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## Object Picking Probability - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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### Complexity=10, Mode=replace

 1 A bag contains 6 marbles: 2 red marbles, 1 yellow marble, and 3 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble? 2 A bag contains 8 coins: 1 penny, 5 nickels, and 2 dimes. If you take one coin out, put it back, and take another coin out, what is the probability that you'll get 1 nickel followed by 1 penny?

### Complexity=10, Mode=no replace

 1 A bag contains 6 balls: 1 red ball, 1 yellow ball, and 4 blue balls. If you take one ball out, don't put it back, and take another ball out, what is the probability that you'll get 1 red ball followed by 1 yellow ball? 2 A bag contains 6 marbles: 3 red marbles, 2 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 yellow marble followed by 1 red marble?

### Complexity=10

 1 A bag contains 7 jelly beans: 3 red jelly beans, 2 yellow jelly beans, and 2 blue jelly beans. If you take two jelly beans at the same time, what is the probability that you'll get 1 blue jelly bean and 1 red jelly bean? 2 A bag contains 6 jelly beans: 1 red jelly bean, 2 yellow jelly beans, and 3 blue jelly beans. If you take two jelly beans at the same time, what is the probability that you'll get 1 yellow jelly bean and 1 red jelly bean?

### Complexity=10, Mode=replace

1A bag contains 6 marbles: 2 red marbles, 1 yellow marble, and 3 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble?
Solution
Begin by noting that since the marbles are replaced, each draw does not depend on previous draws and thus the draws are independent.
P(1 blue marble on first pick) = 3/6 = 1/2.
P(1 yellow marble on second pick) = 1/6.
P(picking 1 blue marble then picking 1 yellow marble) = (1/2) × (1/6) = 1/12
2A bag contains 8 coins: 1 penny, 5 nickels, and 2 dimes. If you take one coin out, put it back, and take another coin out, what is the probability that you'll get 1 nickel followed by 1 penny?
Solution
Begin by noting that since the coins are replaced, each draw does not depend on previous draws and thus the draws are independent.
P(1 nickel on first pick) = 5/8.
P(1 penny on second pick) = 1/8.
P(picking 1 nickel then picking 1 penny) = (5/8) × (1/8) = 5/64

### Complexity=10, Mode=no replace

1A bag contains 6 balls: 1 red ball, 1 yellow ball, and 4 blue balls. If you take one ball out, don't put it back, and take another ball out, what is the probability that you'll get 1 red ball followed by 1 yellow ball?
Solution
Begin by noting that the balls are not replaced and thus each draw depends on previous draws. Thus the draws are dependent.
P(1 red ball on first pick) = 1/6.
P(1 yellow ball on second pick) = 1/(total balls - 1) = 1/5.
P(picking 1 red ball then picking 1 yellow ball) = (1/6) × (1/5) = 1/30
2A bag contains 6 marbles: 3 red marbles, 2 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 yellow marble followed by 1 red marble?
Solution
Begin by noting that the marbles are not replaced and thus each draw depends on previous draws. Thus the draws are dependent.
P(1 yellow marble on first pick) = 2/6 = 1/3.
P(1 red marble on second pick) = 3/(total marbles - 1) = 3/5.
P(picking 1 yellow marble then picking 1 red marble) = (1/3) × (3/5) = 1/5

### Complexity=10

1A bag contains 7 jelly beans: 3 red jelly beans, 2 yellow jelly beans, and 2 blue jelly beans. If you take two jelly beans at the same time, what is the probability that you'll get 1 blue jelly bean and 1 red jelly bean?
Solution
Ways of choosing 1 blue jelly bean and 1 red jelly bean = blue jelly bean # × red jelly bean # = 2 × 3 = 6.
Ways of choosing any 2 jelly beans = 7 × 6 ÷ 2 = 21.
P(choosing 1 blue jelly bean and 1 red jelly bean simultaneously) =
(Ways of choosing 1 blue jelly bean and 1 red jelly bean) / (Ways of choosing any 2 jelly beans) = 6/21 = 2/7.   