1. Suppose you draw a card from a standard 52-card deck of cards and then roll a die. Are the events “draw a king” and “roll an odd number” independent or dependent? Explain.

Solution: independent

2.  You randomly pick one donut from a box with 6 glazed, 3 jelly-filled, and 3 cream-filled donuts, and you eat the donut. Then you randomly pick a second donut from the box, and eat it. Are the events “picking a jelly-filled donut” and “picking a glazed donut” independent or dependent in this situation? Explain.

Solution: dependent

3.  You randomly pick one donut from a box with 6 glazed, 3 jelly-filled, and 3 cream-filled donuts, and you put it back. Then you randomly pick a second donut from the box, and you put it back. Are the events “picking a jelly-filled donut” and “picking a glazed donut” independent or dependent in this situation? Explain.

Solution: independent

Let's Suppose that we have a bag of 20 marbles consisting of 8 red marbles, 5 blue marbles, and 7 yellow marbles. This is illustrated in the image below:

1. What is the sample space for drawing two marbles out the bag and recording the color?

Solution: 

S = {BB, BR, BY, RR, RB, RY, YY, YR, YB}

where B = blue, R= red, Y= yellow.

2. If 2 of the 20 marbles are randomly selected with replacement (meaning, one marble is drawn, the color is recorded, then put back in the bag), what is the probability of getting two red marbles?

Solution: P(both red with replacement) = 8/20 *8/20 = (8/20)^2 = 0.16. Notice that since the marble was replaced, the probability of drawing a red marble on the second draw is also 8/20- the same probability as drawing the red marble on the first draw.

3. If 2 of the 20 marbles are randomly selected without replacement, what is the probability of getting two red marbles?

Solution: P(both red without replacement) = 8/20 *7/19 = 0.14736842105. Notice that since the marble was not replaced, the probability of drawing a red marble on the second draw becomes 7/19- there are now only 19 total marbles, only 7 of them are red (the one drawn first- which is assumed to be red, is not replaced)

4. If 2 of the 20 marbles are randomly selected without replacement, what is the probability of getting a red and then a blue marble?

Solution: P(red then blue without replacement) = 8/20 *5/19 = 0.105. Notice that since the (red) marble was not replaced, the probability of drawing a blue marble on the second draw becomes 5/19- there are now only 19 total marbles, 5 of which are still blue.

5. If 2 of the 20 marbles are randomly selected without replacement, what is the probability of getting a red and a blue marble, in any order?

Solution:

P(red and blue without replacement) = [latex](\frac{8}{20})\cdot(\frac{5}{19}) +(\frac{5}{20})\cdot(\frac{8}{19}) \approx 0.211[/latex]

**Notice that since we are asking about the probability of getting red and blue in any order (that means, getting red then blue OR blue then red) this is a multiplication and an "or" probability problem, so we add the probabilities P(red then blue) + P(blue then red).

6. If 2 of the 20 marbles are randomly selected with replacement, what is the probability of getting a red and then a blue marble?

Solution:

P(red and blue with replacement) = [latex](\frac{8}{20})\cdot(\frac{5}{20}) = 0.1[/latex]

7. If 2 of the 20 marbles are randomly selected with replacement, what is the probability of getting a red and a blue marble, in any order?

P(red and blue with replacement) = [latex](\frac{8}{20})\cdot(\frac{5}{20}) +(\frac{5}{20})\cdot(\frac{8}{20}) = 0.2[/latex]

A bin contains 60 light bulbs, of which 10 are defective. Consider the following scenarios.

1. If 4 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones.

Solution:

P(all bulbs are good with replacement) = [latex](\frac{50}{60})^4 \approx 0.482[/latex]

2. If 4 light bulbs are randomly selected from the bin without replacement, find the probability that all the bulbs selected are good ones.

Solution:

P(all bulbs are good without replacement) = [latex](\frac{50}{60})(\frac{49}{59})(\frac{48}{58})(\frac{47}{57}) \approx 0.472[/latex]

3. If 4 light bulbs are randomly selected from the bin without replacement, find the probability that the first bulb was good, the second bulb was defective, the third bulb was defective, and the fourth bulb was good. Let's appreciate this sequence of events as GDDG.

Solution:

P(GDDG without replacement) = [latex](\frac{50}{60})(\frac{10}{59})(\frac{9}{58})(\frac{49}{57}) \approx 0.0189[/latex]

4. If 4 light bulbs are randomly selected from the bin with replacement, find the probability that the first bulb was good, the second bulb was defective, the third bulb was defective, and the fourth bulb was good.

Solution:

P(GDDG without replacement) = [latex](\frac{50}{60})(\frac{10}{60})(\frac{9}{60})(\frac{49}{60}) \approx 0.017[/latex]

5. If 2 light bulbs are randomly selected from the bin with replacement, find the probability that exactly one of the bulbs is good.

Solution:

P(exactly one good bulb with replacement) = [latex](\frac{50}{60})(\frac{10}{60}) + (\frac{10}{60})(\frac{50}{60}) \approx 0.278[/latex]

6. If 2 light bulbs are randomly selected from the bin without replacement, find the probability that exactly one of the bulbs is good.

Solution:

P(exactly one good bulb without replacement) = [latex](\frac{50}{60})(\frac{10}{59}) + (\frac{10}{60})(\frac{50}{59}) \approx 0.282[/latex]

**Notice how this solution also uses the addition rule!

The table below describes the smoking habits of a group of asthma sufferers.

• If two people are randomly selected without replacement from the 1124 subjects, find the probability that they are both heavy smokers.

P(both heavy smokers without replacement) = [latex](\frac{73}{1124})(\frac{72}{1123}) \approx 0.004[/latex]

• If two people are randomly selected with replacement, find the probability that they are both women.

P(both women with replacement) = [latex](\frac{579}{1124})^2\approx 0.265[/latex]

• If four people are randomly selected with replacement, find the probability that they are all women.

P(all women with replacement) = [latex](\frac{579}{1124})^4\approx 0.070[/latex]

• If two people are randomly selected with replacement, find the probability that the first person is a man and the second person is a woman. Would this probability be unusual?

P(first is a man and then women with replacement = [latex](\frac{545}{1124})(\frac{579}{1124}) \approx 0.25[/latex]

Not unusual, because 0.25 means will happen 1 in 4 times. In general if probability is larger than 0.05 (or 5%) than it is not considered unusual.

• If two people are randomly selected without replacement, find the probability that the first person is a nonsmoker and the second person is an occasional smoker. Would this probability be unusual?

P(first is a nonsmoker and then occasional smoker without replacement = [latex](\frac{790}{1124})(\frac{7}{1123}) \approx 0.044[/latex]

Unusual since 0.044 is less than 0.05 (which is our cutoff for determining unusualness).