Question

Find the probability of getting at least one head when the coin is tossed twice.

Answer 1

Hint: In this question, we need to determine the probability of getting at least one head when a coin is tossed twice. For this, we will use the general definition of the probability by calculating the favorable number of outcomes and the total number of outcomes.

Complete step-by-step answer:
When a coin is tossed then, there are in total two possible outcomes i.e., HEAD and TAIL.
Let the total number of outcomes be denoted as n(T) and the total number of favorable outcomes be denoted as n(F).
Here in the question, the same coin has been tossed twice so, the total number of outcomes will be now 4 such that HEAD-HEAD, TAIL-TAIL, HEAD-TAIL and TAIL-HEAD. So, n(T)=4.
Now, we are interested in the probability of getting at least one head, which means that either one head or two heads but not zero heads. So, the favorable number of outcomes is 3 such as HEAD-HEAD, TAIL-HEAD and HEAD-TAIL. So, n(F)=3.
The ratio of the number of the favorable number of outcomes to the number of the total number of outcomes results in the probability of the event. Mathematically, $ P = \dfrac{{n(F)}}{{n(T)}} $
So, substituting the value of n(F) and n(T) in the equation $ P = \dfrac{{n(F)}}{{n(T)}} $ to determine the probability of getting atleast one head in tossing a coin twice.
 $
\Rightarrow P = \dfrac{{n(F)}}{{n(T)}} \\
   = \dfrac{3}{4} \\
  $
Hence, the probability of getting at least one head when the coin is tossed twice is $ \dfrac{3}{4} $ .

Note: Alternatively, we can also get the result by subtracting the probability of getting no heads with 1. So, the probability of getting at least one head = 1- probability of getting no heads.
 $
  P = 1 - \dfrac{1}{4} \\
   = \dfrac{3}{4} \\
  $