Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.
A) 1/3 B) 3/5 C) 2/3 D) 5/6
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Answer:
αnswєr: c) 2/3
єхplαnαtíσn:
s = { 1, 2, 3, 4, .....18 }
=> n(s) = 18
є1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}
=> n(є1) = 9
є2 = {3, 6, 9, 12, 15, 18 }
=> n(є2) = 6
є3 =(є1∩є2)={6, 12, 18}є3 =є1∩є2={6, 12, 18}
=> n(є3) = 3
∴є=є1 ∪ є2 = є1+є2−є3∴є=є1 ∪ є2 = є1+є2-є3
=> n(є) = 9 + 6 - 3 =12
whєrє є = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }
∴p(є)=n(є)/n(s)=12/18=2/3
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