Question

I can't understand how to set up the x/f(x) table with this information... Help Please?



An animal either dies ( D ) or survives ( S ) in the course of a surgical experiment. The experiment is to be performed first with two animals. If both survive, no further trials are to be made. If exactly one animal survives, one more animal is to undergo the experiment. Finally, if both animals die, two additional animals are to be tried.

(a) List the sample space to answer the next questions.

(b) Assume that the trials are independent and the probability of survival in each trial is 1/4

Answer 1

Assign probabilities to the elementary outcomes to answer the next question.

(c) Let X denote the number of survivors. Obtain the probability distribution of X by referring to part (b).

Write the x -values in ascending order. Give exact answers (in form of fraction if necessarry).

Answer 2

although i am not a big fan of tree diagrams, it might be helpful in this case

Answer 3

or maybe not
the probability that both survive is \(\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}\)

Answer 4

Do you always multiply the probability like that

Answer 5

good question!
you do only when the trials are "independent" which you are told they are

Answer 6

okay because I am given a table of 6 empty slots. 3 for x and 3 for f(x) and it says they equal 1. So the x values would be 1/16 and the f(x) one would be divided evenly into three?

Answer 7

i have no idea what the table looks like, but lets see what the possibilities for the number of animals surviving are

Answer 8

both could survive the first experiment, so if we say \(x\) is the number of survivors, one possibility is that \(x=2\)

Answer 9

thats what the table looks like

Answer 10

oooh ok i see
lets go slow

Answer 11

okay. so how would we start it

Answer 12

first off, note that is says
"Let X denote the number of survivors"

Answer 13

so \(x\) is not a probability at all, it is a whole number, the number of survivors

Answer 14

how many survivors can there be? in other words, what possible value can \(x\) take?

Answer 15

couldn't it take any? as long as the value of f(x) equals 1

Answer 16

oh no

Answer 17

\(x\) denotes the number of survivors of the experiment
the way the set up for the experiment is, there could be
2 survivors
1 survivor
or 0 survivors

so the only three possible values of \(x\) are \(2,1,0\)

Answer 18

OHHH!

Answer 19

that is why there are three slots on the left

Answer 20

left column i mean

Answer 21

got it so then it says each is independent

Answer 22

now we have to figure out the probability of each value of \(x\)

Answer 23

meaning the probability is 1/4 so do I multiply that by something? I sort of remember it from class but I'm not too sure

Answer 24

we need
1) the probability that exactly 2 survive
2) the probability that exactly 1 survives and
3) the probability than none survive

Answer 25

it is going to take a while, because there are cases to consider

Answer 26

what do you mean? Aren't I supposed to set them p and 2/4 1/4 0/4

Answer 27

heck no

Answer 28

Shoot. Thats what my example said.

Answer 29

the first column should read \(x=2,x=1,x=0\) or put 2, 1, 0 in the first column

Answer 30

thats done

Answer 31

now we have to consider all the ways you can end up with two live animals
there are a few

Answer 32

first off, both could survive the first experiment
the probability of that is as computed \(\frac{1}{16}\) but there are other ways two can survive

Answer 33

If exactly one animal survives, one more animal is to undergo the experiment

Answer 34

so if one survives and then another one survives, you have to surviving animals
the probability that exactly one survives is
\[2\times \frac{1}{4}\times \frac{3}{4}=\frac{3}{8}\]

Answer 35

where did you get the 1/4 and 3/4

Answer 36

survival in one trial is \(\frac{1}{4}\) so not surviving is \(\frac{3}{4}\)

Answer 37

okay so then for the 1 survivor. It would be 1 x 3/4 because there is only one? then for 0 the same

Answer 38

you have two animals, say A and B
the probability that A survives and B does not is \(\frac{1}{4}\times \frac{3}{4}\) and the probability that B survives and A does not is the same, so the probability that exactly one survives in the first round is \(2\times \frac{1}{4}\times \frac{3}{4}=\frac{3}{8}\)

Answer 39

then you do the experiment again, so given that one survived, the probability that another survives, for a total of two surviving, is \(\frac{1}{8}\times \frac{3}{4}=\frac{3}{32}\)

Answer 40

and finally there is yet another way to get two survivors
both die on the first trial, with probability \(\frac{3}{4}\times \frac{3}{4}\) and then the next two live, with probability \(\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}\)

Answer 41

multiply those together, and get
\[\frac{9}{256}\]

Answer 42

now the total ways to get two survivors is the sum of the previous answer namely
\[\frac{9}{256}+\frac{3}{32}+\frac{1}{16}=\frac{49}{256}\]

Answer 43

now we have to figure out the way for exactly one to survive

Answer 44

one way is that one survives the first round, and then in the second round the animal dies
another way is that both die on the first round, and then one survives on the second

Answer 45

any ideas how to do that?

Answer 46

honestly this feels a lot more complicated than i thought. When we did it in class we would just make them equal the one. I see how you got the previous answers now to find that they both die on the first round it would be similar to the one before. Where you multiplied them.

Answer 47

we already know the probability that one survives the first round is \(\frac{3}{8}\) then the second one not surviving is \(\frac{3}{4}\) so the probability that one survives the first round, and then the second one dies is \(\frac{3}{8}\times \frac{3}{4}=\frac{9}{32}\)

Answer 48

probability that none survive the first round and exactly one survives the second round is
\[\frac{3}{4}\times \frac{3}{4}\times \frac{3}{8}=\frac{27}{128}\]

Answer 49

then that would be the answer?
don't they all need the same denominator

Answer 50

add those up and get
\[\frac{9}{32}+\frac{27}{128}=\frac{63}{128}\]

Answer 51

and finally, what is the probability that none survives?

Answer 52

both die on first trial, and both die on second trial

Answer 53

\[\left(\frac{3}{4}\right)^4=\frac{81}{256}\]

Answer 54

Okay I think I get it I just have to keep finding the probabilities. Just like you did with the first ones.

Answer 55

btw it says to write in ASCENDING order

Answer 56

right?

Answer 57

Lol thanks! I read that right now too

Answer 58

so you should probably write 0,1,2 for the first column

Answer 59

then i think the second column \[P(x=0)=f(0)=\frac{81}{256}\] then
\[P(x=1)=f(1)=\frac{63}{128}\] and finally
\[P(x=2)=f(2)=\frac{49}{256}\]

Answer 60

lets see if they add up to 1, or if there is a mistake

Answer 61

they do!

Answer 62

if you multiple the middle one by 2

Answer 63

http://www.wolframalpha.com/input/?i=81%2F256%2B63%2F128%2B49%2F256

Answer 64

looks good to me!

Answer 65

oh i see what you mean, in order to add you multiply the middle one top and bottom by 2 to make the denominator 256
hope this was more or less clear
it was a bit long and complicated

Answer 66

Thanks a bunch. I was but I read back through your messages and it made sense! Thanks @satellite73 (:

Answer 67

yw