5.4b Statistics - Independent and Dependent Events

1) 1. Write whether the events listed in the following question are either “independent” or “dependent”. Find the probability of drawing a queen and then a second queen from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card.

The events of drawing a queen and then a second queen from a standard deck of 52 cards without replacement are dependent. This is because the outcome of the first event affects the probability of the second event. After drawing the first queen, there are now only 51 cards left in the deck, and only 3 of them are queens. Therefore, the probability of drawing a second queen is dependent on whether or not a queen was drawn in the first event.

Explanation

The events of drawing a queen and then a second queen from a standard deck of 52 cards without replacement are dependent. This is because the outcome of the first event affects the probability of the second event. After drawing the first queen, there are now only 51 cards left in the deck, and only 3 of them are queens. Therefore, the probability of drawing a second queen is dependent on whether or not a queen was drawn in the first event.

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2) 1. The owner of a one-man lawn mowing business owns four old and unreliable riding mowers. As long as one of the four is working he can stay productive. From past experience, one of the mowers is unusable 10 percent of the time, one 9 percent of the time, one 27 percent of the time and one 11 percent of the time. Find the probability that all three mowers are unusable on a given day.

The probability that all three mowers are unusable on a given day can be found by multiplying the probabilities of each mower being unusable. The probability of the first mower being unusable is 10%, the second mower is 9%, and the third mower is 27%. Therefore, the probability that all three mowers are unusable is 10% * 9% * 27% = 0.000267. This can also be expressed as a fraction, 26730/100000000.

Explanation

The probability that all three mowers are unusable on a given day can be found by multiplying the probabilities of each mower being unusable. The probability of the first mower being unusable is 10%, the second mower is 9%, and the third mower is 27%. Therefore, the probability that all three mowers are unusable is 10% * 9% * 27% = 0.000267. This can also be expressed as a fraction, 26730/100000000.

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3) 1. You are playing a game that involves drawing 3 numbers from a hat. There are 20 pieces of paper numbered 1 to 20 in the hat. Each number is replaced after it is drawn. What is the probability that each number is greater than 15 or less than 4?

To find the probability of each number being greater than 15 or less than 4, we need to determine the number of favorable outcomes and the total number of possible outcomes.

The favorable outcomes are the numbers that are greater than 15 or less than 4, which are 1, 2, 3, 16, 17, 18, 19, and 20.

The total number of possible outcomes is the total number of numbers in the hat, which is 20.

Therefore, the probability is the number of favorable outcomes divided by the total number of possible outcomes, which is 8/125.

Explanation

To find the probability of each number being greater than 15 or less than 4, we need to determine the number of favorable outcomes and the total number of possible outcomes.

The favorable outcomes are the numbers that are greater than 15 or less than 4, which are 1, 2, 3, 16, 17, 18, 19, and 20.

The total number of possible outcomes is the total number of numbers in the hat, which is 20.

Therefore, the probability is the number of favorable outcomes divided by the total number of possible outcomes, which is 8/125.

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4) 1. Write whether the events listed in the following question are either “independent” or “dependent”. The owner of a one-man lawn mowing business owns four old and unreliable riding mowers. As long as one of the four is working he can stay productive. From past experience, one of the mowers is unusable 10 percent of the time, one 9 percent of the time, one 27 percent of the time and one 11 percent of the time. Find the probability that all three mowers are unusable on a given day.

The events listed in the question are independent because the probability of one mower being unusable does not affect the probability of another mower being unusable. Each mower's usability is determined independently of the others, so the probability of all three mowers being unusable on a given day can be calculated by multiplying the individual probabilities together.

Explanation

The events listed in the question are independent because the probability of one mower being unusable does not affect the probability of another mower being unusable. Each mower's usability is determined independently of the others, so the probability of all three mowers being unusable on a given day can be calculated by multiplying the individual probabilities together.

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5) 1. Write whether the events listed in the following question are either “independent” or “dependent”. You are playing a game that involves drawing 3 numbers from a hat. There are 20 pieces of paper numbered 1 to 20 in the hat. Each number is replaced after it is drawn. What is the probability that each number is greater than 15 or less than 4?

The events listed in the question are independent because the probability of drawing a number greater than 15 or less than 4 does not depend on the previous draws. Each draw is replaced after it is drawn, so the probability of drawing a specific number remains the same for each draw. Therefore, the events are independent.

Explanation

The events listed in the question are independent because the probability of drawing a number greater than 15 or less than 4 does not depend on the previous draws. Each draw is replaced after it is drawn, so the probability of drawing a specific number remains the same for each draw. Therefore, the events are independent.

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6) 1. Write whether the events listed in the following question are either “independent” or “dependent”. You are playing a game that involves drawing 3 numbers from a hat. There are 20 pieces of paper numbered 1 to 20 in the hat. Each number is replaced after it is drawn. What is the probability that each number is greater than 15 or less than 4?

The events are independent because the probability of drawing a number greater than 15 or less than 4 does not depend on the previous draws. Each draw is replaced, so the probability remains the same for each draw.

Explanation

The events are independent because the probability of drawing a number greater than 15 or less than 4 does not depend on the previous draws. Each draw is replaced, so the probability remains the same for each draw.

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7) 1. Write whether the events listed in the following question are either “independent” or “dependent”. Find the probability of drawing a heart and then a diamond from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card.

The events of drawing a heart and then a diamond from a standard deck of 52 cards are dependent because the outcome of the first event affects the outcome of the second event. After drawing the first card, there are now 51 cards remaining in the deck, and the probability of drawing a diamond will be different depending on whether a heart was drawn first.

Explanation

The events of drawing a heart and then a diamond from a standard deck of 52 cards are dependent because the outcome of the first event affects the outcome of the second event. After drawing the first card, there are now 51 cards remaining in the deck, and the probability of drawing a diamond will be different depending on whether a heart was drawn first.

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8) 1. Find the probability of drawing a heart and then a diamond from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card.

When drawing a heart and then a diamond from a standard deck of 52 cards without replacement, we first need to determine the probability of drawing a heart on the first draw. Since there are 13 hearts in a deck of 52 cards, the probability of drawing a heart on the first draw is 13/52. After drawing a heart, there are now 51 cards left in the deck, including 13 diamonds. Therefore, the probability of drawing a diamond on the second draw is 13/51. To find the probability of both events happening, we multiply the probabilities together: (13/52) * (13/51) = 13/204.

Explanation

When drawing a heart and then a diamond from a standard deck of 52 cards without replacement, we first need to determine the probability of drawing a heart on the first draw. Since there are 13 hearts in a deck of 52 cards, the probability of drawing a heart on the first draw is 13/52. After drawing a heart, there are now 51 cards left in the deck, including 13 diamonds. Therefore, the probability of drawing a diamond on the second draw is 13/51. To find the probability of both events happening, we multiply the probabilities together: (13/52) * (13/51) = 13/204.

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9) 1. Find the probability of drawing a heart and then a diamond from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card.

The probability of drawing a heart and then a diamond from a standard deck of 52 cards without replacing the first card before selecting the second card can be calculated by multiplying the probability of drawing a heart (13/52) by the probability of drawing a diamond from the remaining cards (13/51). This is because there are 13 hearts and 52 cards in total, so the probability of drawing a heart is 13/52. After drawing a heart, there are 51 cards left in the deck, including 13 diamonds, so the probability of drawing a diamond is 13/51. Therefore, the overall probability is (13/52) * (13/51) = 13/204.

Explanation

The probability of drawing a heart and then a diamond from a standard deck of 52 cards without replacing the first card before selecting the second card can be calculated by multiplying the probability of drawing a heart (13/52) by the probability of drawing a diamond from the remaining cards (13/51). This is because there are 13 hearts and 52 cards in total, so the probability of drawing a heart is 13/52. After drawing a heart, there are 51 cards left in the deck, including 13 diamonds, so the probability of drawing a diamond is 13/51. Therefore, the overall probability is (13/52) * (13/51) = 13/204.

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10) 1. Find the probability of drawing a heart and then a king from a standard deck of 52-cards when you replace the first card before selecting the second card.

When replacing the first card before selecting the second card, the probability of drawing a heart is 1/4 (as there are 4 hearts in a standard deck) and the probability of drawing a king is 1/13 (as there are 4 kings in a standard deck of 52 cards). Since these events are independent, we can multiply the probabilities: (1/4) * (1/13) = 1/52. Therefore, the probability of drawing a heart and then a king is 1/52.

Explanation

When replacing the first card before selecting the second card, the probability of drawing a heart is 1/4 (as there are 4 hearts in a standard deck) and the probability of drawing a king is 1/13 (as there are 4 kings in a standard deck of 52 cards). Since these events are independent, we can multiply the probabilities: (1/4) * (1/13) = 1/52. Therefore, the probability of drawing a heart and then a king is 1/52.

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11) 1. Find the probability of drawing a heart and then a king from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card. (Assuming the king of hearts is not the heart originally selected)

When drawing a heart and then a king from a standard deck of 52 cards without replacement, the probability can be calculated by multiplying the probability of drawing a heart (13 hearts out of 52 cards) by the probability of drawing a king (4 kings out of 51 remaining cards). Therefore, the probability is (13/52) * (4/51) = 1/51.

Explanation

When drawing a heart and then a king from a standard deck of 52 cards without replacement, the probability can be calculated by multiplying the probability of drawing a heart (13 hearts out of 52 cards) by the probability of drawing a king (4 kings out of 51 remaining cards). Therefore, the probability is (13/52) * (4/51) = 1/51.

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12) 1. Find the probability of drawing a queen and then a second queen from a standard deck of 52-cards when you DO NOT replace the first card before selecting the second card.

The probability of drawing a queen from a standard deck of 52 cards is 4/52, since there are 4 queens in the deck. After drawing the first queen, there are 51 cards left in the deck, with 3 queens remaining. Therefore, the probability of drawing a second queen is 3/51. To find the probability of both events occurring, we multiply the probabilities together: (4/52) * (3/51) = 1/221.

Explanation

The probability of drawing a queen from a standard deck of 52 cards is 4/52, since there are 4 queens in the deck. After drawing the first queen, there are 51 cards left in the deck, with 3 queens remaining. Therefore, the probability of drawing a second queen is 3/51. To find the probability of both events occurring, we multiply the probabilities together: (4/52) * (3/51) = 1/221.

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13) 1. Find the probability of drawing a queen and then a king from a standard deck of 52-cards when you replace the first card before selecting the second card.

When you replace the first card before selecting the second card, the probability of drawing a queen is 4/52 (since there are 4 queens in a deck of 52 cards) and the probability of drawing a king is also 4/52 (since there are 4 kings in a deck of 52 cards). To find the probability of both events happening, we multiply the individual probabilities together: (4/52) * (4/52) = 1/169.

Explanation

When you replace the first card before selecting the second card, the probability of drawing a queen is 4/52 (since there are 4 queens in a deck of 52 cards) and the probability of drawing a king is also 4/52 (since there are 4 kings in a deck of 52 cards). To find the probability of both events happening, we multiply the individual probabilities together: (4/52) * (4/52) = 1/169.

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14) 1. You are playing a game that involves drawing 3 numbers from a hat. There are 20 pieces of paper numbered 1 to 20 in the hat. Each number is replaced after it is drawn. What is the probability that each number is greater than 15 or less than 4?

The probability of each number being greater than 15 or less than 4 can be calculated by finding the probability of each number being greater than 15 and the probability of each number being less than 4, and then multiplying these probabilities together. There are 4 numbers less than 4 (1, 2, 3, and 4) and 5 numbers greater than 15 (16, 17, 18, 19, and 20). The probability of drawing a number less than 4 is 4/20, or 1/5, and the probability of drawing a number greater than 15 is 5/20, or 1/4. Multiplying these probabilities together gives us (1/5) * (1/4) = 1/20. Since we are drawing 3 numbers, we need to multiply this probability by itself 3 times, resulting in (1/20)^3, which simplifies to 1/8000. However, the question asks for the probability that each number is greater than 15 or less than 4, so we need to subtract this probability from 1 to get the final answer of 1 - 1/8000 = 7999/8000. Simplifying this fraction gives us the final answer of 8/125.

Explanation

The probability of each number being greater than 15 or less than 4 can be calculated by finding the probability of each number being greater than 15 and the probability of each number being less than 4, and then multiplying these probabilities together. There are 4 numbers less than 4 (1, 2, 3, and 4) and 5 numbers greater than 15 (16, 17, 18, 19, and 20). The probability of drawing a number less than 4 is 4/20, or 1/5, and the probability of drawing a number greater than 15 is 5/20, or 1/4. Multiplying these probabilities together gives us (1/5) * (1/4) = 1/20. Since we are drawing 3 numbers, we need to multiply this probability by itself 3 times, resulting in (1/20)^3, which simplifies to 1/8000. However, the question asks for the probability that each number is greater than 15 or less than 4, so we need to subtract this probability from 1 to get the final answer of 1 - 1/8000 = 7999/8000. Simplifying this fraction gives us the final answer of 8/125.

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15) 1. You and three friends go to a restaurant and order a sandwich. The menu has 20 types of sandwiches and each of you is equally likely to order any type. What is the probability that each of you orders a different type?

The probability that each of you orders a different type of sandwich can be calculated by using the concept of permutations. Since there are 20 types of sandwiches on the menu, the first person can choose any of the 20 sandwiches. After the first person has made their choice, there are 19 types of sandwiches left for the second person to choose from. Similarly, the third person has 18 choices, and the fourth person has 17 choices. Therefore, the probability can be calculated as (20/20) * (19/20) * (18/20) * (17/20) = 0.6845, or approximately 68.45%.

Explanation

The probability that each of you orders a different type of sandwich can be calculated by using the concept of permutations. Since there are 20 types of sandwiches on the menu, the first person can choose any of the 20 sandwiches. After the first person has made their choice, there are 19 types of sandwiches left for the second person to choose from. Similarly, the third person has 18 choices, and the fourth person has 17 choices. Therefore, the probability can be calculated as (20/20) * (19/20) * (18/20) * (17/20) = 0.6845, or approximately 68.45%.

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5.4b Statistics - Independent And Dependent Events