Q118 / 180 · KCET 2024 · Mathematics

A die is thrown 10 times. The probability that an odd number will come up atleast once is

  • A
    $\frac{11}{1024}$
  • B
    $\frac{1013}{1024}$
  • Correct answer
    C
    $\frac{1023}{1024}$
  • D
    $\frac{1}{1024}$

Solution

Given, $n=10$ Probability of odd number, $p=\frac{1}{2}$ $\therefore \quad q=\frac{1}{2}$ $\therefore$ Required probability $=p(X \geq 1)$ $=1-p(X=0)$ $=1-{ }^{10} C_0\left(\frac{1}{2}\right)^{10-0}\left(\frac{1}{2}\right)^0$ $=1-\frac{1}{2^{10}}$ $=1-\frac{1}{1024}=\frac{1023}{1024}$

Hence, the correct answer is option: 'C'.

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