EXAMPLES

1. Identify the sample space\\
The possible outcomes when a coin is tossed once are: \(S = \{H, T\}\).

2. Count the total number of outcomes\\
\(n(S) = 2\).

3. Probability of getting a head\\
- Favourable outcome for event "Head" is \{H\}.
- Number of favourable outcomes \(n(H) = 1\).
- Using the definition, \[P(\text{Head}) = \frac{n(H)}{n(S)} = \frac{1}{2} = 0.5.\]

4. Probability of getting a tail\\
- Favourable outcome for event "Tail" is \{T\}.
- Number of favourable outcomes \(n(T) = 1\).
- Using the definition, \[P(\text{Tail}) = \frac{n(T)}{n(S)} = \frac{1}{2} = 0.5.\]

5. Conclusion\\
Since the coin is fair, both events are equally likely and each has probability \(\frac{1}{2}\).

1. List the sample space (all possible outcomes). Since the bag has three balls of the same size, each ball is equally likely to be drawn.
$$S = \{\text{red}, \text{blue}, \text{yellow}\}$$
Hence, total number of equally likely outcomes \(n(S) = 3\).

2. (i) Probability of drawing the yellow ball:
- Favourable outcome = {yellow}, so \(n(\text{yellow}) = 1\).
- $$P(\text{yellow}) = \frac{n(\text{yellow})}{n(S)} = \frac{1}{3}$$

3. (ii) Probability of drawing the red ball:
- Favourable outcome = {red}, so \(n(\text{red}) = 1\).
- $$P(\text{red}) = \frac{n(\text{red})}{n(S)} = \frac{1}{3}$$

4. (iii) Probability of drawing the blue ball:
- Favourable outcome = {blue}, so \(n(\text{blue}) = 1\).
- $$P(\text{blue}) = \frac{n(\text{blue})}{n(S)} = \frac{1}{3}$$

5. Since each ball is equally likely, all three probabilities are equal to \(\frac{1}{3}\).

1. Let $S$ denote the event that Sangeeta wins the match and $R$ denote the event that Reshma wins the match.
2. Since the match can be won by either Sangeeta or Reshma and there are no other possibilities, the events $S$ and $R$ are mutually exclusive and exhaustive.
3. Therefore, $$P(S) + P(R) = 1.$$
4. Given $P(S) = 0.62$, substitute this value:
$$0.62 + P(R) = 1.$$
5. Solve for $P(R)$:
$$P(R) = 1 - 0.62 = 0.38.$$
6. Hence, the probability that Reshma wins the match is $0.38$.

1. Total possible outcomes\
Since each person can be born on any of the $365$ days (leap year ignored) and the birthdays are independent, the total number of ordered pairs of birthdays is\
$$\text{Total}=365 \times 365 = 133\,225.$$\
\
2. (ii) Probability that they have the same birthday\
- For the birthdays to be the same, once Savita’s birthday is fixed (any of the $365$ days), Hamida must be born on that same day.\
- Hence the number of favourable outcomes = $365$ (one for each possible common day).\
- Probability\
$$P(\text{same birthday}) = \frac{\text{favourable}}{\text{total}} = \frac{365}{133\,225}=\frac{1}{365}.$$\
\
3. (i) Probability that they have different birthdays\
- The complement of the event "same birthday" is "different birthdays".\
- Number of favourable outcomes for different birthdays = total outcomes $-$ same‑birthday outcomes\
$$\text{Favourable}=133\,225 - 365 = 132\,860.$$\
- Probability\
$$P(\text{different birthdays}) = \frac{132\,860}{133\,225}=\frac{364}{365}.$$\
- Alternatively, using conditional probability: after fixing Savita’s birthday, Hamida can be born on any of the remaining $364$ days, so\
$$P(\text{different}) = \frac{364}{365}.$$\
\
4. Result\
$$P(\text{different birthdays}) = \frac{364}{365}, \qquad P(\text{same birthday}) = \frac{1}{365}.$$

1. Identify total number of outcomes:\
The teacher has written the names of all 40 students on separate cards. Hence the total number of equally likely outcomes is \( n(S) = 40 \).

2. (i) Probability of drawing a girl:\
- Number of favourable outcomes (girls) = \( n(G) = 25 \).
- Using the definition of probability,\
$$ P(\text{girl}) = \frac{n(G)}{n(S)} = \frac{25}{40} = \frac{5}{8}. $$

3. (ii) Probability of drawing a boy:\
- Number of favourable outcomes (boys) = \( n(B) = 15 \).
- Hence,\
$$ P(\text{boy}) = \frac{n(B)}{n(S)} = \frac{15}{40} = \frac{3}{8}. $$

4. Check: Since the events "girl" and "boy" are complementary,\
$$ P(\text{girl}) + P(\text{boy}) = \frac{5}{8} + \frac{3}{8} = 1, $$
confirming the calculations are correct.

1. Total number of marbles
\[\text{Total}=3\text{ (blue)}+2\text{ (white)}+4\text{ (red)}=9\]

2. Probability of drawing a white marble
- Favourable outcomes = 2 (white marbles)
- \[P(\text{white})=\frac{2}{9}\]

3. Probability of drawing a blue marble
- Favourable outcomes = 3 (blue marbles)
- \[P(\text{blue})=\frac{3}{9}=\frac{1}{3}\]

4. Probability of drawing a red marble
- Favourable outcomes = 4 (red marbles)
- \[P(\text{red})=\frac{4}{9}\]

5. Check
- Sum of probabilities = \(\frac{2}{9}+\frac{3}{9}+\frac{4}{9}=\frac{9}{9}=1\), confirming the calculations are correct.

1. Identify the sample space (S).
Since the two coins are different, each toss can give a head (H) or a tail (T). The ordered pairs are:
$$S = \{(H_1,H_2),\;(H_1,T_2),\;(T_1,H_2),\;(T_1,T_2)\}$$
Hence, \(|S| = 4\).

2. Determine the favourable outcomes.
The event "at least one head" includes all outcomes except the one where both are tails:
$$E = \{(H_1,H_2),\;(H_1,T_2),\;(T_1,H_2)\}$$
So, \(|E| = 3\).

3. Apply the probability formula.
$$P(\text{at least one head}) = \frac{|E|}{|S|} = \frac{3}{4}$$

4. Write the answer in simplest form.
The required probability is \(\boxed{\dfrac{3}{4}}\).

1. Identify the sample space\\
The music can stop at any time \(t\) such that \(0 \le t \le 2\) minutes. Hence the sample space \(S\) is the interval \([0,2]\) minutes.

2. Assume uniform distribution\\
Since the stopping time is equally likely at any instant within the 2‑minute interval, the probability density is constant. For a uniform distribution on \([0,2]\), the total probability is 1.

3. Determine the favourable interval\\
The event of interest is "music stops within the first half‑minute" i.e. \(0 \le t \le 0.5\) minutes. The length of this favourable interval is \(0.5\) minute.

4. Compute the probability\\
\[
P(\text{stop within first 0.5 min}) = \frac{\text{length of favourable interval}}{\text{length of total interval}} = \frac{0.5}{2} = \frac{1}{4}.
\]

5. Express the answer\\
The required probability is \(\boxed{\frac{1}{4}}\) (or 0.25).

Hence, the probability that the music stops within the first half‑minute is \(\frac{1}{4}\).

1. Total number of shirts = 100 (all outcomes are equally likely).
2. For Jimmy: He accepts only the good shirts.
- Number of favourable shirts = 88.
- Probability \(P_{\text{Jimmy}} = \dfrac{88}{100} = \dfrac{22}{25} = 0.88\).
3. For Sujatha: She rejects only the shirts with major defects.
- Shirts that are *acceptable* to Sujatha are those that are not majorly defective, i.e., good shirts + shirts with minor defects.
- Number of favourable shirts = 88 (good) + 8 (minor defects) = 96.
- Probability \(P_{\text{Sujatha}} = \dfrac{96}{100} = \dfrac{24}{25} = 0.96\).
4. Answer:
- (i) \(P(\text{acceptable to Jimmy}) = \dfrac{22}{25}\).
- (ii) \(P(\text{acceptable to Sujatha}) = \dfrac{24}{25}\).

1. List of all possible outcomes
The ordered pair \((b,g)\) denotes the number on the blue die \(b\) and the number on the grey die \(g\).
$$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$
Total outcomes \(N = 36\).

2. Probability that the sum is 8
Find all pairs with \(b+g = 8\):
$$\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$$
Number of favourable outcomes \(F = 5\).
$$P(\text{sum}=8) = \frac{F}{N} = \frac{5}{36}$$

3. Probability that the sum is 13
The maximum possible sum with two dice is \(6+6 = 12\). Hence no outcome gives a sum of 13.
$$F = 0 \quad\Rightarrow\quad P(\text{sum}=13) = \frac{0}{36} = 0$$

4. Probability that the sum is \(\le 12\)
Every possible outcome has a sum between 2 and 12 inclusive. Therefore all 36 outcomes satisfy the condition.
$$F = 36 \quad\Rightarrow\quad P(\text{sum}\le 12) = \frac{36}{36} = 1$$