Answer :
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
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...