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Question
A month is selected at random in a year. The probability that it is March or October, is
Options
\[\frac{1}{12}\]
\[\frac{1}{6}\]
\[\frac{3}{4}\]
None of these
Solution
A month is selected at random in a year.
TO FIND: Probability that it is March or October
Total months in a year is 12
Hence total number of favorable outcome is 2 i.e. March or October
`"We know that PROBABILITY" =" Number of favourable event"/"Total number of event"`
Hence probability that the month selected is March or October is equal to `2/12=1/6`
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