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 showing the letters A, B, C, D, E, and A. The total number of possible outcomes is 6.
For the event 'getting A', the outcomes favourable are the two faces marked 'A'. Therefore, the number of favourable outcomes is 2. Using the probability formula, P(A) = 2/6 = 1/3.
For the event 'getting D', the letter 'D' appears on only one face of the die. Thus, the number of favourable outcomes is 1.
Therefore, the probability of getting D is P(D) = 1/6.
Explanation
The solution applies the probability formula P(E) = (Number of favourable outcomes) / (Total number of possible outcomes) as demonstrated in Example 3 of the provided context. The total number of outcomes is 6 because the die has six faces. For part (i), the letter 'A' appears on two faces (A, B, C, D, E, A), making the probability 2/6. For part (ii), the letter 'D' appears on only one face, making the probability 1/6.