MHT CET · Maths · Probability
If a fair coin is tossed 8 times, then the probability that it shows heads more than tails is
- A \(\frac{91}{256}\)
- B \(\frac{97}{256}\)
- C \(\frac{93}{256}\)
- D \(\frac{95}{256}\)
Answer & Solution
Correct Answer
(C) \(\frac{93}{256}\)
Step-by-step Solution
Detailed explanation
Given \(n=8 .\) Here \(p=\frac{1}{2}, q=\frac{1}{2}\)
\(P(x>4)=P(x=5)+P(x=6)+P(x=7)\) \(+~P(x=8) \)
\(={ }^{8} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\) \(\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \)
\(=\frac{56}{256}+\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{93}{256} \)
\(P(x>4)=P(x=5)+P(x=6)+P(x=7)\) \(+~P(x=8) \)
\(={ }^{8} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\) \(\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \)
\(=\frac{56}{256}+\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{93}{256} \)
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