Answer is 1-(125/216) = 0.42
Probability That you will win = 1 - (Your Number would not come) = 1 - (5*5*5/6*6*6) = 91/216 = 0.42
But the problem asks what is the probability of winning
money, not merely getting the desired number on the dice. Every event has some weightage.
So we have to calculate the probability that the man earns more money than he loses.
Probability of getting the no. on all 3 dice = 1/6* 1/6 * 1/6 =
1/216Probability of getting the no. on only 2 dice = (1/6* 1/6 * 5/6) + (1/6* 5/6 * 1/6) +(5/6* 1/6 * 1/6) =
15/216Probability of getting the no. on only 1 die = (1/6* 5/6 * 5/6) + (5/6* 1/6 * 5/6) +(5/6* 5/6 * 1/6) =
75/2161/216 + 15/216 + 75/216 = 91/ 216 = 1 - (125/216)
The point is, in 216 trials the man (ideally) wins 3*1 + 2*15 + 1*75 =
108.
but he loses 1*125 =
125. Essentially, the man never wins money? ABsurd....?? :shocked:
Can someone help?