When six coins are tossed simultaneously, the probability of getting at least 4 heads is
When six coins are tossed simultaneously, the probability of getting at least 4 heads is
- $\frac{11}{64}$
- $\frac{15}{64}$
- $\frac{11}{32}$
- $\frac{15}{32}$
Solution
Total possible chances $=2^6=64$
Required probability
$
\begin{gathered}
=P(\text { Getting } 4 \text { Heads })+P(\text { Getting } 5 \text { Heads }) \\
\{\text { TTHHHH }\} \quad\{\text { THHHHH }\} \\
+P(\text { Getting } 6 \text { Heads })
\end{gathered}
$
\{Нннннн \}
$
\begin{aligned}
& =\frac{6 !}{2 ! 4 !}+\frac{6 !}{64 ! 5 !}+\frac{\frac{6 !}{6 !}}{64}=\frac{15}{64}+\frac{6}{64}+\frac{1}{64} \\
& =\frac{22}{64}=\frac{11}{32}
\end{aligned}
$
Hence, option (3) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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