A fail coin is tossed N times. The probability that head does not turn up in any of the tosses is A. $${\left( {\frac{1}{2}} \right)^{{\text{N}} – 1}}$$ B. $$1 – {\left( {\frac{1}{2}} \right)^{{\text{N}} – 1}}$$ C. $${\left( {\frac{1}{2}} \right)^{\text{N}}}$$ D. $$1 – {\left( {\frac{1}{2}} \right)^{\text{N}}}$$

[amp_mcq option1=”$${\left( {\frac{1}{2}} \right)^{{\text{N}} – 1}}$$” option2=”$$1 – {\left( {\frac{1}{2}} \right)^{{\text{N}} – 1}}$$” option3=”$${\left( {\frac{1}{2}} \right)^{\text{N}}}$$” option4=”$$1 – {\left( {\frac{1}{2}} \right)^{\text{N}}}$$” correct=”option1″]

The correct answer is $\boxed{B. 1 – {\left( {\frac{1}{2}} \right)^{{\text{N}} – 1}}}$.

A fair coin has two sides, heads and tails. When a fair coin is tossed, the probability of getting heads is $\frac{1}{2}$ and the probability of getting tails is also $\frac{1}{2}$.

If a fair coin is tossed $N$ times, the probability of getting heads all $N$ times is $\left( {\frac{1}{2}} \right)^{{\text{N}}}$.

The probability of not getting heads in any of the tosses is the complement of the probability of getting heads all $N$ times. The complement of an event is the event that does not occur. So, the probability of not getting heads in any of the tosses is $1 – \left( {\frac{1}{2}} \right)^{{\text{N}}}$.