Question

A coin is tossed \[1000\] times with the following frequencies:
Head: \[455\], Tail: \[545\]
Compute the probability of each event.

Answer 1

Hint:With the help of the formula of probabilities we will find the value of each event and then we will compare it.
Let us consider, the total number of outcomes of an event space is \[n\]. And, the number of outcomes for a particular event is \[m.\]
So, \[{\text{The probability of a certain event is = }}\dfrac{{{\text{the number of outcomes for a particular event\;}}}}{{{\text{the total number of outcomes of an event space}}}}\]

Complete step-by-step answer:
It is given that; a coin is tossed \[1000\] times.
Also given that the number of heads is \[455\] and number of tails is \[545\].
Now, we have to find the probability of head and probability of tail.
Let us consider, the total number of outcomes of an event space is \[n\]. And, the number of outcomes for a particular event is \[m.\]
So, \[{\text{The probability of a certain event P \;is = }}\dfrac{{{\text{the number of outcomes for a particular event}}}}{{{\text{\;the total number of outcomes of an event space}}}}\]
That is, \[P = \dfrac{m}{n}\]
Here, the coin is tossed for \[1000\] times. So \[n = 1000\]
Head comes for \[455\]times, so \[m = 455\]
So, the probability of head is \[P(H) = \dfrac{{455}}{{1000}}\]
Let us simplify the above equation we get,
\[P(H) = \dfrac{{91}}{{200}}\]
Consider \[n = 1000\]
Tail comes for \[545\] times so \[m = 545\]
So, the probability of tail is \[P(T) = \dfrac{{545}}{{1000}}\]
Let us simplify the above equation we get,
\[P(T) = \dfrac{{109}}{{200}}\]
Hence,
The probability of head is \[P(H) = \dfrac{{91}}{{200}} = 0.455\]
The probability of tail is \[P(T) = \dfrac{{109}}{{200}} = 0.545\]
Let us compare both the probabilities we have $0.455 < 0.545$ that is \[P(H) < P(T)\].
Hence we have found that the probability of getting heads is less than the probability of getting tails.

Note:The probability of any event lies between 0 and 1.
If there are two events in an event space, the total probability will be 1.
So, \[P(H) + P(T) = \dfrac{{91}}{{200}} + \dfrac{{109}}{{200}} = 1\]