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**Hint:**As a man has to toss coin at least n times to have the probability of getting at least one head more than \[{\text{80% }}\]. The probability of having head is \[\dfrac{{\text{1}}}{{\text{2}}}\]and probability of having tail is \[\dfrac{{\text{1}}}{{\text{2}}}\]. Now, we apply binomial probability distribution, to get our solution.

**Complete step by step answer:**

As per the given, a man tosses a fair coin so that the probability of having at least one head is more than \[{\text{80% }}\].

Let the success in the event be \[{\text{p = }}\dfrac{{\text{1}}}{{\text{2}}}\]

Let the failure be \[{\text{q = }}\dfrac{{\text{1}}}{{\text{2}}}\]

Applying binomial distribution,

\[

{\text{P(x = r)}}{{\text{ = }}^{\text{n}}}{{\text{c}}_{\text{r}}}{{\text{(}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)}}^{\text{n}}} \\

\because {\text{p = q = }}\dfrac{{\text{1}}}{{\text{2}}} \\

\]

Now, probability of getting at least one head is

\[{\text{P(r) = 1 - (no head)}}\]

So making the below expression suitable,

\[{\text{P(r) = 1}}{{\text{ - }}^{\text{n}}}{{\text{c}}_{\text{0}}}{{\text{(}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)}}^{\text{n}}}\]

\[

= 1 - {{\text{(}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)}}^{\text{n}}} > 0.8 \\

= {{\text{(}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)}}^{\text{n}}} < 0.2 \\

\]

Hence, It is clearly evident at least value on n is \[{\text{3}}\]

Hence ,

**option(a) is our correct answer.**

**Note:**: Binomial probability refers to the probability of exactly x successes on n repeated trials in an experiment that has two possible outcomes (commonly called a binomial experiment). The binomial distribution assumes a finite number of trials, n. Each trial is independent of the last. This means that the probability of success, p, does not change from trial to trial. The probability of failure, q, is equal to \[{\text{1 - p}}\]; therefore, the probabilities of success and failure are complementary.

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