Asked by Anonymous
                A game has a box containing four marbles - three red and one green. The player draws one marble at a time without replacement until the green marble is drawn. After that, the game is over. 
The game ends in four different ways. What is the probability of drawing a green marble for each of these possible outcomes?
            
        The game ends in four different ways. What is the probability of drawing a green marble for each of these possible outcomes?
Answers
                    Answered by
            Damon
            
    green on first try = 1/4
green on second try:
red first = 3/4
then green = 1/3
so
3/4 * 1/3 = 1/4
green on third try
red first = 3/4
red second = 2/3
green third = 1/2
3/4 * 2/3 * 1/2 = 1/4
green o fourth try
red first = 3/4
red second = 2/3
red third = 1/2
green fourth = 1/1
would you believe 1/4 again :)
    
green on second try:
red first = 3/4
then green = 1/3
so
3/4 * 1/3 = 1/4
green on third try
red first = 3/4
red second = 2/3
green third = 1/2
3/4 * 2/3 * 1/2 = 1/4
green o fourth try
red first = 3/4
red second = 2/3
red third = 1/2
green fourth = 1/1
would you believe 1/4 again :)
                    Answered by
            Anonymous
            
    There is a shooting game that consists of 10 targets. The player must use a bow to hit all the targets in order to win a prize. What is the probability of hitting all 10 targets if the player has 15 arrows?
    
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