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Question
If three coins are tossed simultaneously, then the probability of getting at least two heads, is
Options
\[\frac{1}{4}\]
\[\frac{3}{8}\]
\[\frac{1}{2}\]
\[\frac{1}{4}\]
Solution
GIVEN: Three coins are tossed simultaneously.
TO FIND: Probability of getting at least two head.
When three coins are tossed then the outcome will be
TTT, THT, TTH, THH. HTT, HHT, HTH, HHH
Hence total number of outcome is 8.
At least two heads means that, THH, HHT, HTH and HHH are favorable events
Hence total number of favorable outcome is 4
`"We know that PROBABILITY" = "Number of favourable event"/"Total event number of event"`
Hence probability of getting at least two head when three coins are tossed simultaneously is equal to `4/8=1/2`
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