# Thread: basic probabilty

1. ## basic probabilty

Ninth grade homework giving me trouble again, here is the problem,

Your school cafeteria offers 3 entrees for lunch each day. The three entrees are randomly selected from a list of 20. Your favorite is grilled cheese. What is the probability the cafeteria will serve grilled chees at least once during the week( 5 days).

2. Originally Posted by LArchambeau3
Ninth grade homework giving me trouble again, here is the problem,

Your school cafeteria offers 3 entrees for lunch each day. The three entrees are randomly selected from a list of 20. Your favorite is grilled cheese. What is the probability the cafeteria will serve grilled chees at least once during the week( 5 days).

What have you tried and what is your thinking on the problem? Here is a hint using the compliment of the event may be a good idea!

3. ## cafeteria math

thoughts: find the total count of combinations for any one lunch. 20 Choose 3. 20!/(17!*3!)= 1140

find the count of negative out comes 19 choose 3: 19!/(16!*3!)=969

85% chance of it not happening. 15% it does. Now I am stuck on what to do about 5 days. If i add them together because it is an "or" situation I get 75% but the teacher says it about 55%.

4. The probability that the lunch cafeteria serves Grilled Cheese at least ONCE during the week, can be found by taking the compliment of them not serving any grilled cheese. You got the .85 (85%) and the .15 right. The problem is the next step. Remember that the probability of A happening AND B happening, is P(A)*P(B). In this case our A and B are days of the week. These days are independent; which means what is served on one day, does not affect what is being served on the other day (although it should, who wants Grilled Cheese twice a week).

Therefore, there are two options for any one day, agreed? Either the serve grilled cheese (.15), or they do not (.85). And we have five days that we need to account for therefore: (.85)*(.85)*(.85)*(.85)*(.85), or (.85)^(5). All we have done is looked at each individual day and said, "Ok I want there to be no grilled cheese on Monday AND on Tuesday AND on Wednesday. . .etc". This is a P(A & B & C & D & E) statement.

Of course this isn't the final answer, as we weren't asked to find the probability of their being no grilled cheese, but of their being grilled cheese at least once. Therefore we need to take the compliment of P(A & B & C & D & E) (since the opposite of it happening at LEAST once is that it doesn't happen at all).

5. ## details

Since there is a 15 % chance on each day, why doesn't .15^5 give the right answer? Is it because of the atleast once wording? Or is it more about adding the fact it could happen 5 times, 4times,3times,2times, 1 times in the 5 day week.

6. The reason it doesn't work is because that is the answer to the statement - "What is the probability that they serve grilled cheese each day for a week".

Even if you added up (0.15)^1*(0.85)^4+(0.15)^2*(0.85)^3+... - you wouldn't get the right answer. To solve "at least once. . ." WITHOUT using the compliment of an event, you would have to be familiar with the "binomial formula", and you would have to use it 5 times.

Now you may be familiar with that formula, but using the compliment is so much easier.

7. ## Binomial formula

Vague recollection that the binomial formula exists. You are so right about the compliment being easier. Thanks so much for your help.