Question

Write the probability of getting each number top face when a die was rolled and find the sum of probabilities of all the outcomes.

Answer 1

As we are given that a die is rolled we get the sample space to be S = \left\\{ {1, 2, 3, 4, 5, 6} \right\\} and the considering a event that a number faces the top we can get its probability by using the formula P(A) =$\dfrac{{n(A)}}{{n(S)}}$. We need to find the probability for every number and adding all the probabilities we get the required sum .

Complete step by step solution:
We are given that a die is rolled
From this we get the sample space to be S = \left\\{ {1, 2, 3, 4, 5, 6} \right\\}
From this $n(S) = 6$
Now we are asked the probability of each number facing the top so now ,
Let , A be the event that the number 1 faces the top
Hence the probability of A, P(A) =$\dfrac{{n(A)}}{{n(S)}}$
=$\dfrac{1}{6}$
Let , B be the event that the number 2 faces the top
Hence the probability of B, P(B) =$\dfrac{{n(B)}}{{n(S)}}$
=$\dfrac{1}{6}$
Let , C be the event that the number 3 faces the top
Hence the probability of C, P(C) =$\dfrac{{n(C)}}{{n(S)}}$
=$\dfrac{1}{6}$
Let , D be the event that the number 4 faces the top
Hence the probability of D , P(D) =$\dfrac{{n(D)}}{{n(S)}}$
=$\dfrac{1}{6}$
Let , E be the event that the number 5 faces the top
Hence the probability of E , P(E) =$\dfrac{{n(E)}}{{n(S)}}$
=$\dfrac{1}{6}$
Let , F be the event that the number 6 faces the top
Hence the probability of F , P(F) =$\dfrac{{n(F)}}{{n(S)}}$
=$\dfrac{1}{6}$
Now we need to find the sum of all the probabilities.
$\Rightarrow P(A) + P(B) + P(C) + P(D) + P(E) + P(F) \\\ \Rightarrow \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{6}{6} = 1 \\\$

Hence we get the sum of the probabilities to be 1.

Note:
A) A probability of 1 means that an event is certain.
B) An event with a higher probability is more likely to occur.
C) Probabilities are always between 0 and 1.
D) The probabilities of our different outcomes must sum to 1.