20. In a game at a charity fundraiser, players bet $1 and have a chance to either win $2 ($1 plus the original bet) or lose their $1. The probability of winning at this game is. 493. What is the average amount a player can expect to lose on each play of the game?

Answer: A = - 1.4cents

Average amount Expected to lose is 1.4cent

Step-by-step explanation:

Given;

In the game,

i. There is 0.493 probability of winning additional $1

ii. There is (1-0.493 = 0.507) probability of losing $1

Therefore,

The average amount a player is expected to lose is give by;

A = amount Expected to win - amount Expected to lose.

A = 0.493 * $1 - 0.507 * $1

A = $0.493 - $0.507

A = - $0.014

A = - 1.4cents

Average amount Expected to lose is 1.4cent.

Therefore, the more games you play, the more the chances of losing ...

Full question

In a game at a charity fundraiser, players bet $1 and have a chance to either win $2 ($1 plus the original bet) or lose their $1. The probability of winning at this game is. 493. What is the average amount a player can expect to lose on each play of the game?

A ...7 cents

B ...07 cents

C. 14 cents

D. 7 cents

E. 1.4 cents

Answer : 1.4 cents

explanation:

The probability of winning is 0.493 therefore the probability of losing is 0.507 (1-0.493)

Multiply each outcome by their probability

1*0.493 + (-1) * 0.507=0.014

It adds up to a negative outcome

-1.4 cents