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three-story apartment building

This topic has 2 expert replies and 0 member replies
vikram4689 Legendary Member
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three-story apartment building

Post Sun Aug 19, 2012 4:09 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    In a certain three-story apartment building, there are 11 tenants on the first floor, 3 tenants on the second floor, and 7 tenants on the third floor. If two tenants are selected at random to be co-chairs of the tenants' association, what is the probability that the two co-chairs will be from adjacent floors?


    (A) 7/60

    (B) 9/70

    (C) 11/60

    (D) 11/70

    (E) 9/35

    how can it be solved using INDIVIDUAL PROBABILITY. i mean don't solve using desired pairs/total pairs

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    Post Sun Aug 19, 2012 6:31 am
    vikram4689 wrote:
    In a certain three-story apartment building, there are 11 tenants on the first floor, 3 tenants on the second floor, and 7 tenants on the third floor. If two tenants are selected at random to be co-chairs of the tenants' association, what is the probability that the two co-chairs will be from adjacent floors?
    (A) 7/60
    (B) 9/70
    (C) 11/60
    (D) 11/70
    (E) 9/35
    Always ask, "What needs to happen for the desired outcome to occur?"
    We get: P(co-chairs from adjacent floors) = P(one from 1st floor then one from 2nd floor OR one from 2nd floor then one from 3rd floor OR one from 2nd floor then one from 1st floor OR one from 3rd floor then one from 2nd floor).
    = (11/21)(3/20) + (3/21)(7/20) + (3/21)(11/20) + (7/21)(3/20)
    = 33/420 + 21/420 + 33/420 + 21/420
    = 108/420
    = 9/35
    = E

    Cheers,
    Brent

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    Post Sun Aug 19, 2012 7:28 am
    vikram4689 wrote:
    In a certain three-story apartment building, there are 11 tenants on the first floor, 3 tenants on the second floor, and 7 tenants on the third floor. If two tenants are selected at random to be co-chairs of the tenants' association, what is the probability that the two co-chairs will be from adjacent floors?
    (A) 7/60
    (B) 9/70
    (C) 11/60
    (D) 11/70
    (E) 9/35
    We can also solve this question using counting techniques (as is the case with many probability questions).
    In this case, it might be useful to use the complement. That is P(co-chairs from adjacent floors) = 1 - P(co-chairs not from adjacent floors)

    P(co-chairs not from adjacent floors) = (# of outcomes where selections are not adjacent)/(total # outcomes)

    As always, calculate the denominator first.
    total # outcomes = total number of ways to select 2 tenants from 21 tenants
    This can be accomplished in 21C2 ways (this is combination notation)


    # of outcomes where selections are not adjacent
    There are two ways this can happen:
    1) The two people are selected from floor #1 and/or floor #3
    2) The two people are selected from floor #2

    1) The two people are selected from floor #1 and/or floor #3
    To determine the number of possible outcomes, take the 18 tenants from floors 1 and 3, and select 2 of them.
    This can be accomplished in 18C2 ways

    2) The two people are selected from floor #2
    To determine the number of possible outcomes, take the 3 tenants from floor #2, and select 2 of them.
    This can be accomplished in 3C2 ways


    So, P(co-chairs not from adjacent floors) = (18C2 + 3C2)/(21C2)
    = (153 + 3)/(210)
    = 156/210
    = 26/35


    And finally, P(co-chairs from adjacent floors) = 1 - P(co-chairs not from adjacent floors)
    = 1 - 26/35
    = 9/35
    = E

    Cheers,
    Brent

    If anyone is interested, we have a free video on calculating combinations in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

    _________________
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    Come see all of our free resources

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