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Did I do this counting problem correct?

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    The question says:
    A chain of stereo stores is offering a special price on a complete set of components (reciever, compact disc player, speakers). A purchaser is offered a chocie of manufactuer for each component:
    Reciever: Kenwood, Onkyo, Pioneer, Sony, Yamaha
    Compact disc player: Onkyo, Pioneer, Sony, Panasonic
    Speakers: Boston, Infinity, Polk.

    A switchboard dispaly in the store allows a customer to hook together any selection of components (consisting of one of each type.) Use the product rules to answer the following questions:

    a. In how many ways can one component of each type be selected?

    b. in how manyways can components be selected if both the reciever and the compact disc player are to be Sony.

    c. In how many ways can components be selected if none is to be Soney?

    d. In how many ways can a selection be made if at least one Sony component is to be included?

    e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component?
    Exactly one Sony component?

    2. Relevant equations
    Use the product rule.

    3. The attempt at a solution
    a) you must choose one from each group. Each choice is independent of the other choices of equipment. so of the first group (receivers) we have 5 choices, and of those five choices we have 5 different ways of choosing one of those items. and the same for the rest of the groups (4 ways to chose cd players, and 3 ways to chose the speakers). So we just multiply them together to get how many choices there are.
    5*4*3 = 60

    b)I think I know what it is asking and if I am correct....if the receiver and the cd player have to be sony then the only thing changing is the speakers and there are 3 so there are altogether 3 different choices.

    c) 4*3*3 because if you take out the sony for the receivers you still have 4 left and if you take the sony out of the cd players you have 3 left and the other has 3 left.

    d)when it says AT LEAST 1 must be sony then you must account for the combos that have 2 sonys as well.
    So in this case if you chose the first category to be a sony we must account for the possibility that a cd player could be a sony so we take the whole number of items in the cd player category times the number of items in the last category. 1*4*3

    and if we choose the 2 category to be a sony we must account for the possibility that a receiver can be a sony as well so we get 5*1*3.

    So if there are 2 possible combinations, 1*4*3 and 5*1*3 would the answer be
    1*4*3 + 5*1*3?

  2. jcsd
  3. Jan 31, 2008 #2


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    For d), no. You still haven't counted the possibilities where both are Sony. The rest look ok. But you haven't tried e) yet. BTW shouldn't d) be a) minus c)??
  4. Jan 31, 2008 #3
    Thanks for the responce:

    well D = a - c = 60 - 36 = 24

    So if they were both soney's it would be
    so that would be:
    3 + (1)(4)(3) + (5)(1)(3) = 30;

    The question says if AT LEAST 1, meaning, I don't have to count for if they are both do I? that would be counting what if they were ALL soney right?
    Because I'm counting:
    First case: Soney is selected for the reciever (at least 1 is satsified)
    second case: Soney is selected for the disk palyer (at least 1 is satsified)

    if I include another count, i would be double counting because (1)(4)(3) in that middle 4, soney is one of the options the user can pick.
    In (5)(1)(3) that 5, soney is one of the options the user could pick.

    they don't match up...am I calculating them bothing being soney wrong? is it not (1)(1)(3)?

    For e i have:

    The probability of a combination containing only 1 sony is .35. you can do it the same as before where we count how many choices contain only 1 sony (21) and we divide it by the total number of combinations (60) or....
    there are 2 catagories that we can make have the 1 sony. If we choose the receiver to have the sony we then need the cd player not to be a sony so we multiply those odds
    (1/5)*(3/4) = (3/20)
    and then iff we choose the cd player to have the sony we do the same with the receiver category not to be a sony.
    (4/5)*(1/4) = (1/5)
    we then add these together
    (3/20)+(1/5) = .35
    Last edited: Jan 31, 2008
  5. Jan 31, 2008 #4


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    You are already double counting if you add 1*4*3 and 5*1*3. Think about it. They both include two Sony cases. Which is what I meant to say, sorry to be vague. The two one Sony cases are 1*3*3 and 4*1*3 and the double Sony case is 1*1*3. That does add up to a)-c). e) looks good. I think you've got it.
    Last edited: Jan 31, 2008
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