This is due today (10 points) urpi and manco are playing a game where they flip a fair coin four times and try to predict the outcomes. using the sample space of possible outcomes listed below, answer each of the following questions. 1) what is p(a), the probability that the second flip is heads? 2) what is p(b), the probability that the fourth flip is heads? 3) what is p(c), the probability that the second flips is heads and the fourth flip is heads? 4) are events a and b independent?

Remark

There are 16 results of throwing a coin 4 times.

Answer 4 first.

These are independent events. No matter what the first result or the second or the third result is outcome can only be 1/2 for any given throw. One event does not influence any other.

One

What this means is that it does not matter what else might be true as long as the second event is heads. So you count THTT and as both being true. If you use your table, I believe there are 8 results where heads is in the second position.

Two

The last position will also contain heats 8 times. Just count them up. Some of them will be duplicated from your first answer. It does not matter. For example, will appear in both lists.

Three

For Heads in the second AND the last position you should get 4. They are THHH THTH HHTH

Answers

P(one) = 8/16. You get 8 answers out of 16 to get a heads in position 2.

P(Two) = 8/16. You get 8 answers out of 16 to get a heads in position 4

P(Three) = 4/16. You get 4 answers out of 16 to get a heads in positions 2 and 4

mxb67

step-by-step explanation:

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.