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A die is thrown, write a sample space (S) and n(S). If event A is getting a number greater than 4, write event A and n(A).

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Last updated date: 21st Jun 2024
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Answer
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Hint: We know that sample space is a collection of all possible outcomes when an event occurs. We throw a die and look for possible outcomes which we know will be {1, 2, 3, 4, 5, 6}. n(S) means that the number of possible outcomes. Once, we have S, we will convert the number of elements it has which will help us to find n(S). After this, in a similar manner, we will find for event A also by considering the numbers greater than 4.

Complete step-by-step answer:
We are asked to find the sample space S and n(S). Before we start finding, we should define what actual sample space means. Sample space is just the collection of all possible outcomes when one thing happens and is denoted by S. n(S) tells us about the number of possible outcomes that occur during that event.
Now, we are given that the die is thrown. We know that a die has 6 surfaces with 1, 2, 3, 4, 5, 6 marked on those 6 surfaces (one on each side). When we roll a die, either one of these numbers can possibly show up.
So, our sample space consists of all three possible numbers. So, we get the sample space as,
\[S=\left\{ 1,2,3,4,5,6 \right\}\]
n(S) tells us about the number of outcomes when rolling a die can give the number from 1 -6. So, a total of 6 outcomes are possible in total.
\[\Rightarrow n\left( S \right)=6\]
Next, we are asked that event A is getting a number greater than 4. We have to look for n(A) and the sample space for A.
For sample space of A, we look for total outcomes when the die is rolled in 1, 2, 3, 4, 5, and 6. In this case, numbers greater than 4 are 5 and 6.
Hence, the sample space of A consists of just 5 and 6.
\[A=\left\{ 5,6 \right\}\]
Now, we can clearly see that A has 2 elements only. So, it means n(A) is 2.

Note: In the event A, we have mentioned a number greater than 4. So, we cannot consider 4 into it as 4 is not greater than 4. For sample space, we have to look for all possible outcomes that can occur. So if one event occurs twice on throwing a die, we will write it just once as we look just for the possible outcome not going for its frequency.