Respuesta :
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 [tex]P(4)=\frac{16}{36}, P(-2)= \frac{14}{36}[/tex] and [tex]P(-6)=\frac{6}{36}[/tex].
Therefore, the probability of losing money in a single play of the game will be the sum of the probability of the losses.
Thus, [tex]P(loss in one play)=P(X=-6)+P(X=-2)=\frac{6}{36}+ \frac{14}{36}=\frac{20}{36}= \frac{5}{9}[/tex].
Thus, the approximate probability of losing money in one play of the game is [tex]\frac{5}{9}[/tex] or 55.56% (approximately).
Answer:
0.56 or 56%
Step-by-step 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) = [tex]\frac{6}{36}[/tex]
and P(x = -2) = [tex]\frac{14}{36}[/tex]
and P(x = 4) = [tex]\frac{16}{36}[/tex]
P (loosing money in one play of game) = P (x ≤ -2)
and P(x ≤ -2) = P ( x = -6) + P ( x = -2)
= [tex]\frac{6}{36}[/tex] + [tex]\frac{14}{36}[/tex] = [tex]\frac{20}{36}[/tex]
= 0.555555556 ≈ 0.56
Probability of losing money in one play of the game is 0.56 or 56%
