MHT CET · Maths · Probability
One coin is thrown 100 times, then the probability of getting head in odd number is
- A \(\frac{1}{8}\)
- B \(\frac{1}{5}\)
- C \(\frac{1}{2}\)
- D \(\frac{3}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(n=100, p=\frac{1}{2}, q=\frac{1}{2}\)
\(p(x=r\) where \(r\) is odd \()=\sum_{r=\text { odd }}^n C_r p^r q^{n-r}\)
\(\begin{aligned} & =\sum_{r=\text { odd }}^{100} C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{100-r} \\ & =\left(\frac{1}{2}\right)^{100} \sum_{r=\text { odd }}{ }^{100} C_r \\ & =\left(\frac{1}{2}\right)^{100} \times 2^{99}=\frac{1}{2}\end{aligned}\)
\(p(x=r\) where \(r\) is odd \()=\sum_{r=\text { odd }}^n C_r p^r q^{n-r}\)
\(\begin{aligned} & =\sum_{r=\text { odd }}^{100} C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{100-r} \\ & =\left(\frac{1}{2}\right)^{100} \sum_{r=\text { odd }}{ }^{100} C_r \\ & =\left(\frac{1}{2}\right)^{100} \times 2^{99}=\frac{1}{2}\end{aligned}\)
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