Question 177723: A local church is studying the amount of offerings in an envelope from their early Sunday mornings services. The church studied 500 envelopes and found the following:
Less than $5 200
$5 to $9.99 100
$10 to $19.99 75
$20 to $49.99 75
$50 or more 50
a. what is the probability of opening one envelop and finding $20 or more in it? ______
b. What is the probability of opening one envelop and finding less than $10 in it? ______
c. What is the probability of opening two envelopes and finding less than $5 in
each one of them? ________
Found 2 solutions by stanbon, gonzo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A local church is studying the amount of offerings in an envelope from their early Sunday mornings services. The church studied 500 envelopes and found the following:
Less than $5 200
$5 to $9.99 100
$10 to $19.99 75
$20 to $49.99 75
$50 or more 50
a. what is the probability of opening one envelop and finding $20 or more in it? (75+50)/500 = 125/500 = 1/4
------------------------------------------
b. What is the probability of opening one envelop and finding less than $10 in it? (200+100)/500 = 3/5
------------------------------------------
c. What is the probability of opening two envelopes and finding less than $5 in
each one of them? 200C2/500C2 = 0.159519...
------------------------------------------
Cheers,
Stan H.
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! a = 200 envelopes out of 500 contained less than $5.00.
b = 100 envelopes out of 500 contained $5.00 to $9.99.
c = 75 envelopes out of 500 contained $10.00 to $19.99.
d = 75 envelopes out of 500 contained $20.00 to $49.99.
e = 50 envelopes out of 500 contained $50.00 or more.
---
the probability of any event occurring is the number of times that event can occur divided by the total number of events possible.
---
given that definition, then:
p(a) = 200/500
p(b) = 100/500
p(c) = 75/500
p(d) = 75/500
p(e) = 50/500
---
what is the probability of opening one envelope and finding $20 or more in it?
this would be p(d or e) = p(d) + p(e) = 125/500.
---
What is the probability of opening one envelope and finding less than $10 in it?
this would be p (a or b) = p(a) + p(b) = 300/500.
---
What is the probability of opening two envelopes and finding less than $5 in
each one of them?
this would be p(a and x) = p(a) * p(x) = (200/500) * (199/499) = (200*199)/(500*499)
---
what is p(x)?
here's what happened.
you had 500 envelopes.
you withdrew one envelope and that means you had 499 left to pick from.
the one you picked had less than $5.00 in it.
since you originally had 200 envelopes out of 500 with less than $5.00 in it, that means that you now have 199 envelopes out of 499 with less then $5.00 in it.
---
x = 199 envelopes out of 499 contained less than $5.00 in it.
p(x) = 199/499
---
the probability of 2 events occurring in succession is the probability of the first event occurring times the probability of the second event occurring.
this is why probability of getting 2 envelopes with less than $5.00 in it is p(a) * p(x).
---
your answers are:
the probability of opening one envelope and finding $20.00 or more is 125/500
the probability of opening one envelope and finding less than $10.00 is 300/500.
the probability of opening two envelopes and finding less than $5.00 in each is (200/500)*(199/499).
---
you can stop here if you are satisfied. otherwise continue for some more explanation.
---
to help you see this, take a smaller example:
suppose you have 4 envelopes:
1 envelope contains $1.00
1 envelope contains $2.00
1 envelope contains $3.00
1 envelope contains $4.00
---
since there are 4 total envelopes,
probability of opening 1 envelope and getting $1.00 is 1/4
probability of opening 1 envelope and getting $2.00 is 1/4
probability of opening 1 envelope and getting $3.00 is 1/4
probability of opening 1 envelope and getting $4.00 is 1/4
---
what is the probability of getting $1.00 or $2.00 on the first draw.
this would be 2/4 = 1/2 because your chances of success are 2 out of 4.
this is the same as p($1.00) + p($2.00) = 1/4 + 1/4 = 2/4 = 1/2.
---
what is the probability of opening 1 envelope and getting $1.00 and then opening a second envelope and getting $2.00?
---
once you opened the first envelope and got $1.00, you are left with 3 envelopes, 1 of which contains $2.00.
---
p($2.00) once you withdrew $1.00 is 1/3.
P($1.00 and then $2.00) = p($1.00) * p($2.00) = 1/4 * 1/3 = 1/12
---
how does this work?
well, 1/4 of the time you will get the first envelope containing $1.00.
1/3 of the time you will get the second envelope containing $2.00 assuming you got $1.00 the first time.
1/4 of 1/3 of the time you will get $1.00 opening the first envelope and then $2.00 opening the second envelope.
this is the same as (1/4)*(1/3) = (1/12)
---
since i used smaller numbers of envelopes, i can show you how this happens.
let A = envelope with $1.00 in it.
let B = envelope with $2.00 in it.
let C = envelope with $3.00 in it.
let D = envelope with $4.00 in it.
let the first time a letter shows up be the opening of the first envelope with the amount of money in it indicated by the letter.
let the second time a letter shows up be the opening of the second envelope with the amount of money in it indicated by the letter.
here's all the possibilities with first and second draw.
AB
AC
AD
BA
BC
BD
CA
CB
CD
DA
DB
DC
---
total number of possible ways to draw the first and second envelope is 12
only 1 of this draws A first and then B.
probability is therefore 1/12 of drawing A first and then B.
this is the same as p(A) * p(B) = 1/4 * 1/3 = 1/12.
---
|
|
|
