Let a die be rolled n times. Let the probabil | vedclass

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Question

$P$ (odd number 7 times) $= P$ (odd number 9 times)

${ }^{ n } C _7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{ n -7}={ }^{ n } C _9\left(\

Vedclass · Accepted Answer

$60$

Vedclass · Answer

$P$ (odd number 7 times) $= P$ (odd number 9 times)

${ }^{ n } C _7\left(\frac{1}{2}ight)^7\left(\frac{1}{2}ight)^{ n -7}={ }^{ n } C _9\left(\frac{1}{2}ight)^9\left(\frac{1}{2}ight)^{ n -9}$

${ }^{ n } C _7={ }^{ n } C _9$

$\Rightarrow n =16$

Required

$P ={ }^{16} C _2 	imes\left(\frac{1}{2}ight)^{16}$

$=\frac{16 \cdot 15}{2} 	imes \frac{1}{2^{16}}=\frac{15}{2^{13}}$

$\Rightarrow \frac{60}{2^{15}} \Rightarrow k =60$

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 $\frac{ k }{2^{15}}$, then $k$ is equal to:

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