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Let's solve the problems one by one.

### Problem 15:
Two teams, A and B, play a game with the following probabilities:
- Probability that A wins = 50% = 0.5
- Probability that B wins = 40% = 0.4
- Probability of a draw = 1 - (0.5 + 0.4) = 0.1

They play the game three times. Let's solve each part:

#### (i) A wins all three games:
For A to win all three games, we multiply the probability of A winning each game:
\[
P(\text{A wins all 3}) = (0.5) \times (0.5) \times (0.5) = (0.5)^3 = 0.125
\]

#### (ii) Teams will win alternately:
This means the order could be A, B, A or B, A, B (alternating win-loss pattern). There are two possible patterns. For each pattern, the probability is:
\[
P(\text{ABA}) = P(A) \times P(B) \times P(A) = 0.5 \times 0.4 \times 0.5 = 0.1
\]
Since there are two patterns (ABA and BAB), the total probability is:
\[
P(\text{alternating wins}) = 2 \times 0.1 = 0.2
\]

#### (iii) Two games will end up as a draw:
For this, the exact sequence is: two draws and one non-draw outcome (either A or B wins).

The total possibilities for two draws can be calculated by combinations:
- Probability of two draws: \( P(\text{draw})^2 = (0.1)^2 = 0.01 \)
- The remaining game can be won by either A or B, with probabilities of 0.5 or 0.4:
\[
P(\text{two draws, one non-draw}) = 3 \times 0.01 \times (0.5 + 0.4) = 3 \times 0.01 \times 0.9 = 0.027
\]

#### (iv) All three games will end up as a draw:
For all games to be draws, the probability is simply the draw probability raised to the power of 3:
\[
P(\text{all draws}) = (0.1)^3 = 0.001
\]

---

### Problem 16:
This involves calculating the probability that a student will pass both examinations, given that:
- Probability of passing at least one exam = 72% = 0.72
- Probability of passing each exam is 40% = 0.4

Let \( P(\text{pass both exams}) = x \). We can use the inclusion-exclusion principle:
\[
P(\text{at least one}) = P(\text{pass Exam 1}) + P(\text{pass Exam 2}) - P(\text{pass both})
\]
Substituting the given values:
\[
0.72 = 0.4 + 0.4 - x
\]
\[
x = 0.8 - 0.72 = 0.08
\]
So, the probability that a student will pass both examinations is 0.08, or 8%.

---

### Problem 17:
The expected value of petrol sales depends on whether the day is rainy or clear. We are given:
- Expected sales on a rainy day = Rs. 80,000
- Expected sales on a clear day = Rs. 95,000
- Probability of a clear day = 0.76
- Probability of a rainy day = 0.24

The expected value (E) is calculated as:
\[
E(\text{petrol sales}) = (P(\text{clear}) \times \text{sales on clear day}) + (P(\text{rainy}) \times \text{sales on rainy day})
\]
\[
E(\text{petrol sales}) = (0.76 \times 95,000) + (0.24 \times 80,000)
\]
Let's calculate this:

\[
E(\text{petrol sales}) = (0.76 \times 95000) + (0.24 \times 80000) = 72,200 + 19,200 = 91,400
\]

So, the expected value of petrol sales is Rs. 91,400.

---

Let me know if you need further details or explanations.

### Related Questions:
1. How would the probabilities change in Problem 15 if the probability of a draw were higher?
2. What is the expected value of petrol sales if the probabilities of rain and clear days were equal?
3. How would you calculate the probability if teams A and B played 5 games instead of 3?
4. Can you calculate the probability that a student passes only one of the exams in Problem 16?
5. How would the expected value in Problem 17 change if the average sales on a rainy day were Rs. 90,000?

### Tip:
When calculating probabilities involving multiple independent events, always multiply the individual probabilities of the events occurring in sequence.

1. Two teams A and B play a game three times. The probability that A wins is 50% and B wins is 40%. Find the probabilities for different outcomes of the games.
2. A table shows the marks obtained by a group of students in two exams. Given that the passing rate for each exam is 40%, find the probability that a student passes both exams.
3. A petrol pump sells petrol at different rates based on the weather. Find the expected value of petrol sales on Wednesday.

Solve probability and expected value problems involving game outcomes between two teams, student exam passing rates, and expected petrol sales based on weather conditions. Detailed step-by-step solutions provided for Grades 10-12.

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