Question

A gold miner is mining for gold in Texas, California and Nevada. It is predicted there is a 52% chance of finding gold in Texas, a 31% chance in California and a 20% chance in Nevada. What is the probability that:

A) Gold is found in California, but not Nevada?
B) Gold is found in Nevada, but not Texas?
C) Gold is not found in Texas, California nor Nevada?
D) Gold is found in at least one of these states?

Round to three decimal places. Write a decimal such as 0.25 as just .25.

Answer 1

no one wants to do this one i see. let me draw a little venn diagram to make it easier

Answer 2

Somewhere is should be stated that you are assuming Independence.

Answer 3

no it doesn't state that

Answer 4

then you can't do the problem

Answer 5

it is certainly assuming independence. let's do A)

\[P(C)=.31, P(N)=.2\]
\[P(C\cap N^c)=P(C)-P(N\cap C)\] and
\[P(N\cap C)=P(C)\times P(N)\] assuming independence. first off is this notation clear?

Answer 6

yeah makes sense

Answer 7

one can also go directly to this

\[P(C\cap N^{c})=P(C)\cdot P(N^{c})\]

by independence

Answer 8

@satellite73 we understand your formulas we just dont know how to go about solving them.

Answer 9

i am trying to find a nice venn diagram template. i am sure i had one but i cannot seem to find it. in any case we can compute
\[P(C\cap N^c)=.31-.2\times .31=.248\]

Answer 10

is that a

Answer 11

similarly for
\[P(N\cap T^c)=.2-.2\times .52=.098\]

Answer 12

those are a and b respectively

Answer 13

so A is .248 and B is .098

Answer 14

.096

Answer 15

for wat?

Answer 16

.2-.2*.52=.096

Answer 17

zarkon is right. it should be
\[.2-.2\times .52=.096\]

Answer 18

ok thanks and wats C and D?

Answer 19

c) is a little tricky, unless i am missing something simple.
\[P(T\cup N\cup C)=P(T)+P(N)+P(C)\]
\[-P(T\cap N)-P(T\cap C)
-P(N\cap C)+P(N \cap T\cap C)\]

Answer 20

@zarcon if i am making too much of this jump in and correct it please

Answer 21

\[P(T^{c}\cap C^{c} \cap N^{c})\]

\[=P(T^{c})P(C^{c})P(N^{c})=.48\cdot .69\cdot.8=.26496\]

again by independence.

Answer 22

buy i think you have to compute
\[.52+.31+.20-.52\times .31-.52\times .2-.31\times .2+.52\times .31\times .20\]

Answer 23

ah yes zarcon way much easier sorry

Answer 24

and D is
\[1-c\]

Answer 25

yes ;)

Answer 26

I have a Ph.D. in mathematical statistics...I know what I am doing :)

Answer 27

So the final answers are- A) .248, B) .096, C) .265 and D) .735

Answer 28

@Zarkon are those answers right?

Answer 29

yes

Answer 30

thank you!
stay around we got one more question for you

Answer 31

Four Olympic skaters are attempting to win a medal. The skaters success rates for skaters A, B, C and D are: 80%, 83%, 52% and 86% respectively. Find the probability that:

A) Skater A and D are successful, but C and B are not.
B) Skater B and C are successful, but A and D are not.
C) None of the skaters win a medal.
D) At least one of the four skater wins a medal.

Round to four decimal places. Write a decimal such as 0.25 as just .25.

Answer 32

Again, assuming independence (which seems really weird in this case)

\[P(A\text{ and }D\text{ and }C^{c}\text{ and }B^{c})\]

\[P(A)P(D)P(C^{c})P(B^{c})\]

Answer 33

these problems are done just like the last problem you posted

Answer 34

well we still need help, please

Answer 35

tell me what

\[P(A)=\]

\[P(D)=\]
\[P(C^{c})=\]
\[P(B^{c})=\]

Answer 36

idk

Answer 37

"The skaters success rates for skaters A, B, C and D are: 80%, 83%, 52% and 86% respectively. "

Answer 38

are you telling me that you don't know at least the first two because they are written in the text I quoted from you.

Answer 39

no we do but we thought you already knew

Answer 40

we got A its .056 right?

Answer 41

yes

Answer 42

so wat are b, c and d

Answer 43

last one I'm giving...

b) .0120848

Answer 44

but i need help

Answer 45

tell me what you have and I'll tell you if it is correct. Again, these problems are done exactly like the problem you posted before

Answer 46

the things is we don't know how to do it, we didn't quite understand your explanation last time

Answer 47

for (c) you don't want anyone to get a metal so you need find the probability of the complement that do win the metal

\[P(Z^{c})=1-P(Z)\]

and multiply all four of them together

Answer 48

what is PZ

Answer 49

uhhh i just dont understand I suck at math

Answer 50

um..ok..

take each of the probabilities given and subtract them from 1

them multiply those 4 numbers together

Answer 51

.0022848

Answer 52

yes...that is the answer, though you really have no idea why.

Answer 53

the answer to part D is just 1 minus the answer to part C

Answer 54

hahah no unfortunatly but .9977152

Answer 55

correct

Answer 56

it was wrong but thanks anyway

Answer 57

under the assumption of independence it is not wrong