CBSE Class 10 Mathematics Probability Assignment Set 17

Long Answer Type Questions

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 2. A box contains 20 balls bearing numbers 1, 2, 3, ..., 19, 20 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is (i) a prime number (ii) an even number (iii) divisible by 3 or 5 (iv) neither divisible by 3 nor by 5.
Answer:
Total = 20.
(i) Primes: 2, 3, 5, 7, 11, 13, 17, 19 = 8. \( P = \frac{8}{20} = \frac{2}{5} \).
(ii) Even numbers = 10. \( P = \frac{10}{20} = \frac{1}{2} \).
(iii) Multiples of 3: 3, 6, 9, 12, 15, 18 (6); multiples of 5: 5, 10, 15, 20 (4); 15 is common, so the numbers divisible by 3 or 5 = 6 + 4 - 1 = 9. \( P = \frac{9}{20} \).
(iv) \( P(\text{neither}) = 1 - \frac{9}{20} = \frac{11}{20} \).
In simple words: For 'or', do not count a number twice.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 3. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3, ..., 12 (see the figure). What is the probability that it will point to
(i) 10? (ii) an odd number? (iii) a number which is a multiple of 3?
Answer:
Total number of outcomes = 12.
(i) \( P(10) = \frac{1}{12} \).
(ii) Odd numbers: 1, 3, 5, 7, 9, 11 = 6. \( P = \frac{6}{12} = \frac{1}{2} \).
(iii) Multiples of 3: 3, 6, 9, 12 = 4. \( P = \frac{4}{12} = \frac{1}{3} \).
In simple words: Count the favourable numbers out of 12.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 4. 240 students reside in a hostel. Out of them 40% go for the morning walk, 25% have joined the Gym club and 15% go to play cricket. The rest of the students have joined the laughing club. (i) Find the number of students who go to play cricket. (ii) What is the probability that a student has joined the laughing club? (iii) What is the probability that a student has not joined any club?
Answer:
(i) Cricket: \( 15\% \text{ of } 240 = 36 \) students.
(ii) Laughing club: \( 100\% - (40\% + 25\% + 15\%) = 20\% \) of 240 = 48 students. \( P = \frac{48}{240} = \frac{1}{5} \).
(iii) Every student has joined one of the four clubs (the rest joined the laughing club), so no student is without a club. \( P = \frac{0}{240} = 0 \), an impossible event.
In simple words: Convert the percentages to numbers before finding probabilities.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 5. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16.
Answer:
Total pairs (x, y) = \( 4 \times 4 = 16 \).
Products less than 16: x = 1: y = 1, 4, 9 (3 pairs); x = 2: y = 1, 4 (2 pairs); x = 3: y = 1, 4 (2 pairs); x = 4: y = 1 (1 pair). Total = 8. (Note \( 4 \times 4 = 16 \) is not less than 16.)
\( P = \frac{8}{16} = \frac{1}{2} \).
In simple words: List the pairs systematically for each x.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 6. A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one-digit number (ii) a number divisible by 5 (iii) an odd number less than 30 (iv) a composite number between 50 and 70.
Answer:
Total cards = \( 70 - 6 + 1 = 65 \).
(i) One-digit numbers: 6, 7, 8, 9 = 4. \( P = \frac{4}{65} \).
(ii) Multiples of 5: 10, 15, ..., 70 = 13. \( P = \frac{13}{65} = \frac{1}{5} \).
(iii) Odd numbers less than 30: 7, 9, 11, ..., 29 = 12. \( P = \frac{12}{65} \).
(iv) Composite numbers between 50 and 70: 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69 = 15 (the primes 53, 59, 61, 67 are excluded). \( P = \frac{15}{65} = \frac{3}{13} \).
In simple words: The cards run from 6 to 70, so there are 65 of them.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 7. Two different dice are thrown together. Find the probability that the numbers obtained have (i) an even sum (ii) an even product.
Answer:
Total outcomes = 36.
(i) The sum is even when both numbers are odd or both are even: \( 3 \times 3 + 3 \times 3 = 18 \) outcomes. \( P = \frac{18}{36} = \frac{1}{2} \).
(ii) The product is odd only when both numbers are odd: \( 3 \times 3 = 9 \) outcomes. So the product is even in \( 36 - 9 = 27 \) outcomes. \( P = \frac{27}{36} = \frac{3}{4} \).
In simple words: For the product, count the odd case and subtract.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 8. Two customers, Shyam and Ekta, are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Answer:
There are 5 days (Tuesday to Saturday), so each can visit on any of 5 days. Total number of equally likely outcomes \( = 5 \times 5 = 25 \).
(i) Same day: (T, T), (W, W), (Th, Th), (F, F), (S, S) = 5. \( P = \frac{5}{25} = \frac{1}{5} \).
(ii) Consecutive days: (T, W), (W, Th), (Th, F), (F, S) and the same pairs in the reverse order (W, T), (Th, W), (F, Th), (S, F) = 8. \( P = \frac{8}{25} \).
(iii) Different days: \( P = 1 - \frac{1}{5} = \frac{4}{5} \).
In simple words: Write the sample space as pairs (Shyam's day, Ekta's day).

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 10. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer:
Let there be x blue balls. Total = x + 5. \( \frac{x}{x + 5} = 2 \times \frac{5}{x + 5} \Rightarrow x = 10 \). There are 10 blue balls.
In simple words: Equal denominators allow a direct comparison of the numerators.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 11. A box contains 12 balls of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Answer:
\( P(\text{black}) = \frac{x}{12} \).
After adding 6 black balls there are 18 balls, of which \( x + 6 \) are black: \( P = \frac{x + 6}{18} \).
\( \frac{x + 6}{18} = 2 \times \frac{x}{12} \Rightarrow x + 6 = 3x \Rightarrow x = 3 \).
In simple words: Write the probability before and after adding the balls and equate as stated.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 12. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \( \frac{2}{3} \). Find the number of blue marbles in the jar.
Answer:
Green marbles \( = \frac{2}{3} \times 24 = 16 \).
Blue marbles \( = 24 - 16 = 8 \).
In simple words: Find the green marbles from the probability, then subtract from 24.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.