Let a die be rolled 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 , then is equal to [2023]

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Solution

Q.
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 [2023]

1 30

2 15

3 60

4 90

Ans.
(3)

$Given, P (getting odd numbers 7 times) = P (getting odd numbers 9 times)$

$\Rightarrow {}^{n}{C_{7}} {(\frac{1}{2})}^{7} {(\frac{1}{2})}^{n - 7} = {}^{n}{C_{9}} {(\frac{1}{2})}^{9} {(\frac{1}{2})}^{n - 9}$

$\Rightarrow {}^{n}{C_{7}} = {}^{n}{C_{9}}$ $∴$ $n = 9 + 7 = 16$

$Now, P (getting even number twice) = {}^{16}C_{2} {(\frac{1}{2})}^{14} {(\frac{1}{2})}^{2} =$ $\frac{{}^{16}{C_{2}}}{2^{16}}$

$= \frac{16!}{2! \times 14!} \times \frac{1}{2^{16}} = \frac{16 \times 15}{2 \times 2^{16}} = \frac{15 \times 4}{2^{15}} = \frac{k}{2^{15}}$

$So, k = 60$

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