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Hint: Probability of event E = $\dfrac{n(E)}{n(S)}$= $\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the Probability. Find the number of elements in the sample space and the number of elements in Event E by listing all the possible cases and picking the favourable cases among them. Then use the above formula to find the Probability.
Complete step-by-step answer:
Let E be the event that on tossing the coin Head occurs.
Here S = {H,T}
and E = {H}
So, n(S) = 2 and n(E) = 1
Hence $P(E)=\dfrac{1}{2}$
Hence the Probability that Head occurs = $\dfrac{1}{2}$
Let F be the event that on tossing a coin Tail occurs
So, we have F = {T}
Hence n(F) = 1
Hence $P(F)=\dfrac{1}{2}$
Hence the Probability that tail occurs = $\dfrac{1}{2}$
Note:
[1] The Probability of an event always lies between 0 and 1
[2] The sum of probabilities of an event E and its complement E’ = 1
i.e. $p(E)+p(E')=1$
Hence, we have $p(E')=1-p(E)$. This formula is applied when it is easier to calculate p(E’) instead of p (E).
In the question we have F = E’. Hence $P(F)=1-P(E)=1-\dfrac{1}{2}=\dfrac{1}{2}$ which is the same as above.
[3] Probability is the likelihood of an event and not the guarantee. It means that if the Probability of an event is 0.5 it does not mean that the event occurs half the time. It simply ascertains that the likelihood of occurrence of the event is half the times.
Complete step-by-step answer:
Let E be the event that on tossing the coin Head occurs.
Here S = {H,T}
and E = {H}
So, n(S) = 2 and n(E) = 1
Hence $P(E)=\dfrac{1}{2}$
Hence the Probability that Head occurs = $\dfrac{1}{2}$
Let F be the event that on tossing a coin Tail occurs
So, we have F = {T}
Hence n(F) = 1
Hence $P(F)=\dfrac{1}{2}$
Hence the Probability that tail occurs = $\dfrac{1}{2}$
Note:
[1] The Probability of an event always lies between 0 and 1
[2] The sum of probabilities of an event E and its complement E’ = 1
i.e. $p(E)+p(E')=1$
Hence, we have $p(E')=1-p(E)$. This formula is applied when it is easier to calculate p(E’) instead of p (E).
In the question we have F = E’. Hence $P(F)=1-P(E)=1-\dfrac{1}{2}=\dfrac{1}{2}$ which is the same as above.
[3] Probability is the likelihood of an event and not the guarantee. It means that if the Probability of an event is 0.5 it does not mean that the event occurs half the time. It simply ascertains that the likelihood of occurrence of the event is half the times.
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