Winning at Coins

There is a $\frac{1}{10}$ probability of picking the card labeled $10$ and winning $\$ 20$ . There is a $\frac {9}{10}$ probability of picking any other card and winning nothing. Thus, by the Linearity of Expectation Value, our answer is $\$ \left( \dfrac {9}{10} \cdot 0 + \dfrac {1}{10} \cdot 20 \right) = \$ \boxed {2}.$

There are $10$ cards , so the probability to get number $10$ card is $\frac {1}{10}$

because it just $1$ card from $10$ cards

cause the probability to get number $10$ is $\frac {1}{10}$

so the expected amount is $\frac {1}{10} \times 20$ dollars = $\boxed {2}$ dollars.

On average, if you pick up the cards 10 times, you'll win $20 once. In 10 tries you get $20, so on average you'd win $2 "per try".

total sum=20 the ways can be attained=10 so 20/10=2

mathematical expectation is the sum of probabilities multiplied by the rate..... here, p(selecting the card labeled 10) is 1/1o therefore, expectation E= 1/10 * 20 = 2.

In ten tries you are expected to get the "10"card once, winning you 20 dollars 20 divided by 10 equals 2

prob of winning=1/10 and winning amount =20$..we can expect 1/10th of amount which is equal to 2$

(1/10) * 20 = 2

You are expected to win $2. this is because you have a 1/10 chance of winning $20 and a 9/10 chance of winning $0. you add the answer to 1/2 X 20 and 9/10 x 0. the answer to this is 2. so the average winning is $2.