Let a die be rolled n times.

Question

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $$\frac{k}{2^{15}}$$

Accepted Answer

Given that,

$\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times)

$$
\begin{aligned}
& \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\
& \Rightarrow{ }^n \mathrm{C}_7={ }^n \mathrm{C}_9 \\
& \Rightarrow n=7+9=16
\end{aligned}
$$

$$
\begin{aligned}
& \text { Hence, } \mathrm{P}(\text { Even number twice })=\frac{k}{2^{15}} \\
& \Rightarrow{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2}=\frac{k}{2^{15}} \\
& \Rightarrow \frac{16 \times 15}{2} \times \frac{1}{2^{16}}=\frac{k}{2^{15}} \Rightarrow k=60
\end{aligned}
$$

Solutions · 1