Let's go through each part of the question step by step:
(a) Are the events [tex]\( T \)[/tex] and [tex]\( E \)[/tex] independent or not?
- Yes, the events [tex]\( T \)[/tex] (getting a tail on the coin) and [tex]\( E \)[/tex] (getting an even number on the die) are independent. This is because the outcome of the coin toss does not affect the outcome of the die roll.
(b) How many elements does the sample space have?
- The sample space for tossing a fair coin has 2 possible outcomes: {Head, Tail}.
- The sample space for rolling a standard die has 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
- Therefore, the total number of elements in the sample space when both the coin is tossed and the die is rolled is [tex]\( 2 \times 6 = 12 \)[/tex].
(c) Find the probability [tex]\( P(T) \)[/tex].
- The probability of getting a tail (T) on the coin is given by the ratio of the favorable outcomes (1 tail) to the total possible outcomes (2 outcomes: Head or Tail).
- Hence, [tex]\( P(T) = \frac{1}{2} \)[/tex].
(d) Find the probability [tex]\( P(E) \)[/tex].
- The probability of getting an even number on the die (E) is given by the ratio of the favorable outcomes (2, 4, 6 - 3 outcomes) to the total possible outcomes (6 outcomes: 1, 2, 3, 4, 5, 6).
- Hence, [tex]\( P(E) = \frac{3}{6} = \frac{1}{2} \)[/tex].
(e) Find the probability [tex]\( P(T \text{ and } E) \)[/tex].
- Since the events T and E are independent, the probability of both T and E happening together is the product of their probabilities.
- [tex]\( P(T \text{ and } E) = P(T) \times P(E) \)[/tex].
- So, [tex]\( P(T \text{ and } E) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)[/tex].
(f) Find the probability [tex]\( P(T \text{ or } E) \)[/tex].
- The probability of either T or E happening is given by the sum of their individual probabilities minus the probability of both happening together.
- [tex]\( P(T \text{ or } E) = P(T) + P(E) - P(T \text{ and } E) \)[/tex].
- So, [tex]\( P(T \text{ or } E) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4} \)[/tex].
Let's fill in the boxes with the values obtained:
(b) How many elements the sample space has? [tex]\( \boxed{12} \)[/tex]
(c) Find the probability [tex]\( P(T) \)[/tex] = [tex]\( \boxed{\frac{1}{2}} \)[/tex].
(d) Find the probability [tex]\( P(E) \)[/tex] = [tex]\( \boxed{\frac{1}{2}} \)[/tex].
(e) Find the probability [tex]\( P(T \text{ and } E) \)[/tex] = [tex]\( \boxed{\frac{1}{4}} \)[/tex].
(f) Find the probability [tex]\( P(T \text{ or } E) \)[/tex] = [tex]\( \boxed{\frac{3}{4}} \)[/tex].
