A bag contains seven green marbles and nine yellow marbles. You randomly select three marbles.

Question

A bag contains seven green marbles and nine yellow marbles. You randomly select three marbles. What is the probably that all three marbles are green when (a) you replace each marble before selecting the next marble, and (b) you do not replace each marble before selecting the next marble? Write each probability as a decimal round to the nearest thousandth. Then compare the probabilities.

Answer 1

Probability = the number of wanted outcomes/the number of possible outcomes

The probability of drawing a green marble is

number of green marbles/ total number of marbles in the bag

7/16

If you replace the marble each time, each time your chance of drawing green is 7/16. If this is done 3 times, the probability is

(7/16)(7/16)(7/16) = 0.084

For not replacing green, there is no impact on the first draw, the probability of getting green on the first draw is still 7/16. If you do not replace, this means that there are no longer 16 marbles in the bag for the second draw, there are now 15. And since you already drew one green one, there are no longer 7 green ones, there are now 6. So probability for green on the second draw is 6/15. If you draw a 3rd time and there isn't replacement, there are no longer 15 marbles in the bag for the second draw, there are now 14. And since you already drew one green one, there are no longer 6 green ones, there are now 5. So probability for green on the third draw is 5/14. Combining the 3 draws together:

(7/16)(6/15)(5/14) = 0.063

The probability of drawing 3 with replacement is higher than without.