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Question: A child has a die whose six faces show the letters as given below A B C D E A The die is thrown ...

A child has a die whose six faces show the letters as given below
A B C D E A
The die is thrown once. What is the probability of getting
(a) A?
(b) D?

Explanation

Solution

Hint:If the die was thrown once, it has six possible outcomes- A, B, C, D, E. To find the probability of an event, we need to divide the number of favorable outcomes by the total number of all possible outcomes. Thus, probability of an event E is given by
P(E)=n(E)n(S)P(E)=\dfrac{n(E)}{n(S)}
where,
n(E) is the number of favorable outcomes for event E to occur
n(S) is the total number of all possible outcomes (also called sample space)
For this experiment n(S)=5n(S)=5

Complete step by step answer:
(a) Here the event E is getting A which is possible in two ways since A is on two faces. Therefore,
n(E)=2 P(E)=25 \begin{aligned} & n(E)=2 \\\ & P(E)=\dfrac{2}{5} \\\ \end{aligned}
(b) Here the event E is getting D which is possible only when dice shows D. Therefore,
n(E)=1 P(E)=15 \begin{aligned} & n(E)=1 \\\ & P(E)=\dfrac{1}{5} \\\ \end{aligned}
Thus, answer for (a) is 25\dfrac{2}{5} and for (b) is also 15\dfrac{1}{5}

Note: A sample space and an event has much better explanation when represented in the form of a set. Then, sample S is a set having all the possible outcomes of an experiment as its elements while event E is the set of all possible outcomes. Event E is a subset of sample space S. It is very important to define the terms experiment, event and sample space very precisely in probability theory.

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