Correct Answer - Option 4 :
\(\frac{{31}}{{32}}\)
Concept:
When 'r' is a random variable, then by Binomial distribution
The probability of (r = n) in 'n' observations is given by
\({\bf{P}}\left( {{\bf{r}} = {\bf{n}}} \right) = {\;^{\bf{n}}}{{\bf{C}}_{\bf{r}}}\;{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}\) and p + q = 1
where p and q are the probability of success and failure.
Calculation:
Given:
p = Probability of getting head
q = Probability of not getting head = Probability of getting tail
n = 5, p = q = 0.5
Now, we know that
\({\bf{P}}\left( {{\bf{r}} = {\bf{n}}} \right) = {\;^{\bf{n}}}{{\bf{C}}_{\bf{r}}}\;{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}\)
Probability of getting at least head = 1 - Probability of getting zero head
∴ \({\rm{P}}\left( {{\rm{r}} > 1} \right) = {\rm{\;}}1 - {\rm{P}}\left( {{\rm{r}} = 0} \right) = 1{ - ^5}{{\rm{C}}_0}{\left( {0.5} \right)^0}{\left( {0.5} \right)^5} = 1 - \frac{1}{{32}} = \frac{{31}}{{32}}\)
Alternate Solution:
The unbiased coin is tossed 5 times.
Sample space = 25 = 32
Probability of getting at least head = 1 - Probability of getting zero head
Now, Probability of getting zero head = Probability of getting tail 5 times
Getting only tail = 1 (T, T, T, T, T)
Probability of getting only tail = \(\frac{1}{{32}}\)
Probability of getting at least head = \(1 - \frac{1}{{32}} = \frac{{31}}{{32}}\)