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Help with 2 simple probability probs

  1. Sep 14, 2004 #1
    help with 2 "simple" probability probs

    Ok, I'm having a hard time with probability.

    First problem.
    A restaurant offers a choice of 4 appetizers, 14 entrees, 6 desserts, and 5 beverages. How many different meals are possible if a diner intends to order only three courses? (consider the beverage to be a course)

    Maybe I'm making this problem harder than it is. I want to do 4*14*6*5 = answer, but that would only be for a 4 course meal, not a three course meal...right?

    Second problem

    Residents of a condominium have an automatic garage opener that has a row of 8 buttons. Each garage door has been programmed to respond to a particular set of buttons being pushed. If the condominium has 250 families, can residents be assured that no two garage doors will open on the same signal? If so, how many additional families can be aded before the 8-button code becomes inadequate? *Note* The order in which the buttons are pushed is irrelevant.

    I was thinking maybe the answer would be like 250^8, but that doesn't seem right to me. Since the order doesnt matter, 8 buttons such as 12345678 would be same thing as 87654321.

    So I was thinking maybe answer was 8! ?
  2. jcsd
  3. Sep 14, 2004 #2


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    Homework Helper

    First problem: You need to consider all the ways it is possible to get 3 courses.

    Second problem: A button will either be depressed or it will not. That gives two choices for each of 8 buttons. Also note that the case with no buttons pressed should be excluded.
  4. Sep 18, 2004 #3
    1st problem:
    4*14*6 + 14*6*5 + 4*14*5

    2nd problem:
  5. Sep 18, 2004 #4
    Actually, Tile was right, 00000000 should be excluded, so it should be

    Simply consider it's an 8-digit binary number.
  6. Sep 18, 2004 #5
    It's almost midnight where I'm from and I am feeling very sleepy. I can't think straight but this doesn't look right. The way I see it there are 4 ways of choosing 3 beverages from 4 beverages. So the answer is = 4*14*6*5*4.
  7. Sep 18, 2004 #6
    = n(ABC)+n(ABD)+n(ACD)+n(BCD)

    -- AI
  8. Sep 18, 2004 #7
    That looks like a handful. :bugeye: Not entirely elegant, is it? But it is good to memorise this just in case you are not able to solve it logically in an exam.
  9. Sep 18, 2004 #8
    No need to memorize .... it comes directly from a theorem called as the addition theorem ...... its more important to understand what i did in my first step rather what i have done in the following steps .... cuz they are derived from the theorem itself ...

    -- AI
    P.S -> I put the answers in the hope that it will encourage the enquiring mind as to what and how those steps came ...... instead of simply rote learning them .....
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