crazypriya wrote:
From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize. From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.
Total number of marbles % likelihood that Jennifer wins
7
10
20
21
30
35
Dear
crazypriya,
I'm happy to help.
I don't know that there's a good way to do this other than plugging in one at at time. Let
N = total number of marbles.
If N = 7, then there's a 100% chance of picking two red marbles. That doesn't work.
If N = 10, then there's a 7/10 chance of getting red on the first pick, and 6/9 = 1/3 of getting red on the second pick. That's a (7/10)*(1/3) = 7/30 chance, which is approx. 23.33%. That doesn't work.
If N = 20, then there's a 7/20 chance of getting red on the first pick, and 6/19 of getting red on the second pick. That's a (7/20)*(6/19) = 21/190 chance, which is slightly more than 10 percent. That doesn't work.
If N = 21, then there's a 7/21 = 1/3 chance of getting red on the first pick, and 6/20 = 3/10 of getting red on the second pick. That's a (1/3)*(3/10) = 1/10, or 10% chance. This works.
I disagree with the order of answers given in the spoiler. The total number of marbles is 21, and the % probability that Jennifer wins is 10.
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)