99 points no fake answers
a standard six-sided dice is repeatedly rolled until it lands on a two. what is the probability that the first two is rolled after the first six rolls?
a) 33%
b) 41%
c) 53%
d) 67%

suppose you buy one ticket for $1 out of a lottery of 1,000 tickets where the for the one winning ticket is to be $500. what is your expected value?
a) $0.00
b) $-1.00
c) $-0.40
d) $-0.50

Step-by-step explanation

so the probability of not getting 2 first is 5/6, and this happens six times so we do (5/6)^6 and this is 0.33489797668 , or 33%, so A.

The formula for the expected value would be:

Expected Value = (Probability of Winning)*(Prize if won) + (Probability of not winning)*(Prize if not won)

The price you get if you don't win is 0, so we can ignore the 2nd term. Now, the probability of winning is 1/1000, because you bought 1 ticket out of 1000. Since the prize is $500:

Exp. Value = (1/1000)*500 = $0.50 , so its D

I hope this helps!

1). A 33%

2). D $-0.50

Step-by-step explanation:

1).  The probability of not getting a 2 first is 5/6 that is 1-1/6=5/6 = 0.8333

number of occurrence is 6

probability = (5/6)^6

= 0.8333^6

= 0.33489797668 ,

= approximately 0.33 to two decimal place

= 33%

The formula for the expected value would be:

Expected Value = (Probability of Winning)*(Prize if won) + (Probability of not winning)*(Prize if not won)

The price you get if you don't win is 0,

so expected value = (probability of winning)*(prize if won)

So, the probability of winning is 1/1000, because you bought 1 ticket out of 1000.

Since the prize is $500:

Exp. Value = (1/1000)*500 = $0.50 , so its D

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

i cant answer this, theres no table. sorry man

step-by-step explanation: