Probability

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Probability

by davo45 » Tue Sep 13, 2011 12:06 am
In Township K, 1/5 of the housing units are equipped with cable television . If 1/10 of the housing units , including 1/3 of those that are equipped with cable television , are equipped with vidieocassette recorders ,what fraction of the housing units have neither cable television nor videocassette recorders ?

(A)23/30
(B)11/15
(C)7/10
(D)1/6
(E)2/15

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by cans » Tue Sep 13, 2011 1:11 am
1/5 -> c
1/10 -> v
1/15 -> v and c
thus neither v nor c = 1 - 1/5 - 1/10 + 1/15 = 23/30
IMO A
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by knight247 » Tue Sep 13, 2011 1:17 am
Let the number of ppl in the township be 330
Number of ppl with cable=1/5th of 330=66
Number of ppl with vcr=1/10th of 330=33

1/3rd of those equipped with cable also have vcrs. So 1/3rd of 66=22. Meaning 22 people have both vcrs and cable.

This problem is best understood with a venn diagram which I will post later.

Total=Ppl with Cable+Ppl with vcr-ppl with both vcr and cable+ppl with neither
330=66+33-22+Neither
Neither=253

Fraction of ppl who have neither=253/330=23/30 Hence [spoiler]A[/spoiler]
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by GMATGuruNY » Tue Sep 13, 2011 3:09 am
davo45 wrote:In Township K, 1/5 of the housing units are equipped with cable television . If 1/10 of the housing units , including 1/3 of those that are equipped with cable television , are equipped with vidieocassette recorders ,what fraction of the housing units have neither cable television nor videocassette recorders ?

(A)23/30
(B)11/15
(C)7/10
(D)1/6
(E)2/15
Total = Group 1 + Group 2 - Both + Neither.

The big idea with overlapping groups is to SUBTRACT THE OVERLAP. When we count all of the housing units with cable (Group 1) and all of the housing units with VCRs (Group 2), the overlap -- the number of households with BOTH cable and VCRS -- gets counted twice. So we need to subtract the overlap so that these households are not double-counted.

Let total households = 30.
Group 1 = households with cable = (1/5)*30 = 6.
Group 2 = households with VCRs = (1/10)*30 = 3.
Both = 1/3 of the households with cable = (1/3)*6 = 2.

Plugging these values into the formula:
30 = 6 + 3 - 2 + N
N = 23.

Neither/Total = 23/30.

The correct answer is A.
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by sl750 » Tue Sep 13, 2011 5:15 am
Or you could use the matrix method. Assume 30 or 100 people

VCR Yes VCR No
Cable Yes 2 4 6
Cable No 1 23 24
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3 27 30