Question

A die is thrown 200 times and the outcomes with the frequencies are listed below:

Outcome	1	2	3	4	5	6
Frequency	40	38	43	29	28	22

Find the probabilities of getting a number more than 1 and less than 6 in a toss.
A. 0.65
B. 0.55
C. 0.69
D. none of these

Answer 1

We take the two events of total and probable. We find the number of outcomes of those events. If there are subevents, we express them as the addition form. Then we find the probability of the event using the theorem of probability.

Complete step-by-step solution:
There are two types of events that we need to consider. One is the total event of tossing the die 200 times. Let’s consider that event as event S.
So, number of outcomes of event S is $n\left( S \right)=200$.
The second event being the one where outcome of the die is more than 1 and less than 6 in a toss. The possible outcome numbers are 2, 3, 4, 5.
Let’s consider this event as event A which has four exclusive subevents ${{A}_{2}},{{A}_{3}},{{A}_{4}},{{A}_{5}}$ where ${{A}_{n}}$ defines the subevent of A having outcome number on the dice as n.
All these subevents are exclusive and independent of each other.
Number of outcomes of event ${{A}_{n}}$ is $n\left( {{A}_{2}} \right)=38,n\left( {{A}_{3}} \right)=43,n\left( {{A}_{4}} \right)=29,n\left( {{A}_{5}} \right)=28$.
Now we need the probabilities of getting a number more than 1 and less than 6 in a toss.
So, number of outcomes of event A is $n\left( A \right)=n\left( {{A}_{2}} \right)+n\left( {{A}_{3}} \right)+n\left( {{A}_{4}} \right)+n\left( {{A}_{5}} \right)$.
We put the value to get $n\left( A \right)=38+43+29+28=138$.
We know probability of event A will be $\dfrac{n\left( A \right)}{n\left( S \right)}$.
So, the probability is $\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{138}{200}=0.69$.
The correct option is (C).

Note: The equation $n\left( A \right)=n\left( {{A}_{2}} \right)+n\left( {{A}_{3}} \right)+n\left( {{A}_{4}} \right)+n\left( {{A}_{5}} \right)$ only possible because of the events being independent and exclusive to each other. Otherwise we would have used Bayes’ theorem to find out the probability differently.
We can also use the complement theorem where instead of taking outcomes as a number more than 1 and less than 6 in a toss, we would have taken outcomes as the complement numbers like 1 and 6. Finding their probability of outcome we would have subtracted from 1 to get the answer of the solution as the total probability is always 1.