Suppose that baseball team A is better than baseball team B. Team A is better by enough that it has a 2/3 probability of beating team B in any one game, and this probability remains the same for each game, regardless of the outcomes of previous games.

Explain what it means to say that team A has a 2/3 probability of beating team B in any one game.

a. If these two games play each other 99 times, team A will win 66 of the 99 games.

b. All answers are equivalent.

c. If these two teams play each other 3 times, team A will win 2 of the three games.

d. If these two teams play each other many, many times under identical conditions, team A will win 2/3 of the games in the long run.

It would be correct to say that if team A plays team B for 3 games, A guaranteed to win exactly twice.

a. True

b. False

It would be correct to say that if team A plays team B for 30 games, A is guaranteed to win exactly twenty games.

a. True

b. False

If team A plays team B for 30 games, is it very likely that team A will win exactly 20 times? Explain.

a. Yes, since the probability has been calculated to be 2/3, we expect team A to win 20 games.

b. Yes, team A will be likely to win twenty of thirty games because the long-run probability is 2/3.

c. This cannot be determined without seeing each game's outcome.

d. No, the proportion of games that team A wins will likely be close to but not exactly 2/3 (such as 18, 19, 20, 21, 22 wins), so there could still be a low probability that A wins exactly 2/3 of the 30 games.

Suppose that a birth is equally likely to be a boy or a girl, and the outcome of one birth does not change this probability for future births.

Select the best simulation below for how a coin could be used to approximate the probability that a couple with four children would have two boys and two girls.

a. There is no best way to calculate the probability of having 2 boys and 2 girls.

b. Let heads represent a boy and tails a girl. Flip the coin four times and record the number of boys (heads) in those four children (tosses). Divide the number of outcomes with 2 boys and 2 girls by 4. This is the probability that a couple will have four children with 2 boys and 2 girls.

c. Let heads represent a boy and tails a girl. Flip the coin four times and record the number of boys (heads) in those four children (tosses). Repeat this 1,000 times, and look at what proportion of those 1,000 repetitions results in 2 boys and 2 girls. This is the probability that a couple will have four children with 2 boys and 2 girls.

d. Let heads represent a boy and tails a girl. Flip the coin four times and record the number of boys (heads) in those four children (tosses). Repeat this 10 times, and look at what proportion of those 10 repetitions results in 2 boys and 2 girls. This is the probability that a couple will have four children with 2 boys and 2 girls.

It turns out that the probability is 0.375 that a couple with four children would have two boys and two girls. Which is the best interpretation of what this probability means?

a. In 1000 couples with 4 children, you would expect 375 couples to have 2 boys and 2 girls.

b. In a very large number of couples with four children, roughly 37.5% of the couples will have 2 boys and 2 girls (assuming each birth is equally likely to be a boy or a girl).

c. Both A and B are correct statements.

d. The probability cannot be 0.375 because that would be a fractional number of couples which cannot happen.

Based on the probability given in part b (p= 0.375), what is the probability that a couple with four children does not have 2 boys and 2 girls?

the absolute tolerance is +/-0.001 _____________________________

Is the reasoning below a good explanation as to why it makes sense (at least in hindsight) that a couple with four children is more likely not to have two of each sex than to have two of each sex?This fact can be correctly explained by saying that having exactly two of each gender is pretty specific. Even though a specific 3-1 or 4-0 split is less likely than a 2-2 split, there are more ways to obtain a result other than a 2-2 split.

a. True

b. False

Suppose that there are 50 people in a room. It can be shown (under certain assumptions) that the probability is approximately 0.97 that at least two people in the room have the same birthday (month and date, not necessarily year).

Select which statement would best explain what 0.97 means to a person who has never studied probability or statistics.

a. In many, many rooms each containing 50 people, 97% of the rooms will have at least two people with the same birthday.

b. If you have 100 rooms with 50 people, then 97 of the rooms should have at least two people with the same birthday.

c. Each of these statements are equivalent statements about probability.

d. Neither of these statements correctly explain what the probability means.

Now consider the question of how likely it is that at least one person in the room of 50 matches your particular birthday. (Do not attempt to calculate this probability.) Is this event more, less, or equally likely as the event that at least two people share any birthday? In other words, do you think this probability will be smaller than 0.97, larger than 0.97, or equal to 0.97 and why?

a. The probability cannot be predicted since we don't know how the people were selected to be in the room.

b. Equal to 0.97 since the probability has been determined previously.

c. Greater than 0.97 since your birthday is as likely to occur as any other.

d. Less than 0.97, since there are many pairs of people in the room that do not involve you and only 49 pairs of people that involve you.

Imagine you had a hat with 365 slips of paper in it, each with a number from 1 to 365 listed on it to represent each day of the year. Select from the simulations below the best simulation method you could use to confirm that the probability is 0.97 (as given in Part (a)).

a. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have at least 50 numbers recorded. If any of the 49 numbers are the same, then you have a 'match.' Repeat this process 1000 times and see in what proportion of the sets of 50 numbers there is a match.

All simulations are equivalent.

b. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have at least 50 numbers recorded. If any of the 49 numbers are the same as the first number drawn, then you have a 'match.' Repeat this process 100 times and find the proportion of the 10 sets of 50 numbers for which there is a match.

c. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have at least 50 numbers recorded. If any of the 49 numbers are the same as the first number drawn, then you have a 'match.' Repeat this process 10 times and find the proportion of the 10 sets of 50 numbers for which there is a match.

Imagine you had a hat with 365 slips of paper in it, each with a number from 1 to 365 listed on it to represent each day of the year.Select from the simulations below the best simulation method you could use to confirm the probability of someone in the room matchingyour particular birthdaywill be smaller than 0.97, larger than 0.97, or equal to 0.97.

a. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have 50 numbers recorded. If any of the 50 numbers are the same as the first number drawn, then you have a 'match'. Repeat this process 1000 times and find the proportion of the 1000 sets of 50 numbers for which there is a match.

b. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have 50 numbers recorded. If any of the 50 numbers are the same, then you have a 'match'. Repeat this process many, many times and see in what proportion of the sets of 50 numbers there is a match.

All simulations are equivalent.

c. Draw a slip of paper out of the hat and write down the number on a piece of paper. Replace this slip, mix up the slips, and repeat the draw 49 more times until you have 50 numbers recorded. If any of the 50 numbers are the same as the first number drawn, then you have a 'match'. Repeat this process 10 times and find the proportion of the 10 sets of 50 numbers for which there is a match.