Question

determine if each situation represents an independent or dependent probability. then find the probability. 1. a bag contains 2 striped cubes, 3 dotted cubes, 4 white cubes and 3 red cubes. what is the probability of drawing a white cube, not replacing it, and then drawing a dotted cube? 2. a bag contains 5 striped cubes, 3 dotted cubes, 4 white cubes and 3 red cubes. what is the probability of drawing a red cube, replacing it, and then drawing a white cube? 3. a bag contains 2 striped cubes, 5 dotted cubes, 5 white cubes and 3 red cubes. what is the probability of drawing two striped cubes in succession, without replacing the first cube drawn? 4. a pocket contains 3 pennies, 2 nickels, 1 quarter and 4 dimes. what is the probability of randomly choosing a dime, replacing it, and then drawing a penny? 5. a pocket contains 3 pennies, 2 nickels, 3 quarters and 4 dimes. what is the probability of randomly choosing a quarter two times in a row if the first coin drawn is not replaced? 6. at her party, luz will allow her guests to choose their party favor at random from a shopping bag that contains 6 i-tunes™ gift cards, 5 starbucks™ gift cards and 4 amc™ gift cards. when a guest selects a party favor, the item will be theirs to take home and no new items will be added to the shopping bag. what is the probability of a guest drawing an i-tunes™ gift card, keeping it, and then another guest drawing a second i-tunes™ gift card?

Explanation:

Problem 1

Step1: Find total number of cubes

Total cubes = 2 (striped) + 3 (dotted) + 4 (white) + 3 (red) = 12.

Step2: Probability of white cube

Probability of white cube, \( P(W) = \frac{4}{12} = \frac{1}{3} \).

Step3: Probability of dotted cube (no replacement)

After removing 1 white cube, total cubes = 11. Probability of dotted cube, \( P(D) = \frac{3}{11} \).

Step4: Combined probability (dependent)

Since no replacement, events are dependent. \( P = \frac{1}{3} \times \frac{3}{11} = \frac{1}{11} \).

Step1: Total number of cubes

Total cubes = 5 + 3 + 4 + 3 = 15.

Step2: Probability of red cube

\( P(R) = \frac{3}{15} = \frac{1}{5} \).

Step3: Probability of white cube (with replacement)

Since replaced, total cubes remain 15. \( P(W) = \frac{4}{15} \).

Step4: Combined probability (independent)

Events are independent (replacement). \( P = \frac{1}{5} \times \frac{4}{15} = \frac{4}{75} \).

Step1: Total number of cubes

Total cubes = 2 + 5 + 5 + 3 = 15.

Step2: Probability of first striped cube

\( P(S_1) = \frac{2}{15} \).

Step3: Probability of second striped cube (no replacement)

After removing 1 striped cube, total cubes = 14, striped cubes = 1. \( P(S_2) = \frac{1}{14} \).

Step4: Combined probability (dependent)

\( P = \frac{2}{15} \times \frac{1}{14} = \frac{1}{105} \).

Answer:

Dependent, probability is \(\frac{1}{11}\)

Problem 2