Question

One coin is tossed, the probability of getting head is:
 \[\left( a \right)0\]
\[\left( b \right)1\]
 \[\left( c \right)\dfrac{1}{2}\]
\[\left( d \right)-\dfrac{1}{3}\]

Answer 1

Hint: To solve this question, we will, first of all, write the sample space possible or the total number of outcomes possible. Then we will calculate the number of favorable outcomes that are the number of times when the head comes while tossing a coin. Finally, we will use the formula of probability as \[\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}.\]

Complete step-by-step solution
Given that one coin is tossed. We have to calculate the probability of getting a head on this coin tossed. First of all, we will write all possible sample space in this event. The event is the tossing of a coin. A coin has two faces, one head, and one tail. Therefore, the possibility of faces when the coin is tossed is given by
Sample Space = {H, T}
where H represents the head and T represents the tail.
Therefore, the total sample space or the total number of outcomes is \[\left\{ H,T \right\}=2.\]
The probability is calculated by using the below stated formula.
\[\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]
The number of favourable outcomes = {H} = 1.
The probability of getting “a” head is \[\dfrac{1}{2}.\]
Therefore, the probability of getting “a” head is \[\dfrac{1}{2}.\] Hence, option (c) is the right answer.

Note: Students should also know that the probability of getting a tail is also \[\dfrac{1}{2}\] as there are two outcomes possible that is {H, T} and hence the probability of getting each is half, i.e. \[\dfrac{1}{2}.\] There is no other possibility in this case.