Math Practice Online > free > lessons > Florida > 9th grade > Object Picking Probability

These sample problems below for Object Picking Probability were generated by the MathScore.com engine.

## Sample Problems For Object Picking Probability

### Complexity=10, Mode=replace

 1 A bag contains 9 balls: 7 red balls, 1 yellow ball, and 1 blue ball. If you take one ball out, put it back, and take another ball out, what is the probability that you'll get 1 red ball followed by 1 blue ball? 2 A bag contains 6 marbles: 1 red marble, 3 yellow marbles, and 2 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?

### Complexity=10, Mode=no replace

 1 A bag contains 8 balls: 6 red balls, 1 yellow ball, and 1 blue ball. 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 blue ball followed by 1 red ball? 2 A bag contains 10 marbles: 2 red marbles, 3 yellow marbles, and 5 blue marbles. 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 blue marble followed by 1 yellow marble?

### Complexity=10

 1 A bag contains 9 coins: 5 pennies, 2 nickels, and 2 dimes. If you take two coins at the same time, what is the probability that you'll get 1 nickel and 1 dime? 2 A bag contains 10 marbles: 4 red marbles, 2 yellow marbles, and 4 blue marbles. If you take two marbles at the same time, what is the probability that you'll get 1 red marble and 1 blue marble?

### Complexity=10, Mode=replace

1A bag contains 9 balls: 7 red balls, 1 yellow ball, and 1 blue ball. If you take one ball out, put it back, and take another ball out, what is the probability that you'll get 1 red ball followed by 1 blue ball?
Solution
Begin by noting that since the balls are replaced, each draw does not depend on previous draws and thus the draws are independent.
P(1 red ball on first pick) = 7/9.
P(1 blue ball on second pick) = 1/9.
P(picking 1 red ball then picking 1 blue ball) = (7/9) × (1/9) = 7/81
2A bag contains 6 marbles: 1 red marble, 3 yellow marbles, and 2 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) = 2/6 = 1/3.
P(1 yellow marble on second pick) = 3/6 = 1/2.
P(picking 1 blue marble then picking 1 yellow marble) = (1/3) × (1/2) = 1/6

### Complexity=10, Mode=no replace

1A bag contains 8 balls: 6 red balls, 1 yellow ball, and 1 blue ball. 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 blue ball followed by 1 red 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 blue ball on first pick) = 1/8.
P(1 red ball on second pick) = 6/(total balls - 1) = 6/7.
P(picking 1 blue ball then picking 1 red ball) = (1/8) × (6/7) = 3/28
2A bag contains 10 marbles: 2 red marbles, 3 yellow marbles, and 5 blue marbles. 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 blue marble followed by 1 yellow 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 blue marble on first pick) = 5/10 = 1/2.
P(1 yellow marble on second pick) = 3/(total marbles - 1) = 3/9 = 1/3.
P(picking 1 blue marble then picking 1 yellow marble) = (1/2) × (1/3) = 1/6

### Complexity=10

1A bag contains 9 coins: 5 pennies, 2 nickels, and 2 dimes. If you take two coins at the same time, what is the probability that you'll get 1 nickel and 1 dime?
Solution
Ways of choosing 1 nickel and 1 dime = nickel # × dime # = 2 × 2 = 4.
Ways of choosing any 2 coins = 9 × 8 ÷ 2 = 36.
P(choosing 1 nickel and 1 dime simultaneously) =
(Ways of choosing 1 nickel and 1 dime) / (Ways of choosing any 2 coins) = 4/36 = 1/9.