Question

An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is

1. \(\frac{1}{{32}}\)
2. \(\frac{{13}}{{32}}\)
3. \(\frac{{16}}{{32}}\)
4. \(\frac{{31}}{{32}}\)

Answer 1

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 = \(1 - \frac{1}{{32}} = \frac{{31}}{{32}}\)