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Mathematics 13 Online
OpenStudy (anonymous):

HELP ASAP!!!!!!!! FAN + MEDAL!!!! Imagine you are the owner of a pizza shop that offers three different types of crust, seven different toppings, and three different cheeses. Create an ad for your shop with 1 crusts, 3 toppings, 1 cheese. In your ad emphasize a special where any three topping pizza with one type of cheese and one type of crust only $9.99. Complete this statement and include it on your ad: "For this special we offer _______ different pizza choices for you to choose from!"

OpenStudy (anonymous):

Answer the following questions: Of the possible combinations you offer, which is your favorite? What is the probability that a person would order your favorite pizza at random?

OpenStudy (anonymous):

hang on

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

what are the choices?

OpenStudy (anonymous):

if there are any

OpenStudy (anonymous):

no chioces

OpenStudy (anonymous):

wait no you do 3*7*3 because the 7=the toppings 3 and 3 = the cheeses and the crust

OpenStudy (anonymous):

multiply it together what doyou get?

OpenStudy (anonymous):

63 combos?

OpenStudy (anonymous):

that should be it

OpenStudy (anonymous):

that should be the amount of pizza combos that you can make

OpenStudy (anonymous):

ok thanks can you help me figure out the rest?

OpenStudy (anonymous):

ill try :P im not the best but sure

OpenStudy (anonymous):

ok :) i just need help with the last question

OpenStudy (anonymous):

k

OpenStudy (anonymous):

What is the probability that a person would order your favorite pizza at random?

OpenStudy (anonymous):

ohh probability my worst enemy xD Ill try

OpenStudy (anonymous):

my favorite pizza is pepperoni with mozzarella cheese and garlic crust *not really my fave but ill stick with it* lol

OpenStudy (anonymous):

ok well you have 1 combination that is your favorite right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

well there can ONLY be 1 combo thats your fav

OpenStudy (anonymous):

yup and?

OpenStudy (anonymous):

i guess we have to do 1/63?

OpenStudy (anonymous):

ok so what would be the probability that a random person will order the exact combo

OpenStudy (anonymous):

that looks correct let me just make sure :)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Ok from the looks of it to me, 1/63 is correct because there is 1 combo ONLY and that 1 out of the 63 combos someone will order that

OpenStudy (anonymous):

May sound kinda confusing but that outa do it :P

OpenStudy (anonymous):

sorry if im wrong probability is the hardest for me

OpenStudy (anonymous):

me too its ok

OpenStudy (anonymous):

okays

OpenStudy (anonymous):

lets make sure @kidrah69

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

im gonna check my question real quick brb

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

@whpalmer4 @Holly00d1248 @FibonacciChick666 @King.Void. @Daniellelovee @urbanmorgans @Phebe

OpenStudy (anonymous):

hmm no one yet?

OpenStudy (anonymous):

can you help with this question? What is the probability that a person would order your favorite pizza at random? We think its 1/63

OpenStudy (anonymous):

@iGreen @Preetha

OpenStudy (anonymous):

brb gonna get a mouse and try something real quick

OpenStudy (anonymous):

kk

OpenStudy (holly00d1248):

They offer 6 different choices

OpenStudy (anonymous):

yep 6

OpenStudy (holly00d1248):

think of it this way start with the first crust and theres only one chease to choose from and three topings to choose from so that means theres 3 different pizzas you can mke now the second crust is the same way seeing it now @Anon101

OpenStudy (anonymous):

ok so what now?

OpenStudy (anonymous):

i don't really get probability...

OpenStudy (holly00d1248):

thats your answer i think

OpenStudy (anonymous):

1/63?

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

theres alot of stuff going on in this post, what do you need?

OpenStudy (anonymous):

i just need to understand What is the probability that a person would order your favorite pizza at random?

OpenStudy (holly00d1248):

@Legends Can you help buddy

OpenStudy (amistre64):

3 different types of crust, 7 different toppings, 3 different cheeses. Create an ad for your shop with 1 crusts, 3 toppings, 1 cheese. In your ad emphasize a special where any three topping pizza with one type of cheese and one type of crust only $9.99. how do we determine the number of pizzas we can make? and does each pizza need to be 3 toppings?

OpenStudy (amistre64):

can i have say: garlic crust, cheddar cheese, and say 3 pepperonis as the 3 toppings? or do i have to pick 3 different toppings?

OpenStudy (anonymous):

maybe its asking to do a special for a pizza with 1 crust 3 toppings and 1 cheese for $9.99

OpenStudy (anonymous):

i think it has to be different toppings

OpenStudy (amistre64):

well, we have 3*3 different crust and cheese varieties if pepperoni, sausage, and olives is the same choice regardless of order ... we have 7 choose 3 toppings for each of the 9 cheese crust varieties right?

OpenStudy (legends):

Im sorry @Holly00d1248 I dont get this one bro.

OpenStudy (anonymous):

ya see the 3*7*3 seems to be correct

OpenStudy (anonymous):

what he said ^^

OpenStudy (amistre64):

its not correct

OpenStudy (anonymous):

mm

OpenStudy (anonymous):

i dont get it

OpenStudy (anonymous):

right there with ya now lol

OpenStudy (amistre64):

there are 3*7*3 different one topping, one crust, one cheese

OpenStudy (amistre64):

\[\binom31\binom73\binom31 \]

OpenStudy (anonymous):

so 3*7*3 1*3*1

OpenStudy (amistre64):

7 choose 3 = (7 pick 3)/3! have you learned your combinatorics yet?

OpenStudy (anonymous):

i know permutations

OpenStudy (amistre64):

same thing i beleive how do we choose 3 out of 7?

OpenStudy (anonymous):

7*6*5?

OpenStudy (amistre64):

almost, if order matters, then that is correct. if order does not matter, we are 3! to many.

OpenStudy (amistre64):

not 'too many' but we have a multiple of 3! to many .... if order doesnt matter 7 pick 3 = 3!, times (7 group 3) we want 7 group 3

OpenStudy (anonymous):

what does 7 group 3 mean?

OpenStudy (amistre64):

7 group 3 = 7.6.5/3.2.1 = 7.5

OpenStudy (amistre64):

it means we want groups of 3 so that order doent matter

OpenStudy (anonymous):

ohh so it mean 7!/3!?

OpenStudy (amistre64):

not quite \[\frac{7!}{3!(7-3)!}\]

OpenStudy (amistre64):

\[7*6*5 = \frac{7!}{4!}\] \[\frac{7*6*5}{3!} = \frac{7!}{3!~4!}\]

OpenStudy (anonymous):

do i have to solve it?

OpenStudy (amistre64):

7 pick 3 is counting arrangements, order of elements matter. it is too large of a count. 7*6*5 is too big since it is counting (abc) as different then (bca) we need to reduce it, but how? well there are 3! ways to arrange 3 elements. which means we are 3! times to much

OpenStudy (anonymous):

ohh ok

OpenStudy (amistre64):

if i have 33, and im 3 times too much ... divide by 3

OpenStudy (anonymous):

so... 33/3=11?

OpenStudy (amistre64):

in my example yes, but thats just an example apply it to your case

OpenStudy (amistre64):

7*6*5 ordered arrangements; is 3! times, too much 7*6*5/3! is the number of ways to order our toppings

OpenStudy (anonymous):

so 210/3?

OpenStudy (amistre64):

3! = 6 by the way ...

OpenStudy (amistre64):

3*6*5 / 6 = 7*5

OpenStudy (anonymous):

oh sorry

OpenStudy (amistre64):

well, if my fingers will behave ... 7*6*5/6 = 7*5

OpenStudy (anonymous):

35

OpenStudy (anonymous):

the fin al equation would be 210/6=35 right?

OpenStudy (amistre64):

good 35 ways to order 9 (crust/cheese) combinations :) 35*9 different pizzas can be ordered

OpenStudy (amistre64):

and yes

OpenStudy (anonymous):

35*9=315 different combos

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