Question

A fair coin is tossed a number of times. If the probability of having at least one head is more than $90\% $, then n is greater than or equal to
A. 2
B. 3
C. 4
D. 5

Answer 1

Hint: Here we have to find the probability for the given question. We have to use one of the properties of probability i.e., $P\left( {A'} \right) = 1 - P\left( A \right)$. So, first we find out the probability of getting head when the coin is tossed n times and equating it to the minimum value we can determine the value of n.

Complete step by step solution:
 Given, the probability of having at least one head is more than $90\% $.
I.e., $P\left( {atleast\,one\,head} \right) > 90\% $
or
$P\left( {atleast\,one\,head} \right) > \dfrac{{90}}{{100}}$
$ \Rightarrow \,1 - P\left( {no\,head} \right) > \dfrac{{90}}{{100}}$ ----- (1)
Probability of getting head in coin $ = \dfrac{1}{2}$
By using binomial distribution, $n = n$, $p = $ probability of getting head.
$p = \dfrac{1}{2}$ and $q = 1 - p \Rightarrow q = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
 $P\left( {X = r} \right) = {}^n{C_r}\,{P^{n - r}}{q^r}$
$ \Rightarrow \,P\left( {X = 0} \right) = {}^n{C_0}\,{\left( {\dfrac{1}{2}} \right)^{n - 0}}{\left( {\dfrac{1}{2}} \right)^0}$
$ \Rightarrow P\left( {X = 0} \right) = 1\,{\left( {\dfrac{1}{2}} \right)^n} \times 1$
$ \Rightarrow P\left( {X = 0} \right) = {\left( {\dfrac{1}{2}} \right)^n}$
By (1)
$ \Rightarrow 1 - P\left( {X = 0} \right) > \dfrac{{90}}{{100}}$
$ \Rightarrow 1 - {\left( {\dfrac{1}{2}} \right)^n} > \dfrac{9}{{10}}$
$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^n} < 1 - \dfrac{9}{{10}}$
$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^n} < \dfrac{1}{{10}}$
$ \Rightarrow {2^n} > 10$
So, least value of n is 4

Option ‘C’ is correct

Note: In the probability the student has to know the difference between at least and at most. The word “at least” means it will be the minimum value and then it can be exceeded to the maximum. The word “at most” means it is the maximum value and it can’t exceed further, we have to consider the minimum value also.