 0.002*12130 = 24.26

We'll use this later so let A = 24.26

The chances that you don't win are 1 - 0.002 = 0.998 which multiplies with -35 to indicate you lost $35 in playing the game. So we get B = 0.998*(-35) = -34.93

Lastly, add the values of A and B to get the expected value:

A+B = 24.26 + (-34.93) = -10.67 is the expected value.

On average, you expect to lose about $10.67 for any time you play the game.

Part (b)

0.125 is the probability of winning so 1-0.125 = 0.875 is the probability of losing.

Let's say you get really unlucky and lose 12 times in a row. Assuming each trial (aka case when you play the game) is independent, this would mean the probability of such an event is (0.875)^12 = 0.2014172, which is approximate.

Subtract that from 1 to get the probability of winning at least once

1 - (0.875)^12 = 1 - 0.2014172 = 0.7985828

which is also approximate. If we rounded to three decimal places, then it would be 0.799; I'm picking three decimals since 0.125 is to three decimal places. Round however you need to if otherwise.

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}