CBSE Class 10 Probability Sure Shot Questions Set 06

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LOGICAL REASONING BASED QUESTIONS

1. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? [1 Mark]

Answer:
1. Tossing a coin has only two possible outcomes, namely head and tail.
2. Both outcomes are equally likely and mutually exclusive.
3. Neither team can control or predict the outcome beforehand, making it a completely unbiased and fair method.

Teacher's Note:
a) This question tests the fundamental concept of equally likely outcomes in probability.
b) Ensure students mention both symmetry of outcomes and unpredictability.

2. The probability that it will rain today is 0.84. What is the probability that it will not rain today? [1 Mark]

Answer:
1. Let E be the event that it will rain today, so P(E) = 0.84.
2. Let not E be the event that it will not rain today, so P(not E) = 1 - P(E).
3. P(not E) = 1 - 0.84 = 0.16.

Teacher's Note:
a) Uses the complement probability formula P(E) + P(not E) = 1.
b) Watch out for decimal subtraction errors.

3. What is the probability that an ordinary year has 53 Sundays? [1 Mark]

Answer:
1. An ordinary year has 365 days, which means 52 weeks and 1 remaining day.
2. This 1 remaining day can be any of the 7 days of the week: {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}.
3. Total possible outcomes = 7, and favorable outcome for Sunday = 1.
4. Required probability = \(\frac{1}{7}\).

Teacher's Note:
a) Relates days in a year to weekly cycles.
b) Remind students that an ordinary year always has 52 Sundays and one extra day.

4. Find the probability of getting 53 Fridays in a leap year. [1 Mark]

Answer:
1. A leap year has 366 days, which means 52 weeks and 2 remaining days.
2. The 2 remaining days can be: (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday).
3. Total possible outcomes = 7.
4. Favorable outcomes for Friday are (Thursday, Friday) and (Friday, Saturday), so total favorable outcomes = 2.
5. Required probability = \(\frac{2}{7}\).

Teacher's Note:
a) A leap year contains 366 days and has two extra days.
b) List consecutive pairs of days carefully to avoid counting errors.

5. Find the probability of getting 53 Fridays or 53 Saturdays in a leap year. [1 Mark]

Answer:
1. A leap year has 52 weeks and 2 remaining days, with 7 sample space pairs.
2. Favorable pairs containing either Friday or Saturday are (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), and (Sunday, Monday).
3. Total favorable outcomes = 4.
4. Required probability = \(\frac{4}{7}\).

Teacher's Note:
a) Uses the union of events in calendar probability.
b) Note that (Friday, Saturday) satisfies both conditions and is counted only once.

6. Find the probability of getting 53 Mondays or 53 Tuesday in an ordinary year. [1 Mark]

Answer:
1. An ordinary year has 52 weeks and 1 remaining day.
2. Total possible outcomes for the single remaining day = 7 {Mon, Tue, Wed, Thu, Fri, Sat, Sun}.
3. Favorable outcomes for Monday or Tuesday = 2 (Monday and Tuesday).
4. Required probability = \(\frac{2}{7}\).

Teacher's Note:
a) Simple application of single extra day sample space.
b) Ensure students do not confuse ordinary year extra days with leap year extra days.

7. Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective. [1 Mark]

Answer:
1. Total number of bulbs = 400, Number of defective bulbs = 15.
2. Number of non-defective bulbs = 400 - 15 = 385.
3. Probability of not defective = \(\frac{385}{400}\)
4. Simplifying by dividing numerator and denominator by 5 gives \(\frac{77}{80}\).

Teacher's Note:
a) Formula used is P(E) = (Number of favorable outcomes) / (Total number of outcomes).
b) Always simplify fractions to their lowest terms.

8. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize? [1 Mark]

Answer:
1. Number of prizes = 10, Number of blanks = 25.
2. Total outcomes = 10 + 25 = 35.
3. Favorable outcomes (getting a prize) = 10.
4. Probability = \(\frac{10}{35}\) = \(\frac{2}{7}\).

Teacher's Note:
a) Total outcomes is the sum of prizes and blanks.
b) Common mistake is taking total outcomes as 25 instead of 35.

9. 250 lottery tickets were sold and there are 5 prizes on these tickets. If Mahesh purchased one lottery ticket, what is the probability that he wins a prize? [1 Mark]

Answer:
1. Total tickets sold = 250, Number of prize tickets = 5.
2. Mahesh purchased 1 ticket.
3. Probability of winning a prize = \(\frac{5}{250}\) = \(\frac{1}{50}\).

Teacher's Note:
a) Direct application of probability formula.
b) Reduce fraction to simplest form.

10. The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times. (i) What is the probability that on a given day it was correct? (ii) What is the probability that it was not correct on a given day? [2 Marks]

Answer:
1. Total days = 250, Correct forecasts = 175, Incorrect forecasts = 250 - 175 = 75.
2. (i) P(correct) = \(\frac{175}{250}\) = \(\frac{7}{10}\) or 0.7.
3. (ii) P(not correct) = \(\frac{75}{250}\) = \(\frac{3}{10}\) or 0.3.

Teacher's Note:
a) Two-part question testing experimental probability and complement events.
b) Both fraction and decimal answers are acceptable.

11. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it ? (ii) She will not buy it ? [2 Marks]

Answer:
1. Total pens = 144, Defective pens = 20, Good pens = 144 - 20 = 124.
2. (i) She buys if it is good, so P(buy) = \(\frac{124}{144}\) = \(\frac{31}{36}\).
3. (ii) She does not buy if it is defective, so P(not buy) = \(\frac{20}{144}\) = \(\frac{5}{36}\).

Teacher's Note:
a) Translate the buyer's conditions into mathematical events (good pen = buy, defective = not buy).
b) Verify that the sum of both probabilities equals 1.

12. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green? [3 Marks]

Answer:
1. Total marbles = 5 + 8 + 4 = 17.
2. (i) P(red) = \(\frac{5}{17}\).
3. (ii) P(white) = \(\frac{8}{17}\).
4. (iii) P(not green) = (Total - green) / Total = (17 - 4) / 17 = \(\frac{13}{17}\).

Teacher's Note:
a) Three-part problem requiring careful calculation of total outcomes.
b) For 'not green', sum all other colors or subtract green from total.

13. Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). [2 Marks]

Answer:
1. Total days in a year = 365.
2. (ii) Probability of same birthday = \(\frac{1}{365}\).
3. (i) Probability of different birthdays = 1 - P(same birthday) = 1 - \(\frac{1}{365}\) = \(\frac{364}{365}\).

Teacher's Note:
a) Uses the principle of complementary events.
b) Often part (ii) is solved first logically and then part (i) is found by subtraction.

14. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. [1 Mark]

Answer:
1. Defective pens = 12, Good pens = 132.
2. Total pens = 12 + 132 = 144.
3. Probability of getting a good pen = \(\frac{132}{144}\)
4. Simplifying by dividing by 12 gives \(\frac{11}{12}\).

Teacher's Note:
a) Always calculate total outcomes first by adding defective and non-defective items.
b) Reduce fractions completely using common factors.

15. A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a Rs 5 coin? [2 Marks]

Answer:
1. Number of 50p coins = 100, Re 1 coins = 50, Rs 2 coins = 20, Rs 5 coins = 10.
2. Total coins = 100 + 50 + 20 + 10 = 180.
3. (i) P(50p coin) = \(\frac{100}{180}\) = \(\frac{5}{9}\).
4. (ii) P(not a Rs 5 coin) = 1 - P(Rs 5 coin) = 1 - \(\frac{10}{180}\) = \(\frac{170}{180}\) = \(\frac{17}{18}\).

Teacher's Note:
a) Total number of items is the sum of all individual denominations.
b) Complementary approach simplifies the negative condition in part (ii).

16. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish? [1 Mark]

Answer:
1. Number of male fish = 5, Number of female fish = 8.
2. Total number of fish = 5 + 8 = 13.
3. Required probability = \(\frac{5}{13}\).

Teacher's Note:
a) Direct calculation of probability from given frequencies.
b) Check that the denominator is the sum of male and female fish.

17. A number x is selected from the numbers 1, 2, 3 and then a second number y is randomly selected from the number 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9? [3 Marks]

Answer:
1. The possible values of x are 1, 2, 3 and y are 1, 4, 9.
2. Total number of possible outcomes = 3 × 3 = 9.
3. The pairs (x, y) are: (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9).
4. Products xy are: 1, 4, 9, 2, 8, 18, 3, 12, 27.
5. Products less than 9 are: 1, 4, 2, 8, 3 (Total 5 favorable outcomes).
6. Required probability = \(\frac{5}{9}\).

Teacher's Note:
a) List all sample space elements systematically as coordinate pairs.
b) Be careful not to include products equal to 9 when strict inequality is given.
c) Total outcomes come from the product rule of counting.

18. A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. What is the probability that it crashed inside the lake shown in the figure? [3 Marks]

Answer:
1. Total area of the rectangular region = 9 km × 4.5 km = 40.5 km².
2. Lake: width = 9 − 6 = 3 km and height = 4.5 − 2 = 2.5 km, so area of the lake = 3 km × 2.5 km = 7.5 km².
3. Probability of crashing inside the lake = \(\frac{\text{Area of the lake}}{\text{Total area of the region}}\).
4. Probability = \(\frac{7.5}{40.5}\) = \(\frac{15}{81}\) = \(\frac{5}{27}\).

Teacher's Note:
a) Geometric probability is defined as the ratio of favorable area to total area.
b) Ensure all dimensions are in the same unit before calculating areas.
c) Simplify fractions to their lowest terms in the final answer.

19. There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy? [2 Marks]

Answer:
1. Total number of students = Total number of possible outcomes = 40.
2. (i) Number of girls = 25. Probability of getting a girl = \(\frac{25}{40}\) = \(\frac{5}{8}\).
3. (ii) Number of boys = 15. Probability of getting a boy = \(\frac{15}{40}\) = \(\frac{3}{8}\).

Teacher's Note:
a) The sum of probabilities of all elementary events is equal to 1.
b) Reduce fractions to simplest form.

20. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that (i) it is acceptable to Jimmy? (ii) it is acceptable to Sujatha? [3 Marks]

Answer:
1. Total number of shirts = 100.
2. (i) Jimmy accepts only good shirts. Number of good shirts = 88. Probability = \(\frac{88}{100}\) = \(\frac{22}{25}\).
3. (ii) Sujatha rejects only major defect shirts, meaning she accepts good shirts and minor defect shirts. Number of acceptable shirts = 88 + 8 = 96. Probability = \(\frac{96}{100}\) = \(\frac{24}{25}\).

Teacher's Note:
a) Carefully read the condition for acceptability for each person.
b) Sujatha rejects major defects, so minor defects are acceptable to her.

21. 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? [4 Marks]

Answer:
1. Number of days in the week from Tuesday to Saturday = 5 (Tuesday, Wednesday, Thursday, Friday, Saturday).
2. Total possible outcomes for both visiting = 5 × 5 = 25.
3. (i) Both visit on the same day outcomes: (T, T), (W, W), (Th, Th), (F, F), (S, S). Probability = \(\frac{5}{25}\) = \(\frac{1}{5}\).
4. (ii) Consecutive days outcomes: (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F) - total 8 outcomes. Probability = \(\frac{8}{25}\).
5. (iii) Probability of different days = 1 - Probability of same day = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\).

Teacher's Note:
a) Count the number of days carefully from Tuesday to Saturday inclusive.
b) Use complementary probability for different days to save time.

22. Two customers are visiting a particular shop in the same week (Monday 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? [4 Marks]

Answer:
1. Number of days from Monday to Saturday = 6.
2. Total possible outcomes = 6 × 6 = 36.
3. (i) Same day outcomes = 6 ((M, M), (Tu, Tu), etc.). Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
4. (ii) Consecutive days outcomes: (M, Tu), (Tu, M), (Tu, W), (W, Tu), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F) - total 10 outcomes. Probability = \(\frac{10}{36}\) = \(\frac{5}{18}\).
5. (iii) Probability of different days = 1 - Probability of same day = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\).

Teacher's Note:
a) Monday to Saturday has 6 days, unlike the previous question which had 5.
b) List consecutive pairs in both orders (e.g., Mon-Tue and Tue-Mon).

23. 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. [3 Marks]

Answer:
1. Let the number of blue balls be x.
2. Number of red balls = 5.
3. Total number of balls = 5 + x.
4. Probability of drawing a red ball = \(\frac{5}{5 + x}\).
5. Probability of drawing a blue ball = \(\frac{x}{5 + x}\).
6. Given that \(\frac{x}{5 + x} = 2 \times \frac{5}{5 + x}\).
7. Solving for x: x = 10.
8. Therefore, the number of blue balls is 10.

Teacher's Note:
a) Set up the algebraic equation directly using probability definitions.
b) The denominators cancel out as they are identical on both sides.

24. A box contains 12 balls out 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. [4 Marks]

Answer:
1. Total number of balls initially = 12, number of black balls = x.
2. Initial probability of drawing a black ball = \(\frac{x}{12}\).
3. When 6 more black balls are added, total black balls = x + 6, and total balls = 12 + 6 = 18.
4. New probability of drawing a black ball = \(\frac{x + 6}{18}\).
5. Given condition: \(\frac{x + 6}{18} = 2 \times \frac{x}{12}\).
6. Simplify: \(\frac{x + 6}{18} = \frac{2x}{12}\) ⇒ \(\frac{x + 6}{18} = \frac{x}{6}\).
7. Cross multiply: 6(x + 6) = 18x ⇒ 6x + 36 = 18x ⇒ 12x = 36 ⇒ x = 3.

Teacher's Note:
a) Both numerator and denominator change when new balls are added.
b) Solve the linear equation carefully by cross-multiplication.

25. 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. [3 Marks]

Answer:
1. Total number of marbles = 24.
2. Probability of getting a green marble = \(\frac{2}{3}\).
3. Number of green marbles = \(\frac{2}{3} \times 24\) = 16.
4. Number of blue marbles = Total marbles - Number of green marbles = 24 - 16 = 8.
5. Hence, the number of blue marbles in the jar is 8.

Teacher's Note:
a) Multiply total outcomes by the probability to find the count of favorable items.
b) Subtract from total to find the complementary color count.

26. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it ? (ii) She will not buy it ? [2 Marks]

Answer:
1. Total number of pens = 144, Defective pens = 20, Good pens = 144 - 20 = 124.
2. (i) Probability that she will buy it (good pen) = \(\frac{124}{144}\) = \(\frac{31}{36}\).
3. (ii) Probability that she will not buy it (defective pen) = \(\frac{20}{144}\) = \(\frac{5}{36}\).

Teacher's Note:
a) The fundamental probability formula is the ratio of favorable outcomes to total possible outcomes.
b) Always simplify fractions to their lowest terms as shown in the final steps.

27. A jar contains 54 marbles, each of which is blue, green or white. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{1}{3}\) and that of getting a blue marble is \(\frac{4}{9}\). Find the number of white marbles in the jar. [3 Marks]

Answer:
1. Total number of marbles = 54.
2. Probability of getting a green marble = \(\frac{1}{3}\), so number of green marbles = \(\frac{1}{3}\) × 54 = 18.
3. Probability of getting a blue marble = \(\frac{4}{9}\), so number of blue marbles = \(\frac{4}{9}\) × 54 = 24.
4. Number of white marbles = Total marbles - (Green marbles + Blue marbles) = 54 - (18 + 24) = 54 - 42 = 12.

Teacher's Note:
a) Multiply the given probability by the total sample size to find the count of specific items.
b) Ensure the sum of probabilities for mutually exclusive events does not exceed one.
c) Double check subtraction when finding the remaining category count.

28. A letter is chosen at random from the letters of the word ASSASSINATION. Find the probability that the letter chosen is a (i) vowel (ii) consonant (iii) A (iv) S (v) N. [3 Marks]

Answer:
1. The word ASSASSINATION has 13 letters in total.
2. Vowels are A, A, I, A, I, O, I (total 6 vowels). Probability of vowel = \(\frac{6}{13}\).
3. Consonants are S, S, S, S, N, T, N (total 7 consonants). Probability of consonant = \(\frac{7}{13}\).
4. Letter A appears 3 times. Probability of A = \(\frac{3}{13}\).
5. Letter S appears 4 times. Probability of S = \(\frac{4}{13}\).
6. Letter N appears 2 times. Probability of N = \(\frac{2}{13}\).

Teacher's Note:
a) Count duplicate letters carefully by listing all occurrences in the given word.
b) Verify that the sum of probabilities of all individual letter counts equals 1.
c) Distinguish clearly between vowels and consonants as defined in English alphabets.

29. A letter is chosen at random from the letters of the word INDEPENDENCE. Find the probability that the letter chosen is a (i) vowel (ii) consonant (iii) E (iv) N (v) D. [3 Marks]

Answer:
1. The word INDEPENDENCE has 12 letters in total.
2. Vowels are I, E, E, E, E (total 5 vowels). Probability of vowel = \(\frac{5}{12}\).
3. Consonants are N, D, P, N, D, N, C (total 7 consonants). Probability of consonant = \(\frac{7}{12}\).
4. Letter E appears 4 times. Probability of E = \(\frac{4}{12}\) = \(\frac{1}{3}\).
5. Letter N appears 3 times. Probability of N = \(\frac{3}{12}\) = \(\frac{1}{4}\).
6. Letter D appears 2 times. Probability of D = \(\frac{2}{12}\) = \(\frac{1}{6}\).

Teacher's Note:
a) Tabulate the frequency of each character to avoid counting errors in long words.
b) Reduce all fractional answers to their simplest standard form.
c) Check that the number of vowels plus consonants equals the total word length.

30. A letter is chosen at random from the letters of the word MATHEMATICS. Find the probability that the letter chosen is a (i) vowel (ii) consonant (iii) A (iv) T (v) M. [3 Marks]

Answer:
1. The word MATHEMATICS has 11 letters in total.
2. Vowels are A, E, A (total 3 vowels). Probability of vowel = \(\frac{3}{11}\).
3. Consonants are M, T, H, M, T, I, C, S (total 8 consonants). Probability of consonant = \(\frac{8}{11}\).
4. Letter A appears 2 times. Probability of A = \(\frac{2}{11}\).
5. Letter T appears 2 times. Probability of T = \(\frac{2}{11}\).
6. Letter M appears 2 times. Probability of M = \(\frac{2}{11}\).

Teacher's Note:
a) Pay close attention to repeated letters like M, T, and A in the given word.
b) Ensure all subsets partition the complete sample space correctly.
c) Express probabilities strictly as proper fractions.

31. A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant. [1 Mark]

Answer:
1. Total number of English alphabets = 26.
2. Number of vowels = 5 (A, E, I, O, U), so number of consonants = 26 - 5 = 21.
3. Probability that the letter is a consonant = \(\frac{21}{26}\).

Teacher's Note:
a) The standard English alphabet contains 26 letters with 5 vowels and 21 consonants.
b) Complementary probability can also be used as 1 - P(vowel) = 1 - \(\frac{5}{26}\) = \(\frac{21}{26}\).

32. There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize? [3 Marks]

Answer:
1. Total number of envelopes = 1000.
2. Envelopes with Rs 100 prize = 10.
3. Envelopes with Rs 50 prize = 100.
4. Envelopes with Rs 10 prize = 200.
5. Envelopes containing cash prizes = 10 + 100 + 200 = 310.
6. Envelopes containing no cash prize = 1000 - 310 = 690.
7. Probability of picking an envelope with no cash prize = \(\frac{690}{1000}\) = \(\frac{69}{100}\).

Teacher's Note:
a) Subtract the sum of all prize-winning categories from the total sample size to isolate the zero-prize group.
b) Cancel common factors in the numerator and denominator to simplify the final fraction.
c) Read word problems carefully to account for every specified subset.

33. Box A contains 25 slips of which 19 are marked Re 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Re 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Re 1? [3 Marks]

Answer:
1. Box A total slips = 25; slips marked Re 1 = 19; slips marked Rs 5 = 25 - 19 = 6.
2. Box B total slips = 50; slips marked Re 1 = 45; slips marked Rs 13 = 50 - 45 = 5.
3. Total slips in the third box = 25 + 50 = 75.
4. Total slips marked other than Re 1 = (Slips marked Rs 5 from Box A) + (Slips marked Rs 13 from Box B) = 6 + 5 = 11.
5. Probability that the drawn slip is marked other than Re 1 = \(\frac{11}{75}\).

Teacher's Note:
a) Combine totals from both independent boxes to establish the new combined sample space.
b) Identify alternative conditions accurately by focusing on non-target values.
c) Verify arithmetic steps when summing across multiple categories.

34. A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective? [3 Marks]

Answer:
1. Total bulbs = 24, Defective bulbs = 6, Non-defective bulbs = 24 - 6 = 18.
2. Probability that the first bulb is not defective = \(\frac{18}{24}\) = \(\frac{3}{4}\).
3. If a defective bulb is drawn and not replaced, remaining total bulbs = 24 - 1 = 23.
4. Remaining defective bulbs = 6 - 1 = 5.
5. Probability that the second bulb is defective = \(\frac{5}{23}\).

Teacher's Note:
a) Conditional probability without replacement alters both numerator and denominator for subsequent draws.
b) Deduct one from both the target category and the total pool when an item is removed.
c) State each fractional probability clearly in simplified form.

35. A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a (i) triangle (ii) square (iii) square of blue colour (iv) triangle of red colour [4 Marks]

Answer:
1. Total triangles = 8 (Blue = 3, Red = 8 - 3 = 5).
2. Total squares = 10 (Blue = 6, Red = 10 - 6 = 4).
3. Total pieces in the game = 8 + 10 = 18.
4. (i) Probability of losing a triangle = \(\frac{8}{18}\) = \(\frac{4}{9}\).
5. (ii) Probability of losing a square = \(\frac{10}{18}\) = \(\frac{5}{9}\).
6. (iii) Probability of losing a blue square = \(\frac{6}{18}\) = \(\frac{1}{3}\).
7. (iv) Probability of losing a red triangle = \(\frac{5}{18}\).

Teacher's Note:
a) Calculate the grand total by adding all individual pieces across shapes and colors.
b) Subdivide components accurately into specific shape-color intersections.
c) Check that probabilities of mutually exclusive partitions add up correctly.
d) Always reduce fractions where common divisors exist.

36. In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she (i) loses the entry fee. (ii) gets double entry fee. (iii) just gets her entry fee. [4 Marks]

Answer:
1. Total possible outcomes when tossing a coin 3 times = \(2^3\) = 8 outcomes: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
2. Outcomes with 1 head = {HTT, THT, TTH} (3 outcomes). Outcomes with 2 heads = {HHT, HTH, THH} (3 outcomes).
3. Outcomes with 3 heads = {HHH} (1 outcome). Outcomes with 0 heads (all tails) = {TTT} (1 outcome).
4. (i) She loses the entry fee when 0 heads show (TTT), total outcomes = 1. Probability = \(\frac{1}{8}\).
5. (ii) She gets double entry fee when 3 heads show (HHH), total outcomes = 1. Probability = \(\frac{1}{8}\).
6. (iii) She just gets her entry fee back when 1 or 2 heads show, total outcomes = 3 + 3 = 6. Probability = \(\frac{6}{8}\) = \(\frac{3}{4}\).

Teacher's Note:
a) Sample space size for multiple coin tosses is determined by the exponential formula \(2^n\).
b) Group outcomes carefully by the exact count of heads to match game rules.
c) Otherwise condition corresponds exclusively to the zero-head outcome (all tails).
d) Verify that the sum of all mutually exclusive outcome probabilities equals 1.

37. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded. (i) How many different scores are possible? (ii) What is the probability of getting a total of 7? [4 Marks]

Answer:
1. The faces of each die are {0, 1, 1, 1, 6, 6}. Total outcomes when two dice are thrown = 6 × 6 = 36.
2. Possible sums from combinations of face values: 0+0=0, 0+1=1, 0+6=6, 1+1=2, 1+6=7, 6+6=12.
3. (i) The set of all possible distinct scores is {0, 1, 2, 6, 7, 12}, giving a total of 6 different possible scores.
4. (ii) Combinations that yield a total score of 7 are when one die shows 1 and the other shows 6.
5. Since die 1 has three 1s and die 2 has two 6s (and vice versa), favorable pairs are: (1 from first die, 6 from second die) = 3 × 2 = 6 ways, and (6 from first die, 1 from second die) = 2 × 3 = 6 ways.
6. Total favorable outcomes for a total of 7 = 6 + 6 = 12.
7. Probability of getting a total of 7 = \(\frac{12}{36}\) = \(\frac{1}{3}\).

Teacher's Note:
a) Custom-marked dice require analyzing individual face frequencies rather than standard 1 to 6 numbering.
b) List all unique pairwise sums to find the count of possible score values.
c) Multiply individual face frequencies when counting ordered outcome pairs.
d) Always reduce final probability fractions completely.

38. A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is (i) a good phone (ii) a bad phone [2 Marks]

Answer:
1. Total mobile phones = 48, Good phones = 42, Phones with minor defects = 3, Phones with major defects = 3.
2. (i) Probability that it is a good phone = \(\frac{42}{48}\) = \(\frac{7}{8}\).
3. (ii) Bad phones include all phones with either minor or major defects = 3 + 3 = 6.
4. Probability that it is a bad phone = \(\frac{6}{48}\) = \(\frac{1}{8}\).

Teacher's Note:
a) Group defective items together to define the bad phone category.
b) Simplify fractions fully by dividing numerator and denominator by their greatest common divisor.

39. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? [1 Mark]
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ? [2 Marks]

Answer:
1. (i) Total number of bulbs = 20, Number of defective bulbs = 4
2. Probability of getting a defective bulb = \(\frac{4}{20}\) = \(\frac{1}{5}\)
3. (ii) Remaining total bulbs = 19, Remaining non-defective bulbs = 16 - 1 = 15
4. Probability that this bulb is not defective = \(\frac{15}{19}\)

Teacher's Note:
a) Conditional probability without replacement reduces the sample space by 1 for subsequent draws.
b) A common mistake is failing to subtract 1 from both the total and the required category.
c) Always simplify fractions to their lowest terms where possible.

40. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig.), and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9? [4 Marks]

Answer:
1. Total possible outcomes = 8 (Numbers 1, 2, 3, 4, 5, 6, 7, 8)
2. (i) Favorable outcomes for 8 = 1; Probability = \(\frac{1}{8}\)
3. (ii) Odd numbers = 1, 3, 5, 7 (Total 4); Probability = \(\frac{4}{8}\) = \(\frac{1}{2}\)
4. (iii) Numbers greater than 2 = 3, 4, 5, 6, 7, 8 (Total 6); Probability = \(\frac{6}{8}\) = \(\frac{3}{4}\)
5. (iv) Numbers less than 9 = 1, 2, 3, 4, 5, 6, 7, 8 (Total 8); Probability = \(\frac{8}{8}\) = 1

Teacher's Note:
a) This is a classic discrete probability problem with equally likely outcomes.
b) Pay close attention to inclusive or exclusive phrasing like "greater than 2" which excludes 2.
c) An event with probability 1 is a certain event.
d) Always reduce fractions to their simplest form.

41. Suppose you drop a die at random on the rectangular region shown in above right-sided figure. What is the probability that it will land inside the circle with diameter 1m? [2 Marks]

Answer:
1. Area of the rectangular region = length × breadth = 3 × 2 = 6 m²
2. Diameter of the circle = 1 m, so radius \(r\) = \(\frac{1}{2}\) m
3. Area of the circle = \(\pi r^2\) = \(\pi \left(\frac{1}{2}\right)^2\) = \(\frac{\pi}{4}\) m²
4. Probability = \(\frac{\text{Area of the circle}}{\text{Area of the rectangle}}\) = \(\frac{\pi / 4}{6}\) = \(\frac{\pi}{24}\)

Teacher's Note:
a) Geometric probability problems are solved by finding the ratio of favorable area to total area.
b) Ensure all dimensions are in the same unit before calculating areas.

42. A child has a die whose six faces show the letters as given below:
A B C D E A
The die is thrown once. What is the probability of getting (i) A? (ii) D? [2 Marks]

Answer:
1. Total number of faces on the die = 6
2. (i) Number of faces showing letter A = 2; Probability of getting A = \(\frac{2}{6}\) = \(\frac{1}{3}\)
3. (ii) Number of faces showing letter D = 1; Probability of getting D = \(\frac{1}{6}\)

Teacher's Note:
a) The sample space has 6 equally likely outcomes, but some outcomes repeat.
b) Count the exact frequency of the target letter from the given sequence.

43. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. [3 Marks]

Answer:
1. Total possible outcomes when a coin is tossed 3 times = \(2^3\) = 8
2. The sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
3. Winning outcomes (all same results) = {HHH, TTT}, which is 2 outcomes
4. Losing outcomes = Total outcomes - Winning outcomes = 8 - 2 = 6 outcomes
5. Probability that Hanif will lose the game = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Teacher's Note:
a) Writing out the complete sample space prevents counting errors.
b) Note that the question asks for the probability of losing, not winning.
c) Complementary probability can also be used: 1 - P(Winning) = 1 - \(\frac{2}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\).

Question. Cards bearing numbers 1, 3, 5, ..., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing:
(i) a prime number less than 15.
(ii) a number divisible by 3 and 5.
Answer: (i) \( \frac{1}{3} \), (ii) \( \frac{1}{9} \)

Question. Red kings, queens and jacks are removed from a deck of 52 playing cards and then well-shuffled. A card is drawn from the remaining cards. Find the probability of getting (i) King (ii) a red card (iii) a spade.
Answer: (i) \( \frac{1}{23} \), (ii) \( \frac{10}{23} \), (iii) \( \frac{13}{46} \)

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) A king of red suit.
(ii) A queen of black suit.
(iii) A jack hearts.
(iv) A red face card.
Answer: (i) \( \frac{1}{26} \), (ii) \( \frac{1}{26} \), (iii) \( \frac{1}{52} \), (iv) \( \frac{3}{26} \)

Question. Two coins are tossed together find the probability of getting:
(i) at least one tail.
(ii) one head
Answer: (i) \( \frac{3}{4} \), (ii) \( \frac{1}{2} \)

Question. Two dice are thrown at the same time. Find the probability of getting different numbers on both the dice.
Hint: \( P_{\text{(different numbers on both dice)}} = 1 – P_{\text{(same number on both dice)}} \)
Answer: \( \frac{5}{6} \)

Question. Two dice are thrown at the same time. Find the probability of getting same number on both the dice.
Answer: \( \frac{1}{6} \)

Question. A pair of dice is thrown once. Find the probability of getting an odd number on each die.
Answer: \( \frac{1}{4} \)

Question. Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 35 is a:
(i) prime number
(ii) multiple of 7
(iii) a prime number less than 15.
Answer: (i) \( \frac{11}{35} \), (ii) \( \frac{1}{7} \), (iii) \( \frac{6}{35} \)

Question. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) the queen of diamond
(ii) an ace of hearts
(iii) a spade.
Hint: (i) There is only one queen of diamond.
(ii) There is only one ace of hearts.
(iii) There are 13 spade cards.
Answer: (i) \( \frac{1}{52} \), (ii) \( \frac{1}{52} \), (iii) \( \frac{1}{4} \)

Question. Find the probability of getting 53 Sundays in a leap year.
Hint: Number of days in a normal year = 365
Number of days in a leap year = 366
Number of weeks, in a normal year = 52, means 52 Sundays
\( [366 – (52 \times 7)] = 2 \) extra days in a leap year.
These two extra days may have sample-space as:
(Monday – Tuesday), (Tuesday – Wednesday), (Wednesday – Thursday), (Thursday – Friday), (Friday – Saturday), (Saturday – Sunday), (Sunday – Monday).
i.e. out of 7 sample spaces, only two are favourable.
\( \Rightarrow P_{\text{(53 Sundays in a leap year)}} = \frac{2}{7} \)
Answer: \( \frac{2}{7} \)

Question. Two coins are tossed simultaneously. Find the probability of getting:
(i) two Heads
(ii) at least one Head
(iii) no Head.
Hint: In a throw of two coins simultaneously the four possible outcomes are
HH, HT, TH, TT
Answer: (i) \( \frac{1}{4} \), (ii) \( \frac{3}{4} \), (iii) \( \frac{1}{4} \)

Question. From a well suffled pack of playing cards, black jacks, black kings and black aces are removed. A card is then drawn from the pack. Find the probability of getting:
(i) a red card
(ii) not a diamond card.
Answer: (i) \( \frac{13}{23} \), (ii) \( \frac{33}{46} \)

Question. A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears.
(i) a two-digit number
(ii) a number which is a perfect square.
Answer: (i) \( \frac{81}{89} \), (ii) \( \frac{8}{89} \)

Question. Cards numbered 1 to 30 are put in a bag. A card is drawn at random from this bag. Find the probability that the number on the drawn card is :
(i) not divisible by 3.
(ii) a prime number greater than 7.
(iii) not a perfect square number. 
Hint: Total possible outcomes = 30
(i) Numbers not divisible by 3 [1 to 30] are :
1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29,
∴ Number of favourable outcomes = 20
\( \Rightarrow \text{Required Probability} = \frac{20}{30} = \frac{2}{3} \)
(ii) Prime numbers greater than 7 are : 11, 13, 17, 19, 23, 29
∴ Required probability = \( \frac{6}{30} = \frac{1}{5} \)
(iii) Perfect squares are 1, 4, 9, 16, 25 \( \Rightarrow \) Total No. = 5
∴ Required pribability = \( \frac{30 - 5}{30} = \frac{25}{30} = \frac{5}{6} \)
Answer: (i) \( \frac{2}{3} \), (ii) \( \frac{1}{5} \), (iii) \( \frac{5}{6} \)

Question. Two different dice are tossed together. Find the probability :
(i) That the numbers on either die is even.
(ii) That the sum of numbers appearing on the two dice is 5.
Hint: Total possible outcomes = 36
(i) Numbers of favourable outcomes = 9
[ \( \because \) (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4) and (6, 6) are desired outcomes.
Required Probability = \( \frac{9}{36} = \frac{1}{4} \)
(ii) Desired (favourable) outcomes are : (1, 4), (2, 3), (3, 2), (4, 1)
Required Probability = \( \frac{4}{36} = \frac{1}{9} \)
Answer: (i) \( \frac{1}{4} \), (ii) \( \frac{1}{9} \)

Question. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the card drawn is :
(i) a king
(ii) of red colour
(iii) a face-card
(iv) a queen 
Hint: A pack of playing cards consists of 52 cards. 2 red queens and 2 black jacks are removed.
Therefore, remaining number of cards = 52 - 4 = 48.
(i) Numbers of kings = 4 \( \Rightarrow \) Required P = \( \frac{4}{48} = \frac{1}{12} \)
(ii) Remaining red cards = 26 – 2 = 24 \( \Rightarrow \) Required P = \( \frac{24}{48} = \frac{1}{2} \)
(iii) Remaining face-cards = 12 – 4 = 8 \( \Rightarrow \) Required P = \( \frac{8}{48} = \frac{1}{6} \)
(iv) Remaining queens = 4 – 2 = 2 \( \Rightarrow \) Required P = \( \frac{2}{48} = \frac{1}{24} \)
Answer: (i) \( \frac{1}{12} \), (ii) \( \frac{1}{2} \), (iii) \( \frac{1}{6} \), (iv) \( \frac{1}{24} \)

Question. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail. 
Hint: Total outcomes = 4 (HH, HT, TH, TT)
Fovourable outcomes = 3 (HT, TH, TT)
\( \Rightarrow \text{Required P} = \frac{3}{4} \)
Answer: \( \frac{3}{4} \)

Question. A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is :
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number 
Hint: Possible outcomes = 49
(i) Odd numbers are : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 \( \Rightarrow \) Favourable outcomes = 25
\( \Rightarrow \text{Required P} = \frac{25}{49} \)
(ii) Multiples of 5 are = 5, 10, 15, 20, 25, 30, 35, 40, 45 \( \Rightarrow P = \frac{9}{49} \)
(iii) Perfect squares are : 1, 4, 9, 16, 25, 36, 49 \( \Rightarrow P = \frac{7}{49} = \frac{1}{7} \)
(iv) Even Primes are : only 2. \( \Rightarrow P = \frac{1}{49} \)
Answer: (i) \( \frac{25}{49} \), (ii) \( \frac{9}{49} \), (iii) \( \frac{1}{7} \), (iv) \( \frac{1}{49} \)