A basket contains 7 green marbles and 5 red marbles. A marble is taken from the basket at random; its color is recorded, then the marble is returned to the basket. A second marble is then taken from the basket at random, and its color is recorded.

What is the probability that the same color is recorded both times?

For this one, I started by doing 7/12*1/2=7/24, which I think would be the chance of green marbles being picked twice. Then, red marbles would be 5/12*1/3=5/36. Then, the probability would be 7/24*5/36=35/864. I'm pretty sure my solution is wrong somewhere, but I'm not sure where.

A basket contains 7 green marbles and 5 red marbles. Two marbles are taken from the basket at random without replacement (i.e., the first marbles is not put back before the second marbles is drawn). What is the probability that both marbles are the same color?

I would think the same way to do this one, but I think the end result would be wrong. Please help and thank you

DragonLord Mar 23, 2020

#1**+2 **

Green twice (with replacement): p_{g} = (7/12)^{2}

Red twice (with replacement): p_{r} = (5/12)^{2}

Green twice ** or** red twice (with replacement): p = p

Green twice (without replacement): p_{g} = (7/12)*(6/11)

Red twice (without replacement): p_{r} = (5/12)*(4/11)

Green twice** or** red twice (without replacement): p = p

Alan Mar 23, 2020

#1**+2 **

Best Answer

Green twice (with replacement): p_{g} = (7/12)^{2}

Red twice (with replacement): p_{r} = (5/12)^{2}

Green twice ** or** red twice (with replacement): p = p

Green twice (without replacement): p_{g} = (7/12)*(6/11)

Red twice (without replacement): p_{r} = (5/12)*(4/11)

Green twice** or** red twice (without replacement): p = p

Alan Mar 23, 2020