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Question: The probability of sample space is (a)0 (b)\(\dfrac{1}{2}\) (c)1 (d)None of these...

The probability of sample space is
(a)0
(b)12\dfrac{1}{2}
(c)1
(d)None of these

Explanation

Solution

Hint: To solve the question given above, first we will find out what is the meaning of the terms probability and sample space. Then we will take up a random experiment and we will find out its sample space and its probability. The probability will be determined by adding the individual probabilities of the outcomes.

Complete step by step solution:
In the question, we have to find out the probability of sample space. Before solving the question, we must know what is probability and what is sample space. The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event ‘A’ is the total number of ways event ‘A’ can occur divided by the total number of possible outcomes. The sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment. Now, we will take up a random experiment. This random experiment is the tossing of a coin.
Now, we have to find out of the sample space of this experiment. We know that when we toss a coin, either a head will show up or a tail will show. We denote head by ‘H’ and tail by ‘T’. Thus the sample space of tossing the coin is H, T. Now the probability of sample space will be equal to the sum of probabilities of individual outcomes. If P(S) is the probability of sample, P(H) is the probability of heads, and P(T) is the probability of tails, then we have:
P(S)=P(H)+P(T)P\left( S \right)=P\left( H \right)+P\left( T \right)
Now the probability P(H) and P(T) both will be equal to 12\dfrac{1}{2} because they can both come one time each when the coin is tossed two times. Thus, we have:
P(S)=12+12P\left( S \right)=\dfrac{1}{2}+\dfrac{1}{2}
⇒P(S)=1\Rightarrow P\left( S \right)=1
Thus, the probability of a sample space is 1.
Hence, option (c) is correct.

Note: Here, to solve the question, we have taken up a single coin. Instead of this we can use any number of coins. The thing that will remain as that the probability of sample space will always come out to 1. In fact any random experiment other than tossing coins will also have a probability of random experiment as 1.