CBSE Class 10 Mathematics Probability Worksheet Set 09

Answer: There are 23 cards. The primes are 3, 5, 7, 11, 13, 17, 19, 23, so 8 favourable outcomes. P = \( \dfrac{8}{23} \).

Teacher's Note:
1. Count the total cards first, because the list starts at 3 and not at 1.
2. Remember that 4 is composite and 2 is not on the list.

Answer: Total balls = 12. (a) P(red) = \( \dfrac{5}{12} \). (b) Not green = 5 + 4 = 9 balls, so P = \( \dfrac{9}{12} \) = \( \dfrac{3}{4} \).

Teacher's Note:
1. Use P(not E) = 1 − P(E) as a quick check: 1 − \( \dfrac{3}{12} \) = \( \dfrac{3}{4} \).
2. Always reduce the fraction to its lowest terms.

Answer: (a) Face cards = 12, so P = \( \dfrac{12}{52} \) = \( \dfrac{3}{13} \). (b) Black cards = 26, of which 2 are kings. Black non-king cards = 24, so P = \( \dfrac{24}{52} \) = \( \dfrac{6}{13} \).

Teacher's Note:
1. Face cards are the jacks, queens and kings only; aces are not face cards.
2. In part (b) the condition removes all red cards and also the black kings.

Answer: Numbers greater than 4 are 5 and 6, so 2 outcomes out of 6. P = \( \dfrac{2}{6} \) = \( \dfrac{1}{3} \).

Teacher's Note:
1. Count only 5 and 6, since 4 itself is not greater than 4.
2. The sample space of one die has 6 equally likely outcomes.

Answer: Favourable pairs: (4,6), (5,5), (5,6), (6,4), (6,5), (6,6), which is 6 outcomes out of 36. P = \( \dfrac{6}{36} \) = \( \dfrac{1}{6} \).

Teacher's Note:
1. List the pairs systematically by the first die to avoid missing any.
2. A sum of exactly 9 is not included in this question.

Answer: Same numbers occur in 6 outcomes (1,1) to (6,6). Different numbers = 36 − 6 = 30 outcomes. P = \( \dfrac{30}{36} \) = \( \dfrac{5}{6} \).

Teacher's Note:
1. Finding the complement is faster than listing all different pairs.
2. P(different) = 1 − P(same) = 1 − \( \dfrac{1}{6} \).

Answer: Outcomes: HH, HT, TH, TT. At most one head means HT, TH, TT, which is 3 outcomes. P = \( \dfrac{3}{4} \).

Teacher's Note:
1. At most one means zero heads or one head.
2. Only HH is excluded, so the complement method also gives 1 − \( \dfrac{1}{4} \).

Answer: There are 100 cards. Perfect squares from 2 to 101 are 4, 9, 16, 25, 36, 49, 64, 81, 100, which is 9 numbers. P = \( \dfrac{9}{100} \).

Teacher's Note:
1. Note that 1 is a perfect square but is not on the cards.
2. The number of cards is 101 − 2 + 1 = 100.

Answer: The word has 11 letters and M occurs twice. P = \( \dfrac{2}{11} \).

Teacher's Note:
1. Count letters with repetition, since each position is a separate outcome.
2. The letters A and T also repeat, but only M is asked for.

Answer: The numbers are 3, 5, …, 35, so there are (35 − 3)/2 + 1 = 17 cards. (a) Primes less than 15: 3, 5, 7, 11, 13, so P = \( \dfrac{5}{17} \). (b) Divisible by 15: only 15 (45 is too large), so P = \( \dfrac{1}{17} \).

Teacher's Note:
1. Divisible by both 3 and 5 means divisible by 15.
2. The number of terms of an AP is (last − first)/difference + 1.

Answer: Total pieces = 18. (a) P(square) = \( \dfrac{10}{18} \) = \( \dfrac{5}{9} \). (b) Red triangles = 8 − 3 = 5, so P = \( \dfrac{5}{18} \).

Teacher's Note:
1. Add the triangles and squares to get the total number of pieces.
2. Subtract the blue triangles from all triangles to get the red ones.

Answer: (a) Neither die shows 5: 5 × 5 = 25 outcomes, so P = \( \dfrac{25}{36} \). (b) At least one 5: 36 − 25 = 11 outcomes, so P = \( \dfrac{11}{36} \). (c) Both show 5: only (5,5), so P = \( \dfrac{1}{36} \).

Teacher's Note:
1. Parts (a) and (b) are complementary events and their probabilities add to 1.
2. Each die has 5 outcomes that are not a 5.

Answer: Cards left = 49. (a) Hearts = 13, so P = \( \dfrac{13}{49} \). (b) Queens left = 3, so P = \( \dfrac{3}{49} \). (c) Clubs left = 13 − 3 = 10, so P = \( \dfrac{10}{49} \).

Teacher's Note:
1. The sample space shrinks to 49, not 52.
2. Only the cards named in the question are removed, so hearts are unaffected.

Answer: There are 8 outcomes. Hanif wins on HHH or TTT, so P(win) = \( \dfrac{2}{8} \) = \( \dfrac{1}{4} \). P(lose) = 1 − \( \dfrac{1}{4} \) = \( \dfrac{3}{4} \).

Teacher's Note:
1. Find the winning probability first and subtract from 1.
2. The three tosses give 2 × 2 × 2 = 8 equally likely outcomes.

Answer: (a) Possible sums: 0, 1, 2, 6, 7, 12, so 6 different scores. (b) A total of 7 needs 1 and 6. P(1) = \( \dfrac{1}{2} \) and P(6) = \( \dfrac{1}{3} \), so P(7) = 2 × \( \dfrac{1}{2} \) × \( \dfrac{1}{3} \) = \( \dfrac{1}{3} \).

Teacher's Note:
1. The faces are not equally distinct, so use the probability of each face value.
2. A total of 7 can happen in two orders: (1,6) and (6,1).

Answer: Even sum occurs when both numbers are even or both are odd: 3 × 3 + 3 × 3 = 18 outcomes. P = \( \dfrac{18}{36} \) = \( \dfrac{1}{2} \).

Teacher's Note:
1. Even plus even and odd plus odd both give an even sum.
2. Exactly half the 36 outcomes give an even sum.

Answer: Total = 100. (i) Evens 2, 4, …, 100 = 50, so P = \( \dfrac{1}{2} \). (ii) Numbers 2 to 13 = 12, so P = \( \dfrac{3}{25} \). (iii) 9 squares, so P = \( \dfrac{9}{100} \). (iv) Primes 2, 3, 5, 7, 11, 13, 17, 19 = 8, so P = \( \dfrac{2}{25} \).

Teacher's Note:
1. Count carefully because the list begins at 2.
2. Reduce each fraction to its simplest form.

Answer: A leap year has 366 days = 52 weeks + 2 extra days. The extra days can be (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon). Sunday appears in 2 of the 7 pairs, so P = \( \dfrac{2}{7} \).

Teacher's Note:
1. The 52 full weeks already contain 52 Sundays, so only the 2 extra days matter.
2. List the seven possible pairs of consecutive days.

Answer: CNG users = 100 − 40 − 35 = 25%, so P = 0.25 = \( \dfrac{1}{4} \). CNG is the best choice because it causes the least air pollution.

Teacher's Note:
1. Convert percentages to a probability by dividing by 100.
2. Linking maths to environment helps in value-based questions.

Answer: (a) Primes on a die: 2, 3, 5, so P = \( \dfrac{3}{6} \) = \( \dfrac{1}{2} \). (b) Divisible by 2: 2, 4, 6, so P = \( \dfrac{3}{6} \) = \( \dfrac{1}{2} \).

Teacher's Note:
1. Remember that 1 is not a prime number.
2. Both events have the same probability here but contain different outcomes.

Answer: Total cards = 89. (a) Two-digit numbers 10 to 90 = 81, so P = \( \dfrac{81}{89} \). (b) Squares: 4, 9, 16, 25, 36, 49, 64, 81 = 8, so P = \( \dfrac{8}{89} \).

Teacher's Note:
1. The count of numbers from 2 to 90 is 90 − 2 + 1 = 89.
2. The squares 1 and 100 lie outside the range.

Answer: Pairs with product 12: (2,6), (6,2), (3,4), (4,3), which is 4 outcomes. P = \( \dfrac{4}{36} \) = \( \dfrac{1}{9} \).

Teacher's Note:
1. Check the factor pairs of 12 that are possible on a die.
2. (1,12) is not possible because a die has no 12.

Answer: Cards left = 52 − 2 − 2 = 48. (i) Kings = 4, so P = \( \dfrac{4}{48} \) = \( \dfrac{1}{12} \). (ii) Red cards = 26 − 2 = 24, so P = \( \dfrac{24}{48} \) = \( \dfrac{1}{2} \). (iii) Face cards = 12 − 4 = 8, so P = \( \dfrac{8}{48} \) = \( \dfrac{1}{6} \). (iv) Queens = 2 black, so P = \( \dfrac{2}{48} \) = \( \dfrac{1}{24} \).

Teacher's Note:
1. Update the sample space and each favourable count after removing the cards.
2. Red queens are removed but red kings and red jacks remain.

Answer: Red face cards = 6, so cards left = 46. (i) Red cards left = 20, so P = \( \dfrac{20}{46} \) = \( \dfrac{10}{23} \). (ii) Queens left = 2, so P = \( \dfrac{2}{46} \) = \( \dfrac{1}{23} \). (iii) Aces = 4, so P = \( \dfrac{4}{46} \) = \( \dfrac{2}{23} \). (iv) Face cards left = 6, so P = \( \dfrac{6}{46} \) = \( \dfrac{3}{23} \).

Teacher's Note:
1. There are 3 red face cards in each of the two red suits, giving 6 in total.
2. Aces are not face cards and are all still in the pack.

Answer: The 8 equally likely outcomes range from BBB to GGG. Only GGG has no boy, so P(at least one boy) = 1 − \( \dfrac{1}{8} \) = \( \dfrac{7}{8} \).

Teacher's Note:
1. Use the complement of having no boy.
2. At least one means one or more.

Answer: There are 8 outcomes. (i) Exactly two heads: HHT, HTH, THH, so P = \( \dfrac{3}{8} \). (ii) At least two heads: the three above plus HHH, so P = \( \dfrac{4}{8} \) = \( \dfrac{1}{2} \). (iii) At most two heads excludes only HHH, so P = \( \dfrac{7}{8} \).

Teacher's Note:
1. Write out all eight outcomes before counting.
2. At most two heads is the complement of exactly three heads.