Question

A biologist has determined that a particular osprey has a 70% chance of catching a fish on any given day. Carry out a simulation of twenty trials using the random number table below to find the probability that the osprey will actually catch a fish on all of the next three days.

Answer 1

945 025 354 793 236

106 746 981 105 012

832 180 250 871 835

793 726 864 496 947

Answer 2

@radar

Answer 3

@amistre64

Answer 4

@phi , @.Sam.

Answer 5

@Hero

Answer 6

SOMEONE PLEASE HELP ME!!!!!!!!!!!!!!!!!

Answer 7

@radar

Answer 8

ive never used a random number table, ive seen them but havent really delved into how to use them. can you explain to me the process? since it should be covered in your material.

Answer 9

i think the table will generate the yes-no trials; then 70%, times the number of (yes, yes, yes) outcomes in 20 would seem to be it

Answer 10

Well my teacher said the answer was 35% but i just don't know how to do the work for it

Answer 11

@radar @amistre64

Answer 12

the key is in knowing how to work the table. spose the table gives us x/20 results that are y,y,y

then .7*x/20 = .35 if the teacher is correct

so x=10 ... the table would most likely produce about 10 instances

Answer 13

since i have no real idea how to use the table to produce random numbers then the process is a mystery to me as how we find x :)

Answer 14

@amistre64 could you help me with other probability questions?

Answer 15

maybe, im not the greatest at them but for like stats 1 i can prolly get by

Answer 16

This is Algebra 2, my teacher is teaching us probability

Answer 17

oh, then most likely i can fake it :)

Answer 18

Okay thank you!!
Mana has a candle box containing 6 red candles and 6 white candles. What is the probability the she will pull out three red candles and two white candles?

Answer 19

THis is what i have

\[6P3(6P2)\div12P5=5/132\]

Answer 20

the hardest part about this might be to consider all the ways that she can pull 5 candles such that 2 are white and 3 are red. algebra 2 i spose hasnt covered binomials

Answer 21

oh good, youve got a little

Answer 22

or \[6C3(6C2)\div12C5=225/792\]

Answer 23

But idk which one

Answer 24

if i brute it some to check: the ways to pull are:

rrrww
rrwrw
rwrrw
wrrrw

rrwwr
rwrwr
wrrwr

rwwrr
wrwrr

wwrrr

so 10 different way to pull this combination, all we have to do is assign values to the pulls now

Answer 25

rrrww

r: 6 out of 12
r: 5 out of 11
r: 4 out of 10
w: 6 out of 9
w: 5 out of 8

agreed so far?

Answer 26

Yes

Answer 27

lets pull another set: the out of parts will remain the same regardless so lets focus on the other part

wrrwr:

w:6
r: 6
r: 5
w:5
r: 4

agreed?

Answer 28

Yes

Answer 29

if we do this for each one, we would see that this 66554 is just getting mixed around but the product of them never changes. 6*6*5*5*4 is the same regardless of the order, right?

Answer 30

Right

Answer 31

so:\[10*\frac{^6P_3~ ^6P_2}{^{12}P_5}\]
is the way im seeing it

Answer 32

or it may be 10 times out of 5! (120) total outcomes

Answer 33

nah, its just the addition of all the possible success

Answer 34

10.6.5.4.6.5
-----------
12.11.10.9.8


.6.5.4.6.5
-----------
12.11..9.8


..5.5
------
6.11...


25/66 is what im getting
25/60

Answer 35

lol, 25/60 was a priorcalc i forgot to delete

Answer 36

I see what you did, but could you try the combination, permutation method to see?

Answer 37

You still there @amistre64

Answer 38

im here, sites getting laggy tho

Answer 39

Okay, could you try the combination, permutation method to see?

Answer 40

Because idk when to use combination or when to use permutation

Answer 41

i have a hard time with it myself .... my teacher had us probing poker hands, it was horrible

Answer 42

6C3 * 6C2
---------- = 25/66 also
12C5


nCk = nPk/k!



6P3 * 6P2 * 5!
-------------
3! 2! 12P5



5.4.3.2.1 20
-------- = -- = 10
3.2.1.2.1 2

Answer 43

soo ..\[\frac{6C3~6C2}{12C5}=10\cdot \frac{6P3~6P2}{12P5}\]
:)

Answer 44

the idea is simply to use what suits you best. I can reason better from the 10 cases of 'pick's, rather than the cases of 'choose'

Answer 45

Did you do 6C3.... times 6P3.....?

Answer 46

no, i just went over how the cases relate to one another.

in essense, use P of you can count the number of times that set can occur ... and multiply the results by that number

use C if you feel more comfortable overall

Answer 47

I am so confused haha

Answer 48

spose we wanted 4 red and 1 white

rrrr w is the basic setup, there are 5C4 ways to arrange this, right?

Answer 49

wouldnt it be 6C4? Since there are 6 red candles

Answer 50

not the direction im headed, im simply trying to count the ways that we can have 5 objects arranged, such that 4 are the same and 1 is the same.

there is a rule for this:

n objects, such that a are the same, b are the same, c are the same, .... k are the same

the number of ways that we can arrange them is:\[\frac{n!}{a!~b!~c!~...~k!}\]
in this case we have 5 objects, 4 are the same and 1 is the same:\[\frac{5!}{4!~1!}=\frac{5!}{4!}=5C4\]

Answer 51

Oh okay

Answer 52

im pretty sure we can reason that the setups:\[\frac{6C3~6C2}{12C5}=\frac{5!}{3!~2!}\frac{6P3~6P2}{12P5}\]

so if you want to use one or the other it would be fine

Answer 53

if you use the P setup, it has a multiplier to deal with

Answer 54

spose we have a different setup:

7 red, and 5 white; such that we want 3 red and 2 white ...

Answer 55

okay...

Answer 56

\[\frac{7C3~5C2}{12C5}\]
\[\frac{5!~7P3~5P2}{3!~2!~12P5}\]
\[\frac{7C3~5C2}{12C5}=\frac{5!}{3!~2!}\frac{7P3~5P2}{12P5}\]

Answer 57

if we do the P setup, we still need to determine how many times it occurs: which is still 5C3 (or 5C2) = 10 times

the C setup is just a 'simplification' of the P setup

Answer 58

Why are you including 5! and 3! in the equation?

Answer 59

spose we have 5 cats, 3 dogs, and 2 rabbits; 10 objects total

what the probability of getting 3 cats, 1 dog, and 2 rabbits? 6 objects total
\[\frac{6!}{3!~1!~2!}~\frac{5P3~3P1~2P2}{10P6}\]

Answer 60

its a counting rule: mentioned above with n objects such that such and such are the same

Answer 61

I guess i understand, but for my question above was my answer, 5/132 or 25/88

Answer 62

you might want to recalculate those options you posted

Answer 63

http://www.wolframalpha.com/input/?i=%286+choose+3%29*%286+choose+2%29%2F%2812+choose+5%29

25/66, not 25/88

Answer 64

and, 5/132 is calced correctly

10*5/132 = 25/66 which is what i would expect of the Ps

Answer 65

Oh okay thanks! So is it 25/66(combination) or 5/132(permutation)

Answer 66

without the multiplier 10, yeah thats what you would get.

to correct the permutations we would have actually use that process correctly :)