Question

A game involving a pair of dice pays you $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36. what is your approximate probability of losing money in one play of the game?

Answer 1

0.56 or 56%

Explanation:

Let x = winning amount in dollars, in a one play of the game.

there taxes values = -6, -2 and 4.

and P(x = -6) =
`(6)/(36)`

and P(x = -2) =
`(14)/(36)`

and P(x = 4) =
`(16)/(36)`

P (loosing money in one play of game) = P (x ≤ -2)

and P(x ≤ -2) = P ( x = -6) + P ( x = -2)

=
`(6)/(36)` +
`(14)/(36)` =
`(20)/(36)`

= 0.555555556 ≈ 0.56

Probability of losing money in one play of the game is 0.56 or 56%

Answer 2

Let the winnings in dollars in the single play of a game be "X".

Thus, in our case the only three values that X can take is +4 (or 4), -2 and -6.

It is also given that
`P(4)=(16)/(36), P(-2)= (14)/(36)` and
`P(-6)=(6)/(36)`.

Therefore, the probability of losing money in a single play of the game will be the sum of the probability of the losses.

Thus,
`P(loss in one play)=P(X=-6)+P(X=-2)=(6)/(36)+ (14)/(36)=(20)/(36)= (5)/(9)`.

Thus, the approximate probability of losing money in one play of the game is
`(5)/(9)` or 55.56% (approximately).