प्रायिकता (Probability)

The sample space is $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.

• (a) Prime number: Prime outcomes are $\{2, 3, 5\}$ ($1$ is neither prime nor composite):

$$n(E) = 3 \implies P(\text{Prime}) = \frac{3}{6} = \mathbf{\frac{1}{2}}$$

• (b) Number greater than $4$: Favorable outcomes are $\{5, 6\}$:

$$n(E) = 2 \implies P(> 4) = \frac{2}{6} = \mathbf{\frac{1}{3}}$$

• (c) Divisible by $3$: Favorable outcomes are $\{3, 6\}$:

$$n(E) = 2 \implies P(\text{Div by 3}) = \frac{2}{6} = \mathbf{\frac{1}{3}}$$

• (d) Number greater than $6$: No face has a number $> 6$ (Impossible event):

$$n(E) = 0 \implies P(> 6) = \frac{0}{6} = \mathbf{0}$$

.

The sample space is $S = \{HH, HT, TH, TT\} \implies n(S) = 4$.

• (a) Exactly one head: Outcomes are $\{HT, TH\}$:

$$n(E) = 2 \implies P(\text{Exactly 1 Head}) = \frac{2}{4} = \mathbf{\frac{1}{2}}$$

• (b) At least one head (1 or 2 heads): Outcomes are $\{HT, TH, HH\}$:

$$n(E) = 3 \implies P(\text{At least 1 Head}) = \mathbf{\frac{3}{4}}$$

• (c) At most one head (0 or 1 head): Outcomes are $\{TT, HT, TH\}$:

$$n(E) = 3 \implies P(\text{At most 1 Head}) = \mathbf{\frac{3}{4}}$$

.

Total outcomes $n(S) = 52$.

• (a) Red face card: There are 6 red face cards (King, Queen, Jack of Hearts and Diamonds):

$$P(\text{Red Face Card}) = \frac{6}{52} = \mathbf{\frac{3}{26}}$$

• (b) An ace: There are 4 aces in the deck:

$$P(\text{Ace}) = \frac{4}{52} = \mathbf{\frac{1}{13}}$$

• (c) A spade: There are 13 spade cards:

$$P(\text{Spade}) = \frac{13}{52} = \mathbf{\frac{1}{4}}$$

• (d) A black queen: There are 2 black queens (Queen of Spades and Queen of Clubs):

$$P(\text{Black Queen}) = \frac{2}{52} = \mathbf{\frac{1}{26}}$$

.

Step 1: Calculate total balls in bag:

$$n(S) = 5 + 8 + 4 + 7 = \mathbf{24\text{ balls}}$$

• (a) Black ball: $n(\text{Black}) = 7$:

$$P(\text{Black}) = \mathbf{\frac{7}{24}}$$

• (b) Not green ball:
Method 1: Balls that are not green $= 5 + 8 + 7 = 20$.

$$P(\text{Not Green}) = \frac{20}{24} = \mathbf{\frac{5}{6}}$$

Method 2: Using complementary probability:

$$P(\text{Green}) = \frac{4}{24} = \frac{1}{6} \implies P(\text{Not Green}) = 1 - \frac{1}{6} = \mathbf{\frac{5}{6}}$$

.

• No, neither value is possible.
• By the fundamental axioms of probability, the probability of any event $E$ must satisfy the invariant inequality:

$$\mathbf{0 \le P(E) \le 1}$$

• A probability can never be negative because the count of favorable outcomes $n(E) \ge 0$.
• A probability can never exceed $1$ because the number of favorable outcomes cannot exceed the total number of possible outcomes ($n(E) \le n(S)$).

• Simple (Elementary) Event: An event having exactly one single sample point (outcome) from the sample space.
Example: In rolling a die ($S = \{1, 2, 3, 4, 5, 6\}$), getting the number $5$ (Event $E = \{5\}$) is an elementary event.
• Compound Event: An event having two or more sample points.
Example: Getting an even number on a die (Event $E = \{2, 4, 6\}$) is a compound event because it is formed by combining three elementary outcomes.