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 (i) A? (ii) D?
The die has six faces with letters A, B, C, D, E, A. When thrown once, the total number of possible outcomes is 6, as each face is equally likely to appear. The event 'getting A' has favourable outcomes equal to the number of faces showing A, which is 2. Therefore, using the formula P(Event) = (Number of favourable outcomes)/(Total number of possible outcomes), we get P(A) = 2/6 = 1/3.
Similarly, for the event 'getting D', the favourable outcomes are the faces showing D. Only one face shows D. Hence, P(D) = 1/6.
Explanation
This question is analogous to Example 3 in the textbook, where a die is thrown and probabilities are calculated based on favourable outcomes. The die has six faces, two of which show A and one shows D. The classical probability formula P(E) = (Number of favourable outcomes)/(Total outcomes) is applied. The total outcomes are 6, as there are six faces. For part (i), favourable outcomes for A are 2, giving probability 2/6 = 1/3. For part (ii), favourable outcomes for D are 1, giving probability 1/6.