Solution: The sum of the probability of an event and its complement is always 1.

\[ P(E) + P(\text{not } E) = 1 \]

Solution: The probability of an impossible event is 0. Such an event is called an impossible event.

\[ P(\text{Impossible event}) = 0 \]

Solution: The probability of a certain event is 1. Such an event is called a sure event.

\[ P(\text{Certain event}) = 1 \]

Solution: The sum of probabilities of all elementary events of an experiment is 1.

\[ \sum P(E_i) = 1 \]

Solution: The probability of any event \( E \) satisfies:

\[ 0 \leq P(E) \leq 1 \]

Solution: Not equally likely. The probability depends on the car’s condition, fuel, etc.

Solution: Not equally likely. The outcome depends on the player’s skill.

Solution: Equally likely. There’s a 50% chance of guessing correctly.

\[ P(\text{Right}) = P(\text{Wrong}) = 0.5 \]

Solution: Equally likely (assuming equal probability for biological sexes).

\[ P(\text{Boy}) = P(\text{Girl}) ≈ 0.5 \]

Solution:

We know that \( P(E) + P(\text{not } E) = 1 \)

\[ P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95 \]

Solution: Since the bag contains only lemon flavored candies:

\[ P(\text{Orange}) = 0 \] (Impossible event)

Solution: Since all candies are lemon flavored:

\[ P(\text{Lemon}) = 1 \] (Certain event)

Solution: Original deck has 52 cards. After removing 13 hearts, 39 cards remain.

Number of aces in remaining cards: 3 (since Ace of Hearts was removed)

\[ P(\text{Ace}) = \frac{3}{39} = \frac{1}{13} \]

Solution: All 13 diamonds remain in the deck of 39 cards.

\[ P(\text{Diamond}) = \frac{13}{39} = \frac{1}{3} \]

Solution: Since all hearts have been removed, all remaining 39 cards are not hearts.

\[ P(\text{Not heart}) = \frac{39}{39} = 1 \]

Solution: Since all hearts have been removed:

\[ P(\text{Ace of Hearts}) = 0 \] (Impossible event)

Solution:

Let \( P(\text{Same}) \) be the probability that two students share a birthday.

Given \( P(\text{Not same}) = 0.992 \)

\[ P(\text{Same}) = 1 – P(\text{Not same}) = 1 – 0.992 = 0.008 \]

Solution: Prime numbers on a die: 2, 3, 5 (3 outcomes)

Total possible outcomes: 6

\[ P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \]

Solution: Numbers between 2 and 6: 3, 4, 5 (3 outcomes)

\[ P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \]

Solution: Odd numbers on a die: 1, 3, 5 (3 outcomes)

\[ P(\text{Odd}) = \frac{3}{6} = \frac{1}{2} \]

Solution:

Total cards in deck: 52

Red kings: King of Hearts and King of Diamonds (2 cards)

\[ P(\text{Red King}) = \frac{2}{52} = \frac{1}{26} \]

Sample Problems:

1. What is the probability of rolling a number greater than 4 on a standard die?
2. If two cards are drawn from a standard deck without replacement, what is the probability both are spades?
3. In a group of 30 people, what is the probability that at least two people share the same birthday?
4. What is the probability of drawing a face card (Jack, Queen, King) from a standard deck?
5. If you roll two dice, what is the probability that the sum is 7?

These problems can be discussed with friends and teachers to understand probability concepts better.