Let $X$ be a random variable denoting the amount of winnings in any given round of the game. There is some probability $p$ of winning where $P(X=99,500)=p$ , while $P(X=-500)=1-p$ , assuming these are the only two possible outcomes. So we have

$\mathbb E[X]=\displaystyle\sum_xxP(X=x)=99,500p-500(1-p)=100$

$100,000p-500=100\implies p=\dfrac3{500}=0.006$

8 0

3 years ago

Section A

sesenic [268]

_______________________________

8 0

3 years ago

What is 0.422 rounded to the nearest thousandth

Maru [420]

.4 is tenths .42 is hundredths and .422 is thousandths

8 0

4 years ago