Question

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 1

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

Step-by-step 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

Answer 2

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...