A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield tails?
Correct Answer: Option D
\(P\left( E \right) = \frac{{Number\ of\ favorable\ outcomes}}{{Numeber\ of\ possible\ outcomes}} = \frac{{n\left( E \right)}}{{n\left( S \right)}}\)
⇒ Probability of getting tail in one coin = ½,
⇒ Probability of not getting tail in one coin = 1- ½ = ½,
Hence,
All the ten tosses are independent of each other.
∴ Required probability \(= {\left( {\frac{1}{2}} \right)^2} \times {\left( {\frac{1}{2}} \right)^8} = {\left( {\frac{1}{2}} \right)^{10}}\)
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