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*similar to problem 5 from section 1.4 of your text

1.

An experiment consists of flipping a coin 5 times and noting the number of times that tails is flipped. Find the sample space S of this experiment.

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The sample space, as we discussed in the tutorial for this section, is all the possible outcomes of the experiment. While that may sound simple enough, this problem is meant to trick you because the experiment of flipping a coin 5 times can be measured in a variety of different ways, but the sample space refers solely to outcomes that represent how you measure.

In this case, you're measuring the number of times the coin lands tails. If you flip 5 times, that means you could end up with either 0, 1, 2, 3, 4, or 5 instances of tails. Therefore, your answer is S = {0, 1, 2, 3, 4, 5}.

*You may occasionally be asked to measure this experiment by noting whether the coin lands heads or tails each time. In that case, you would end up with a sample space that includes outcomes like HHTHT or TTTTT. Folks often try to answer the above question with that kind of sample space, which is incorrect, so be mindful of exactly what the sample space is measuring when you answer.

*similar to problem 8 from section 1.4 of your text

2.

Assume that X = {A, B, C, D, F} and Y = {1, 2, 3, 4, 5}. A code consists of 3 different symbols selected from X followed by 3 not necessarily different symbols from Y. How many different codes are possible?

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The key to solving this problem is the multiplication principle: the idea that the total number of outcomes in an experiment such as this is the number of options for each thing you're choosing multiplied together. Here's what I mean:

This code is first formed by choosing 3 letters from a group of 5. We're told they must all be different, so when you choose the first letter, you have 5 options to choose from, but when you choose the second letter, you only have 4 to choose from (because the first letter you chose is no longer available to you). Similarly, you only have 3 letters to choose from when you pick the third letter. That means the number of three-letter combinations for this code is 5 x 4 x 3, which is 60.

Now that said, when you choose the 3 numbers from set Y, they don't have to be different. That means each time you choose a number (first, second, and third) all 5 numbers are available to you. So the number of three-number combinations for this code is 5 x 5 x 5, or 5^3, which is 125.

Finally, each combination of letters can be paired with each combination of numbers to form the full code, so the total number of possible codes is 60 x 125, which is 7,500.

*similar to problem 17 from section 1.4 of your text

3.

Assume that product codes are formed from the letters F, N, P, and S and consist of 5 not necessarily distinct letters arranged one after the other. For example, FFNPS is a product code. How many different product codes are there? How many different product codes do not contain P? How many different product codes contain exactly one S?

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As with all problems like this, we'll use the multiplication principle, but pay very close attention to exactly how many things we're choosing and the options available when we make each choice.

We have 4 letters available to us (F, N, P, and S), and we need to fill 5 spaces in this product code. Because the letters don't need to be different, all 4 letters are available each time we fill one of the spots. That means the number of all possible product codes is 4 x 4 x 4 x 4 x 4, or 4^5, which is 1,024.

Next, we're asked to find the number of codes that do not contain the letter P. For this, simply repeat the process above, but bear in mind we only have 3 letters to choose from to fill each slot (F, N, and S), which means the number of possible product codes without P is 3 x 3 x 3 x 3 x 3 = 243.

For the last question, let's first assume that the only S in the product code is in the first slot: S _ _ _ _. How many ways can you fill the last four slots? Well, if there can only be one S, and you've used it already, then you only have 3 letters available to you each time, so the number of possibilities is 3 x 3 x 3 x 3 = 81. That said, the sole S in this product doesn't have to be in the first slot. It could be in any of the five slots. So we should multiply those 81 possibilities where the sole S is in the first slot by 5 to capture all the possible scenarios where the S is in either the first, second, third, fourth, or fifth slot. And 81 x 5 = 405.

*similar to problem 26 from section 1.4 of your text

4.

Assume that a box contains 6 balls: 3 red, 2 green, and 1 blue. Balls are drawn in succession without replacement and their colors are noted until a red ball is drawn. Which of the following is not a valid outcome for this experiment and how many outcomes are there in the sample space?

a. GGBR

b. BRGRGR

c. R

d. BGGR

e. GR

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Let's first address how many outcomes are in the sample space of this experiment because, in doing so, we'll need to list them all. Having done that, it'll be obvious which of the outcomes listed above is invalid. (PRO TIP: Make a tree diagram for this experiment by drawing a branch for each outcome as I describe it.)

When you pick the first ball from the box, it'll either be red, green, or blue. If it's red, the experiment ends, so the outcome R is one possible outcome. If it's green or blue, you keep going.

If the ball is blue, you'll have drawn the only blue ball available, so it won't be an option on the next draw (this is what "without replacement" means). From there, if you draw red, the experiment ends, so BR is another possible outcome. If you draw green, you keep going. On the third draw, you'll either draw red and the experiment will end (so, BGR) or green again and you'll keep going. However, now that you've drawn the only blue ball and both green balls, the only option remaining is red, so the outcome BGGR will end the experiment.

Stepping back to the last possibility for the first draw, you might draw green, in which case every color will be available to you on the second draw. If you draw red, the experiment ends, giving you the outcome GR. If you draw green again on the second draw, only blue and red will be available on the third draw. If it's red the experiment ends (so, GGR) and if it's blue, the only option on the fourth draw will be red, also ending the experiment (so, GGBR).

If after first drawing green, you then drew blue, only green and red would be available to you on the third draw. If you draw red, the experiment ends (GBR) and if you draw green, the only option on the fourth draw will be red, also ending the experiment (so, GBGR). If you go back through this explanation and count the number of outcomes we found (or, ideally, count the number of branches in your tree diagram), you'll see there are 9 possible outcomes for this experiment.

And looking through the list of outcomes in the question, we found all except b (which you could also tell is invalid because the experiment continues after a red ball is drawn second, which the conditions of the experiment forbid).

5.

A card is drawn from a standard deck of 52 playing cards. If the card is a spade, a fair coin is flipped 5 times and the number of tails flipped is noted. If it is not a spade, the coin is only flipped 2 times, but this time the result of each flip is noted. The card drawn is not noted in either case. How many elements are in the sample space for this experiment?

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If you haven't figured it out by now, drawing a tree diagram is essential for these kinds of convoluted problems. So to begin, let's determine our options for the first stage. We're told we first draw a card from a standard deck and the particular card drawn is not noted, only whether it's a spade or not. Therefore, we might call each event S or NS (for "spade" or "not spade").

From there, if the card is a spade, we flip a coin 5 times and the number of tails flipped is noted. As discussed in Question 1 above, that results in 6 possibilities: you either get 0, 1, 2, 3, 4, or 5 tails. So we have the outcomes S0, S1, S2, S3, S4, and S5.

If the card wasn't a spade, you flip the coin twice but this time noting the result of each flip. That means our tree diagram branches out from NS to either H or T (for "heads" or "tails") in the second stage. And each of those branches out to another set of H's and T's in the third stage (which is the second coin flip). Following each branch of your diagram, you get the outcomes NSHH, NSHT, NSTT, and NSTH.

All together, that makes 10 outcomes in the sample space for this experiment.

6.

An experiment consists of drawing 3 cards with replacement from a standard deck of 52 playing cards. Suppose after drawing each card the experimenter writes down the color of the card only. How many outcomes are in the sample space? If only the number/letter on the card is noted, how many outcomes are possible? Suppose now that the experimenter records the exact card (e.g. 6 of clubs or Queen of spades). How many outcomes are possible?

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Because the playing cards are drawn "with replacement," all 52 cards are available to you each time you draw. And that makes our calculations a lot easier and a tree diagram kind of unnecessary (though you should still draw one out each time if it helps you understand the process better).

The first question asks for the number of outcomes given that you only note the color of the card drawn at each of the three stages. There are only two options (red or black) each time, so much like calculating the number of product codes in Questions 2 and 3 above, we can use the multiplication principle to solve: 2 x 2 x 2, or 2^3, is 8.

The next question asks for the number of outcomes if you note the number/letter on each card. The process is exactly the same as I just described, only this time we have 13 options at each stage (the numbers 2-10, Jack, Queen, King, or Ace). That means the number of outcomes is 13 x 13 x 13, or 13^3, which is 2,197.

Finally, we're asked to find the number of outcomes if you note each card drawn uniquely (that is, all 52 cards are a possible outcome at each stage). This means the number of outcomes is 52 x 52 x 52, or 52^3, which is 140,608.

7.

A twenty-something single person is planning a ski vacation. Assume that they have 3 possible destinations: Oregon, Colorado, and Utah. There are 3 ski areas in Oregon with 1 available time for 2 of the areas and 6 times for the other area. There are 2 ski areas in Colorado with 3 available times for one area and 4 times for the other area. There are 2 ski areas in Utah with 2 available times for one area and 1 time for the other area. (A “time” refers to a weekend for which there are vacancies at the ski lodge.) A trip plan involves the selection of a location, ski area, and a time. How many possible plans are there?

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To be perfectly honest, I think this question is meant as a way to get you to transcribe a word problem into a tree diagram. In the first stage, your tree divides into three branches for each destination. From there, each destination divides into the number of ski areas available. Lastly, each ski area divides into the number of available times.

But in terms of answering this question, that just means adding together the number of available times at each area (or, if you draw it out, counting the number of branches by the end of the third stage). There are 8 available times in Oregon, 7 in Colorado, and 3 in Utah, for a total of 18 possible plans.

Crucially, this is not a problem where you can use the multiplication principle to multiply together the number of possible destinations by the number of ski areas by the number of times because you don't have the same number of each available to you at each stage. The number of ski areas depends on which destination you choose and the number of times depends on the particular ski area. This one is really just about drawing a tree and counting branches -- don't overthink it!

8.

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An experiment consists of flipping a fair coin 5 times. Suppose that after flipping the coin all 5 times, the experimenter noted the number of heads. How many outcomes are there in this sample space? Suppose now that after each flip the experimenter records the result (i.e. H or T) of the flip. Find the number of outcomes in this sample space. Suppose now that the experimenter notes the result of each flip, but they will end the experiment early if they get two heads or two tails in a row. They will still only flip a maximum of five times, however. Which of the following outcomes are possible in this experiment?

a. HTHHT

b. HHHHH

c. HTT

d. HTHTT

e. THTHT

As we've discussed previously (see Questions 1 and 5 above), whenever you flip a coin multiple times and only note the number of heads or tails flipped, your answer is 1 + [the number of times the coin was flipped] because, in this case anyway, you'll either get heads 0, 1, 2, 3, 4, or 5 times, giving you 6 possible outcomes.

Once you note the specific result for each flip (i.e. H or T), you use the multiplication principle to multiply together the number of events at each stage. In this case, you always have the two possibilities H or T at each of the five stages. Therefore, the number of outcomes in the sample space is 2 x 2 x 2 x 2 x 2, or 2^5, which is 32.

The last problem asks you to recognize how certain outcomes that might have normally been possible are no longer once certain constraints are placed on an experiment. In this case, we're told the experiment ends once you get two heads or tails in a row. Options a and b, then, aren't possible because the experiment continues after getting two heads in a row. Option c rightly ends after getting two consecutive tails, option d ends after five flips (which is also coincidentally after getting two consecutive tails), and option e ends after five flips (having never gotten two consecutive heads/tails).

9.

An experiment consists of flipping a coin and/or rolling a die. Suppose that the experiment consists of both flipping a coin and rolling a die and noting the result in each case. How many outcomes are there in the sample space? Suppose that the experiment consists of either flipping a coin or rolling a die and recording the result in each case. How many outcomes are there in this sample space?

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This problem is actually extremely important because it introduces a concept that will become an important rule in Chapter 2 and carry on throughout the rest of the semester, so read this explanation carefully.

In the first experiment, you flip a coin and roll a die, so there are two stages. Drawing on the multiplication principle, to find the total number of outcomes in the sample space, multiply together the number of events at each stage. In this first stage, there are two (heads or tails) and in the second stage, there are six (1, 2, 3, 4, 5, or 6). Therefore, there are 12 possible outcomes.

In the second experiment, you flip a coin or roll a die, but not both. So there aren't multiple stages because you only take one action here. If you were drawing a tree diagram, you would list 8 possible outcomes: 2 for the coin flip and 6 for the die roll. In other words, you add the number of events together, rather than multiply.

That association between the word "and" representing multiplication and the word "or" representing addition will become crucial in the chapters that follow -- and this problem is at the foundation of why it works. Even if this explanation seems simple to you, make sure you understand it thoroughly. Check in with me at the end of the semester and tell me I was wrong!

10.

An experiment begins by flipping a fair coin and noting the result. If heads occurs, a fair die is rolled twice and the number that comes up on each roll is noted successively. If tails occurs, the same coin is flipped 3 more times and the total number of heads that occur in those 3 flips is noted. How many possible outcomes are there?

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Due to the number of constraints in this problem, the only way to solve is to map it out on a tree diagram and count the branches.

In the first stage, the tree splits into two branches for heads and tails. We're told that if the result is heads, a die is rolled twice and the number on its face is noted each time. There are 6 possibilities for each roll, so per the multiplication principle, we have 6 x6, or 36, outcomes after the coin lands heads.

If the coin lands tails, the coin is flipped 3 more times, but we only count the number of resulting heads. In that case, there are only 4 outcomes: 0, 1, 2, or 3 heads.

Adding together the 36 outcomes sprouting from the heads branch with the 4 outcome sprouting from the tails branch, we find there are 40 total possible outcomes for this experiment.

11.

An experiment consists of rolling a fair die some number of times. Each time the experimenter rolls the die, they record whether the result was even or odd. The experiment ends when either an odd number is rolled or the die is rolled 4 times. How many outcomes are there in the sample space? How many outcomes are in the event “the die is rolled at least 3 times?”

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Since there are only two possibilities at each stage of this experiment (even or odd), and at most four stages, it'll actually be fairly easy to conceptualize of the sample space without making a tree diagram (but if it helps, please do so anyway!).

Crucially, the experiment ends when an odd number is rolled, so every stage will include an outcome that ends the experiment. In the first stage, it's the outcome O.

The experiment continues if the first roll is even but ends if the second roll is odd (EO). Again, a second even roll pushes us into the third stage where the outcome EEO ends the experiment. And if we roll even three times in a row and make it into the fourth and final stage, we'll either roll even or odd (EEEE or EEEO), ending the experiment either way. That gives us 5 possible outcomes.

Now we need to find the number of outcomes in the event "the die is rolled at least 3 times." Looking through the 5 specific outcomes in our sample space, how many include at least 3 rolls? We see the answer is EEO, EEEE, and EEEO -- or 3 outcomes.

12.

A multiple-choice test has 8 questions and there are 4 choices for the answer to each question. An answer sheet has 1 answer for each question. How many different answer sheets are possible? How many different answer sheets are possible if the same answer is not used for every question?

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The wording of this problem might sound a bit strange, but the process for solving it is no different than finding the number of possible product codes or coin flips. Answering each question on the test is like a different stage. And at each stage, you have 4 possible choices. Using the multiplication principle, we can determine that the total number of possible answer sheets (or outcomes) is 4^8, which is 65,536.

Next, we're asked to find the total number of answer sheets if the same answer isn't used for every question. That means we should exclude the answer sheets where the first answer is marked all 8 times (and the second, third, and fourth answers). That's only 4 answer sheets, so we're still left with 65,532 outcomes.

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