voodoochild wrote:
In a $10 game of chance in Las Vegas, there are two identical bowls that contain 100 marbles. Each bowl contains black colored 95 marbles along with 5 different colored marbles: one blue, one yellow, one green, one red and one gold. You get to pick one marble randomly from each bowl. If you get a matching pair of colored marbles (blue-blue, gold-gold), you win $10000. What is the probability that you win the prize in one play of the game?
HEre's what I tried :
Method 1 = total pairs = 100; Total combinations = 100*100 => PRob = 1/100
Method 2 (really crash landed) --
Choose 1 of the black colored balls from the first bowl and then the second one => 95C1 * 95C1
Choose 1 of blue balls => 1C1 * 1C1 ...Now this will be done for all the five colors. Hence, total = 5*1c1 * 1c1 = 5
Total = (95^2 + 5)/(100^2) = CRASHED !
Can any of the experts please help me ?
Thanks
Note here that 'matching pair of colored marbles' implies only the 5 non-black colors. I agree that there is some ambiguity here - is black considered a colored marble or not? Well, if you are to win $10,000 from a $10 game, a pair of black marbles should not help you win that kind of money so I assume that we are only talking about the 5 different colored marbles.
Probability of picking a non black marble from a bowl = 5/100
Probability of picking the same color marble from the other bowl = 1/100
Probability of a matching non black pair = (5/100)*(1/100) = 5/10000 = 1/2000
On a side note, notice that the probability is against the player. If the probability of winning is 1/2000, for one's $10, one should get $20,000 if one wins. That is why, the house always wins and one should not gamble!