Q81 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
If getting a head on a coin when it is tossed Is considered as success, then the probability of having more number of failures when ten faír coins are tossed simultaneously, is
- A$\frac{105}{2^8}$
- B$\frac{73}{2^7}$
- Correct answerC$\frac{193}{2^9}$
- D$\frac{638}{2^{10}}$
Solution
$n=10, p=\frac{1}{2}, q=\frac{1}{2}$
$$
\begin{aligned}
& \therefore \text { Required probability }=P(r \geq 6) \\
& =P(r=6)+P(r=7)+P(r=8)+P(r=9)+P(r=10) \\
& ={ }^{10} C_6\left(\frac{1}{2}\right)^{10}+{ }^{10} C_7\left(\frac{1}{2}\right)^{10}+{ }^{10} C_8\left(\frac{1}{2}\right)^{10}+{ }^{10} C_9\left(\frac{1}{2}\right)^{10} \\
& +\left(\frac{1}{2}\right)^{10}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{2^{10}}\left({ }^{10} C_6+{ }^{10} C_7+{ }^{10} C_8+{ }^{10} C_9+1\right) \\
& =\frac{1}{2^{10}}(210+120+45+10+1)=\frac{386}{2^{10}}=\frac{193}{2^9}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.