Worksheet titled "Chapter - 13 PROBABILITY" featuring 15 problems on calculating probabilities in various scenarios.

Let's solve each of the 15 probability problems step by step and explain the reasoning behind each solution.

---

1. A coin is tossed once, find the probability of getting 'Head'.

- A fair coin has two outcomes: Head (H) or Tail (T).
- Total possible outcomes = 2
- Favorable outcome for Head = 1

$$
P(\text{Head}) = \frac{1}{2}
$$

✔ Answer: $ \frac{1}{2} $

---

2. In a pack of 52 cards, what is the probability of getting a face card?

- A standard deck has 52 cards.
- Face cards are Jack, Queen, King → 3 per suit.
- There are 4 suits → $ 3 \times 4 = 12 $ face cards.

$$
P(\text{Face card}) = \frac{12}{52} = \frac{3}{13}
$$

✔ Answer: $ \frac{3}{13} $

---

3. A dice is tossed once; find the probability of getting a 'prime number'.

- Dice faces: 1, 2, 3, 4, 5, 6
- Prime numbers among these: 2, 3, 5 → 3 primes
- Total outcomes = 6

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

✔ Answer: $ \frac{1}{2} $

---

4. A dice is tossed once; find the probability of getting a number less than 5.

- Numbers less than 5: 1, 2, 3, 4 → 4 outcomes
- Total outcomes = 6

$$
P(<5) = \frac{4}{6} = \frac{2}{3}
$$

✔ Answer: $ \frac{2}{3} $

---

5. In a pack of 52 cards, find the probability of getting 2 of spades.

- Only one 2 of spades in a deck.
- Total cards = 52

$$
P(2\ \text{of spades}) = \frac{1}{52}
$$

✔ Answer: $ \frac{1}{52} $

---

6. In a cricket match, a batsman hits a boundary 5 times out of 30 balls. Find the probability that he did not hit a boundary.

- Boundary hits = 5
- Total balls = 30
- Not hitting boundary = $ 30 - 5 = 25 $

$$
P(\text{Not boundary}) = \frac{25}{30} = \frac{5}{6}
$$

✔ Answer: $ \frac{5}{6} $

---

7. In a bag there are 5 white, 4 black, 3 red balls. One ball is picked up randomly. What is the probability of getting a black ball?

- Total balls = $ 5 + 4 + 3 = 12 $
- Black balls = 4

$$
P(\text{Black}) = \frac{4}{12} = \frac{1}{3}
$$

✔ Answer: $ \frac{1}{3} $

---

8. A coin is tossed 500 times with frequencies: Head – 245, Tail – 255. What is the probability of getting head?

- Experimental probability = $ \frac{\text{Number of heads}}{\text{Total tosses}} $

$$
P(\text{Head}) = \frac{245}{500} = \frac{49}{100} = 0.49
$$

✔ Answer: $ \frac{49}{100} $ or $ 0.49 $

---

9. In 250 consecutive days, weather forecasts were correct 175 times. Find the probability of getting 'not correct' forecast.

- Correct forecasts = 175
- Total days = 250
- Not correct = $ 250 - 175 = 75 $

$$
P(\text{Not correct}) = \frac{75}{250} = \frac{3}{10}
$$

✔ Answer: $ \frac{3}{10} $

---

10. In class IX, total students are 36. Out of which 20 are boys. Find the probability of girls in the class.

- Total students = 36
- Boys = 20 → Girls = $ 36 - 20 = 16 $

$$
P(\text{Girl}) = \frac{16}{36} = \frac{4}{9}
$$

✔ Answer: $ \frac{4}{9} $

---

11. Between 5 and 15, find the probability of having an odd number.

- Numbers between 5 and 15: 6, 7, 8, 9, 10, 11, 12, 13, 14 → total = 9 numbers
- Odd numbers: 7, 9, 11, 13 → 4 odds

$$
P(\text{Odd}) = \frac{4}{9}
$$

✔ Answer: $ \frac{4}{9} $

> Note: "Between 5 and 15" usually means excluding 5 and 15.

---

12. In 1500 families, 814 have 2 children and rest have 1 child. Find the probability of families having 1 child.

- Families with 2 children = 814
- Total families = 1500
- Families with 1 child = $ 1500 - 814 = 686 $

$$
P(\text{1 child}) = \frac{686}{1500} = \frac{343}{750} \approx 0.4573
$$

✔ Answer: $ \frac{343}{750} $

---

13. In word 'INDIA', what is the probability of getting letter 'I'?

- Word: I, N, D, I, A → 5 letters
- Letter 'I' appears twice

$$
P(\text{I}) = \frac{2}{5}
$$

✔ Answer: $ \frac{2}{5} $

---

14. In a bag of 56 apples, 19 were rotten. One apple is chosen at random. Find the probability of getting a fresh apple.

- Rotten = 19 → Fresh = $ 56 - 19 = 37 $

$$
P(\text{Fresh}) = \frac{37}{56}
$$

✔ Answer: $ \frac{37}{56} $

---

15. A dice is tossed once, what is the probability of getting number '7'?

- A standard die has faces: 1, 2, 3, 4, 5, 6 → no 7
- So, favorable outcomes = 0

$$
P(7) = \frac{0}{6} = 0
$$

✔ Answer: $ 0 $

---

✔ Final Answers Summary:

| Q | Answer |
|---|--------|
| 1 | $ \frac{1}{2} $ |
| 2 | $ \frac{3}{13} $ |
| 3 | $ \frac{1}{2} $ |
| 4 | $ \frac{2}{3} $ |
| 5 | $ \frac{1}{52} $ |
| 6 | $ \frac{5}{6} $ |
| 7 | $ \frac{1}{3} $ |
| 8 | $ \frac{49}{100} $ |
| 9 | $ \frac{3}{10} $ |
|10 | $ \frac{4}{9} $ |
|11 | $ \frac{4}{9} $ |
|12 | $ \frac{343}{750} $ |
|13 | $ \frac{2}{5} $ |
|14 | $ \frac{37}{56} $ |
|15 | $ 0 $ |

Let me know if you'd like these answers formatted differently or explained further!