A game has an expected value to you of $1200. It costs $1200 to play, but if you win, you receive $100,000 (including your $1200 bet) for a net gain of $98 comma 800. What is the probability of winning? Would you play this game? Discuss the factors that would influence your decision.

A) the probability of winning is 0.24%.

B) Yes i will play the game

C) Despite the fact that the probability of winning is very low, one should play the game because the expected value of the game is positive.

Step-by-step explanation:

Expected value of X is denoted by;

E(X) = x1•p1 + x2•p2 + xn•pn

Where;

xi is the observation and pi is the probability of xi

Now, let's make p the probability of the winning bet and 1 - p be the probability of losing the game

If the bet is win, the net gain is $98,800 and if the bet is lose, the loss is -$1200.

Hence the probability distribution will be;

For xi = $98,800, pi = p

For xi = -$1,200, pi = 1 - p

So;

E(X) = Σxi.pi

Thus;

1200 = 98800p - 1200(1 - p)

1200 = 98800p - 1200 + 1200p

1200 + 1200 = 100000p

2400 = 100000p

p = 2400/100000

p = 0.024

Thus, the probability of winning is 0.24%.

Despite the fact that the probability of winning is very low, one should play the game because the expected value of the game is positive.

the answer is n