Question

The probability of getting at least one tail in four throws of a coin is:
A. $\dfrac{{15}}{{16}}$
B. $\dfrac{1}{{16}}$
C. $\dfrac{1}{4}$
D. None of these

Answer 1

Hint: The given question revolves around the concepts and principles of probability. In the given case, each coin toss is an independent event. So, we will calculate the probability of getting heads and tails in each coin. Then, we will find the probability of at least one tail occurring in four coin tosses using the sum of probabilities of any event equal to one.

Formula used:
${\text{Probability = }}\left( {\dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}} \right)$
${\text{P}}\left( {{\text{atleast 1 tail}}} \right) + {\text{P}}\left( {{\text{no tail}}} \right) = 1$

Complete step by step solution:
In the problem, we have to calculate the probability of getting at least one tail in four throws of a coin. So, we first calculate the probability of not getting any tail in four throws of a coin.
Then, getting no tail in four throws of a coin means getting four heads in four throws of the coin.
So, we know the probability of getting heads in a coin toss $ = \left( {\dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}} \right) = \dfrac{1}{2}$.
Also, each coin toss is an independent event.
So, the probability of getting no tail in four coin tosses $ = {\left( {\dfrac{1}{2}} \right)^4} = \dfrac{1}{{16}}$
Now, we know that the sum of the probabilities in a probability distribution table consisting of all possible ways of doing a certain thing is one. Hence, the sum of the probability of getting at least one head and not happening of a certain thing is one.
Hence, the sum of the probability of getting at least one tail and the probability of getting no tail is equal to one.
Let ${\text{P}}\left( {{\text{atleast 1 tail}}} \right)$ denote the probability of getting at least one tail and ${\text{P}}\left( {{\text{no tail}}} \right)$ denotes the probability of getting no tail in four coin tosses.
So, we get, ${\text{P}}\left( {{\text{atleast 1 tail}}} \right) + {\text{P}}\left( {{\text{no tail}}} \right) = 1$
$ \Rightarrow {\text{P}}\left( {{\text{atleast 1 tail}}} \right) + \dfrac{1}{{16}} = 1$
$ \Rightarrow {\text{P}}\left( {{\text{atleast 1 tail}}} \right) = 1 - \dfrac{1}{{16}} = \dfrac{{15}}{{16}}$
Hence, the probability of getting at least one tail in four throws of a coin is $\dfrac{{15}}{{16}}$.

So, option (A) is the correct Answer.

Note: The sum of the elementary probabilities of all the possibilities of an event is always equal to one. There are many ways of solving equations as the one formed in the question itself. Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together.