Question

A coin is tossed \[n\] times. The probability of getting at least one head is greater than that of getting at least two tails by \[\dfrac{5}{{32}}\]. Then \[n\] is:
(A) 5
(B) 10
(C) 15
(D) None of these

Answer 1

Hint: According to the question, we need to know which formula should be used. Here, we have to use the binomial distribution formula. After applying the formula for both the cases, we need to solve it by the help of the given relationship between the two cases in the question

Complete step-by-step solution:
We know that we have to solve this question by the binomial distribution formula. The formula is:
\[ \Rightarrow {}^n{C_m}{p^m}{q^n}\]
Here, in this formula, \[n\]is the number of trials, \[m\]is the number of successes, \[p\]is probability of success, \[q\]is the probability of failure.
We know that when we toss a coin, then the probability of getting one head is \[\dfrac{1}{2}\], and the probability of getting a tail is \[\dfrac{1}{2}\].
Now, the question is the probability of getting at least one head. If we take the probability of getting at least one head as \[P(A)\], then we can represent \[P(A)\]as:
\[P(A) = 1 - P(no\,heads)\]
Now, when we further solve this. Here, we will put the binomial distribution formula. When we put the formula in the above given equation, we get that \[m = 0\]because it is the probability of getting no heads. Then, \[p\]is the probability of success, this means the probability of getting a head, and the probability of getting a head is \[\dfrac{1}{2}\]. So, the value of \[p\]is \[\dfrac{1}{2}\]. Now, \[q\]is the probability of failure. If the probability of success is \[\dfrac{1}{2}\], the probability of failure is \[\dfrac{1}{2}\]. So, the value of \[q\]is \[\dfrac{1}{2}\]. Now, when we put the values in the equation, we get:
\[P(A) = 1 - {}^n{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( {\dfrac{1}{2}} \right)^n}\]
We know that the value of \[{}^n{C_0}\]is 1. When we solve this, we get:
\[ \Rightarrow P(A) = 1 - {\left( {\dfrac{1}{2}} \right)^n}\]
Now, the next case is the probability of getting two tails. We can represent it as \[P(B)\]. Now, when we find the probability of getting at least two tails, we get:
\[P(B) = 1 - P(no\,\,tail) - P(1\,tail)\]
When we solve this according to the formula, we get that \[m = 0\]because it is the probability of getting no tail. Then, \[p\]is the probability of success, this means the probability of getting a tail, and the probability of getting a tail is \[\dfrac{1}{2}\]. So, the value of \[p\]is \[\dfrac{1}{2}\]. Now, \[q\]is the probability of failure. If the probability of success is \[\dfrac{1}{2}\], the probability of failure is \[\dfrac{1}{2}\]. So, the value of \[q\]is \[\dfrac{1}{2}\]. Now, for the probability of getting one tail, we get \[m = 1\]and \[n - 1\]. Now, when we put the values in the equation, we get:
\[P(B) = 1 - {}^n{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( {\dfrac{1}{2}} \right)^n} - {}^n{C_1}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{n - 1}}\]
We know that the value of \[{}^n{C_0}\]is 1 and the value of \[{}^n{C_1}\]is \[n\]. When we solve this, we get:
\[ \Rightarrow P(B) = 1 - {\left( {\dfrac{1}{2}} \right)^n} - n{\left( {\dfrac{1}{2}} \right)^n}\]
Now, according to the question, we get that the relationship between both \[P(A)\]and \[P(B)\]is:
\[ \Rightarrow P(A) - P(B) = \dfrac{5}{{32}}\]
Now, we will put the values of \[P(A)\]and \[P(B)\], and we get:
\[ \Rightarrow 1 - {\left( {\dfrac{1}{2}} \right)^n} - \left[ {1 - {{\left( {\dfrac{1}{2}} \right)}^n} - n{{\left( {\dfrac{1}{2}} \right)}^n}} \right] = \dfrac{5}{{32}}\]
\[ \Rightarrow 1 - {\left( {\dfrac{1}{2}} \right)^n} - 1 + {\left( {\dfrac{1}{2}} \right)^n} + n{\left( {\dfrac{1}{2}} \right)^n} = \dfrac{5}{{32}}\]
\[ \Rightarrow n{\left( {\dfrac{1}{2}} \right)^n} = \dfrac{5}{{32}}\]
\[ \Rightarrow \dfrac{n}{{{2^n}}} = \dfrac{5}{{32}}\]
When we compare both the sides, we get \[n = 5\].

Thus the correct answer is option ‘A’.

Note: In Mathematics, binomial distribution is a probability distribution of the number of successes and the number of failures. This distribution consists of parameters like \[n\,and\,p\]. This distribution is very common and is used in statistics.