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 May 4th, 2017, 08:41 AM #1 Newbie   Joined: May 2017 From: malaysia Posts: 1 Thanks: 0 Help with question In a list of 15 households, 9 own homes and 6 do not own homes. Four households are randomly selected from these 15 households. Find the probability that the number of households in these 4 who own homes is a) Exactly 3 b) At most 1 c) Exactly 4
 May 4th, 2017, 10:47 AM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614 a) $\binom{15}{4}=1365$ ways of selecting 4 $\binom{9}{3}\binom{6}{1}=504$ ways of selecting exactly 3 homeowners $p=\dfrac{504}{1365}=\dfrac{24}{65}$ b) at most 1 = 1 or 0 There are $\binom{6}{4}=15$ ways to select 0 homeowners and $\binom{6}{3}\binom{9}{1} = 180$ ways to select 1 homeowner $p=\dfrac{15+180}{1365} = \dfrac 1 7$ c) you can figure out this one

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