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!"
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?
hang on
ok
what are the choices?
if there are any
no chioces
wait no you do 3*7*3 because the 7=the toppings 3 and 3 = the cheeses and the crust
multiply it together what doyou get?
63 combos?
that should be it
that should be the amount of pizza combos that you can make
ok thanks can you help me figure out the rest?
ill try :P im not the best but sure
ok :) i just need help with the last question
k
What is the probability that a person would order your favorite pizza at random?
ohh probability my worst enemy xD Ill try
my favorite pizza is pepperoni with mozzarella cheese and garlic crust *not really my fave but ill stick with it* lol
ok well you have 1 combination that is your favorite right?
yes
well there can ONLY be 1 combo thats your fav
yup and?
i guess we have to do 1/63?
ok so what would be the probability that a random person will order the exact combo
that looks correct let me just make sure :)
ok
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
May sound kinda confusing but that outa do it :P
sorry if im wrong probability is the hardest for me
me too its ok
okays
lets make sure @kidrah69
ok
im gonna check my question real quick brb
kk
@whpalmer4 @Holly00d1248 @FibonacciChick666 @King.Void. @Daniellelovee @urbanmorgans @Phebe
hmm no one yet?
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
@iGreen @Preetha
brb gonna get a mouse and try something real quick
kk
They offer 6 different choices
yep 6
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
ok so what now?
i don't really get probability...
thats your answer i think
1/63?
@amistre64
theres alot of stuff going on in this post, what do you need?
i just need to understand What is the probability that a person would order your favorite pizza at random?
@Legends Can you help buddy
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?
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?
maybe its asking to do a special for a pizza with 1 crust 3 toppings and 1 cheese for $9.99
i think it has to be different toppings
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?
Im sorry @Holly00d1248 I dont get this one bro.
ya see the 3*7*3 seems to be correct
what he said ^^
its not correct
mm
i dont get it
right there with ya now lol
there are 3*7*3 different one topping, one crust, one cheese
\[\binom31\binom73\binom31 \]
so 3*7*3 1*3*1
7 choose 3 = (7 pick 3)/3! have you learned your combinatorics yet?
i know permutations
same thing i beleive how do we choose 3 out of 7?
7*6*5?
almost, if order matters, then that is correct. if order does not matter, we are 3! to many.
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
what does 7 group 3 mean?
7 group 3 = 7.6.5/3.2.1 = 7.5
it means we want groups of 3 so that order doent matter
ohh so it mean 7!/3!?
not quite \[\frac{7!}{3!(7-3)!}\]
\[7*6*5 = \frac{7!}{4!}\] \[\frac{7*6*5}{3!} = \frac{7!}{3!~4!}\]
do i have to solve it?
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
ohh ok
if i have 33, and im 3 times too much ... divide by 3
so... 33/3=11?
in my example yes, but thats just an example apply it to your case
7*6*5 ordered arrangements; is 3! times, too much 7*6*5/3! is the number of ways to order our toppings
so 210/3?
3! = 6 by the way ...
3*6*5 / 6 = 7*5
oh sorry
well, if my fingers will behave ... 7*6*5/6 = 7*5
35
the fin al equation would be 210/6=35 right?
good 35 ways to order 9 (crust/cheese) combinations :) 35*9 different pizzas can be ordered
and yes
35*9=315 different combos
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