When a coin is tossed 6 times, the probability of getting more heads than tails is
When a coin is tossed 6 times, the probability of getting more heads than tails is
- \(\frac{13}{32}\)
- \(\frac{15}{32}\)
- \(\frac{9}{32}\)
- \(\frac{11}{32}\)
Solution
Total outcomes are \(2^6=64\)
Required Probability \(=P\) (Getting Four Heads) \(+P\) (Getting 5 Heads) \(+P\) (Getting 6 Heads)
\(=\frac{\frac{6 !}{4 ! 2 !}}{64}+\frac{\frac{6 !}{5 ! 1 !}}{64}+\frac{\frac{6 !}{6 !}}{64}=\frac{15}{64}+\frac{6}{64}+\frac{1}{64}=\frac{11}{32}\)
Hence, option (d) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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