KCET · Maths · Probability
A die is thrown 10 times. The probability that an odd number will come up atleast once is
- A \(\frac{11}{1024}\)
- B \(\frac{1013}{1024}\)
- C \(\frac{1023}{1024}\)
- D \(\frac{1}{1024}\)
Answer & Solution
Correct Answer
(C) \(\frac{1023}{1024}\)
Step-by-step Solution
Detailed explanation
Given, \(n=10\)
Probability of odd number, \(p=\frac{1}{2}\)
\(\therefore \quad q=\frac{1}{2}\)
\(\therefore\) Required probability \(=p(X \geq 1)\)
\(=1-p(X=0)\)
\(=1-{ }^{10} C_0\left(\frac{1}{2}\right)^{10-0}\left(\frac{1}{2}\right)^0\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}=\frac{1023}{1024}\)
Probability of odd number, \(p=\frac{1}{2}\)
\(\therefore \quad q=\frac{1}{2}\)
\(\therefore\) Required probability \(=p(X \geq 1)\)
\(=1-p(X=0)\)
\(=1-{ }^{10} C_0\left(\frac{1}{2}\right)^{10-0}\left(\frac{1}{2}\right)^0\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}=\frac{1023}{1024}\)
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