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.
Solution
Correct Answer: Option C
S = { 1, 2, 3, 4, .....18 }
=> n(S) = 18
E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}
=> n(E1) = 9
E2 = {3, 6, 9, 12, 15, 18 }
=> n(E2) = 6
\(E3\;=\left(E1\cap E2\right)=\{6,\;12,\;18\}\)
=> n(E3) = 3
\(\therefore E=E1\;\cup\;E2\;=\;E1+E2-E3\)
=> n(E) = 9 + 6 - 3 =12
where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }
\(\therefore P(E)=\frac{n(E)}{n(S)}=\frac{12}{18}=\frac23\)
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