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# The probability for getting head or tail in a throw of a coin is:A. 0B. 1C. 0.5D. 0.25

Last updated date: 22nd Jul 2024
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Hint: For solving this question you should know about the concept of probability. According to the concept of probability the total sum of any event is equal to one. This means there can be many numbers of outcomes but the probability for all of them can be a maximum of one. We find the probability of that considered experiment to get the required answer. So, use this concept to reach the solution of the given problem.

According to our question it is asked to find the probability for getting head or tail in a throw of a coin. But for understanding the concept of probability, we take an example.
Example: Find the chance of throwing head and tail alternatively in three successive tossing of the coin.
So, S for this can be written as:
$S=\left\{ HHH,HHT,HTH,HTT,THH,TTH,THT,TTT \right\}$
Hence the total number of outcomes = 8.
Let E: event of getting head and tail alternatively in three successive trials.
So, the only possible outcomes are $\left\{ HTH,THT \right\}$.
Hence the number of possible outcomes = 2.
So, the probability of an event E is given by:
$P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Therefore $P\left( E \right)=\dfrac{2}{8}=\dfrac{1}{4}$.
Thus, if we see our question, then,
$S=\left\{ H,T \right\}$
So, the total number of outcomes are = 2.
And the probability for getting head or tail in one throw is,
$P\left( E \right)=\dfrac{1}{2}=0.5$.

So, the correct answer is “Option C”.

Note: In probability theory, an experiment or trial is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as sample space. The probability of an event is always lying between 0 and 1, that is $0\le P\left( E \right)\le 1$. The probability of an event E is given by the formula $P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.