CBSE Class 10 Mathematics Probability MCQs Set 09

Answer: (a) 4

Explanation:
1. Total outcomes = 4 x 3 x 2 = 24.
2. With Box A and red fixed, only the door can vary: 4 x 1 x 1 = 4.

Teacher's Note:
1. Fix the given choices and count the free ones.
2. Use the multiplication principle.

Answer: (b) \( \dfrac{4}{11} \)

Explanation:
1. PROBABILITY has 11 letters: P, R, O, B, A, B, I, L, I, T, Y.
2. Vowels: O, A, I, I, which is 4 letters.
3. \( P(\text{vowel}) = \dfrac{4}{11} \).

Teacher's Note:
1. Count repeated letters each time they appear.
2. Vowels are A, E, I, O, U.

Answer: (c) Only (I) and (II)

Explanation:
1. Originally \( P(\text{red}) = \dfrac{3}{5} = 0.6 \).
2. (I) \( \dfrac{4}{6} = 0.67 \) (greater). (II) \( \dfrac{3}{4} = 0.75 \) (greater). (III) \( \dfrac{4}{7} = 0.57 \) (smaller).
3. Only (I) and (II) increase the probability.

Teacher's Note:
1. Recompute the probability after each change.
2. Compare fractions as decimals.

Answer: (d) Blue

Explanation:
1. The colour with the most cards has the highest probability.
2. Blue has 4 cards out of 10, which is the largest number.

Teacher's Note:
1. Most likely means highest probability.
2. Compare the numbers of cards of each colour.

Answer: (a) \( \dfrac{2}{3} \)

Explanation:
1. Let the smaller rectangle be l by b: area lb. The larger is 2l by 1.5b: area 3lb.
2. Area of the region between = 3lb - lb = 2lb.
3. \( P = \dfrac{2lb}{3lb} = \dfrac{2}{3} \).

Teacher's Note:
1. Geometric probability is area of favourable region divided by total area.
2. Subtract the inner area from the outer area.

Answer: (b) \( \dfrac{1}{3} \)

Explanation:
1. Favourable outcomes: \( 3\times 4 = 12 \) out of 36.
2. \( P = \dfrac{12}{36} = \dfrac{1}{3} \).

Teacher's Note:
1. Multiply the number of choices for the two throws.
2. Total outcomes for two throws = 36.

Answer: (c) less than 0

Explanation:
1. Probability lies between 0 and 1, so as a percentage it lies between 0 and 100.
2. A percentage below 0 is impossible.

Teacher's Note:
1. 0 <= P(E) <= 1 always.
2. A negative probability has no meaning.

Answer: (d) 480

Explanation:
1. \( \dfrac{n}{6000} = 0.08 \Rightarrow n = 480 \).

Teacher's Note:
1. Probability = tickets bought / total tickets.
2. Multiply the probability by the total.

Answer: (a) \( \dfrac{6}{23} \)

Explanation:
1. House E students = 23 - (4+8+5+2) = 4.
2. Students not from A, B, C = 2 + 4 = 6.
3. \( P = \dfrac{6}{23} \).

Teacher's Note:
1. Find the missing count first.
2. 'Not from A, B, C' means D or E.

Answer: (b) 14

Explanation:
1. Bad eggs \( = 0.035\times 400 = 14 \).

Teacher's Note:
1. Number = probability x total.
2. Do not confuse bad eggs with good eggs.

Answer: (c) 0 ≤ P(A) ≤ 1

Explanation:
1. The probability of any event lies from 0 (impossible) to 1 (certain).

Teacher's Note:
1. Memorise the range of probability.
2. Impossible = 0, sure = 1.

Answer: (d) \( \dfrac{3}{11} \)

Explanation:
1. MAINTENANCE has 11 letters and N occurs 3 times.
2. \( P = \dfrac{3}{11} \).

Teacher's Note:
1. Count how many times the letter repeats.
2. Total is the number of letters, not distinct letters.

Answer: (a) 25

Explanation:
1. Outcomes: HH, HT, TH, TT. \( P(\text{no head}) = \dfrac{1}{4} \), so \( A=1,\ B=4 \).
2. \( (A+B)^2 = 5^2 = 25 \).

Teacher's Note:
1. List the four outcomes of two coins.
2. Then substitute A and B.

Answer: (b) \( \dfrac{1}{2} \)

Explanation:
1. Outcomes: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG (8).
2. At least two boys: BBB, BBG, BGB, GBB (4).
3. \( P = \dfrac{4}{8} = \dfrac{1}{2} \).

Teacher's Note:
1. Write all 8 equally likely outcomes.
2. Count those with two or three boys.

Answer: (c) Peehu

Explanation:
1. Apoorv: only (6, 6) gives 36, so \( P = \dfrac{1}{36} \).
2. Peehu: only 6 squared gives 36, so \( P = \dfrac{1}{6} \).
3. \( \dfrac{1}{6} > \dfrac{1}{36} \), so Peehu has the better chance.

Teacher's Note:
1. Compute both probabilities and compare.
2. Fewer total outcomes means a larger probability for one favourable outcome.

Answer: (d) \( \dfrac{3}{4} \)

Explanation:
1. Total outcomes = 8. Hanif wins with HHH or TTT, so \( P(\text{win}) = \dfrac{2}{8} = \dfrac{1}{4} \).
2. \( P(\text{lose}) = 1-\dfrac{1}{4} = \dfrac{3}{4} \).

Teacher's Note:
1. Find P(win) and subtract from 1.
2. Three tosses have 8 outcomes.

Answer: (a) y = 3x

Explanation:
1. \( P(\text{wrong}) = 1-\dfrac{x}{y} = \dfrac{2}{3} \Rightarrow \dfrac{x}{y} = \dfrac{1}{3} \).
2. So y = 3x.

Teacher's Note:
1. P(correct) = x/y.
2. Use P(wrong) = 1 - P(correct).

Answer: (b) \( \dfrac{1}{6} \)

Explanation:
1. Pairs with difference 3: (1,4), (2,5), (3,6), (4,1), (5,2), (6,3), which are 6 pairs.
2. \( P = \dfrac{6}{36} = \dfrac{1}{6} \).

Teacher's Note:
1. List ordered pairs; (1,4) and (4,1) are different.
2. Total outcomes = 36.

Answer: (c) \( \dfrac{1}{3} \)

Explanation:
1. Neither red nor black means white: 5 balls out of 15.
2. \( P = \dfrac{5}{15} = \dfrac{1}{3} \).

Teacher's Note:
1. Translate 'neither red nor black' into the remaining colour.
2. Total balls = 15.

Answer: (d) 382

Explanation:
1. Bad eggs = 0.045 x 400 = 18.
2. Good eggs = 400 - 18 = 382.

Teacher's Note:
1. Subtract the bad eggs from the total.
2. Do not stop at the bad-egg count.

Answer: (a) \( \dfrac{12}{13} \)

Explanation:
1. Number of aces = 4, so cards that are not aces = 48.
2. \( P = \dfrac{48}{52} = \dfrac{12}{13} \).

Teacher's Note:
1. Use P(not E) = 1 - P(E).
2. P(ace) = 4/52 = 1/13.

Answer: (b) \( \dfrac{5}{6} \)

Explanation:
1. Same numbers: (1,1), ..., (6,6), which is 6 outcomes.
2. Different numbers: \( 36-6 = 30 \), so \( P = \dfrac{30}{36} = \dfrac{5}{6} \).

Teacher's Note:
1. Count the complement (same numbers) and subtract.
2. Total outcomes = 36.

Answer: (c) \( \dfrac{21}{26} \)

Explanation:
1. The alphabet has 26 letters and 5 vowels, so 21 consonants.
2. \( P = \dfrac{21}{26} \).

Teacher's Note:
1. Consonants = 26 - 5.

Answer: (d) \( \dfrac{3}{4} \)

Explanation:
1. Outcomes: HH, HT, TH, TT.
2. At most one head: HT, TH, TT (3 outcomes).
3. \( P = \dfrac{3}{4} \).

Teacher's Note:
1. 'At most one' means 0 or 1.
2. Do not count HH.

Answer: (a) \( \dfrac{1}{36} \)

Explanation:
1. The product is 1 only for (1, 1), which is one outcome out of 36.

Teacher's Note:
1. Only 1 x 1 equals 1.

Answer: (b) 0

Explanation:
1. The maximum sum of two dice is 6 + 6 = 12.
2. A sum of 13 is an impossible event, so the probability is 0.

Teacher's Note:
1. An impossible event has probability 0.
2. Check the largest possible sum.

Answer: (c) \( \dfrac{2}{5} \)

Explanation:
1. Primes up to 20: 2, 3, 5, 7, 11, 13, 17, 19, which are 8 numbers.
2. \( P = \dfrac{8}{20} = \dfrac{2}{5} \).

Teacher's Note:
1. 1 is not a prime number.
2. List the primes before counting.

Answer: (d) \( \dfrac{3}{7} \)

Explanation:
1. Numbers with square at most 1: -1, 0, 1, which are 3 numbers out of 7.
2. \( P = \dfrac{3}{7} \).

Teacher's Note:
1. Square each number or use
2. x
3. <= 1.
4. Total numbers = 7.

Answer: (a) \( \dfrac{9}{26} \)

Explanation:
1. Distinct letters of TRIGONOMETRY: T, R, I, G, O, N, M, E, Y, which are 9.
2. \( P = \dfrac{9}{26} \).

Teacher's Note:
1. Count distinct letters only.
2. Total is 26 letters.

Answer: (b) \( \dfrac{3}{7} \)

Explanation:
1. x² < 4 holds for -1, 0, 1 only (since 2² = 4 is not less than 4).
2. So 3 favourable numbers out of 7.

Teacher's Note:
1. Strict inequality excludes -2 and 2.

Answer: (c) \( \dfrac{1}{7} \)

Explanation:
1. A leap year has 2 extra days from the 7 consecutive pairs (Mon-Tue, ..., Sun-Mon).
2. Both a Sunday and a Monday are extra only for (Sunday, Monday): 1 pair.
3. \( P = \dfrac{1}{7} \).

Teacher's Note:
1. The two extra days are consecutive.
2. Only one pair has both Sunday and Monday.

Answer: (d) \( \dfrac{5}{12} \)

Explanation:
1. a/b > 1 means a > b. Pairs: a=2: 1; a=3: 2; a=4: 3; a=5: 4; a=6: 5, so 1+2+3+4+5 = 15 pairs.
2. \( P = \dfrac{15}{36} = \dfrac{5}{12} \).

Teacher's Note:
1. a/b > 1 is the same as a > b.
2. Count pairs for each a.

Answer: (d) A is false, but R is true.

Explanation:
1. The products possible are 1, 4, 9, 2, 8, 18, 3, 12, 27. A product of 10 never occurs, so the event is impossible and its probability is 0, not 1. A is false.
2. R is true: an impossible event has probability 0.

Teacher's Note:
1. List the possible products before deciding.
2. An impossible event has probability 0.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Winning and losing are complementary events, so \( P(\text{lose}) = 1-0.345 = 0.655 \). A is true.
2. R is the rule used, so R explains A.

Teacher's Note:
1. Complementary events add up to 1.
2. Subtract from 1 to get the complement.

Answer: (c) A is true, but R is false.

Explanation:
1. There are 5 vowels in 26 letters, so \( P = \dfrac{5}{26} \). A is true.
2. The English alphabet has 21 consonants, not 21 vowels, so R is false.

Teacher's Note:
1. The English alphabet has 5 vowels and 21 consonants.
2. Check each statement separately.

Answer: (c) A is true, but R is false.

Explanation:
1. A leap year has 52 weeks and 2 extra days; 2 of the 7 possible pairs contain a Sunday, so P = 2/7. A is true.
2. A non-leap year has 1 extra day, so P = 1/7, not 5/7. R is false.

Teacher's Note:
1. Leap year: 2 extra days. Non-leap year: 1 extra day.
2. Count the pairs containing Sunday.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Bad eggs = 14, so good eggs = 386. A is true.
2. P(good) = 1 - 0.035 is exactly the complementary rule in R, so R explains A.

Teacher's Note:
1. Good = total - bad.
2. Complement rule: P(not E) = 1 - P(E).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( P(\text{boundary}) = \dfrac{9}{45} = \dfrac{1}{5} \), so \( P(\text{no boundary}) = 1-\dfrac{1}{5} = \dfrac{4}{5} \). A is true.
2. R is the complement rule used, so R explains A.

Teacher's Note:
1. Find P(boundary) first.
2. Subtract from 1.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is the definition of an elementary event, so it is true.
2. R describes the same idea in more detail and explains why such an event has a single outcome.

Teacher's Note:
1. An elementary event has exactly one outcome.
2. It cannot be broken down further.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is a standard result, so it is true.
2. The events are exhaustive (cover everything) and mutually exclusive (do not overlap), which is why their probabilities add to 1. R explains A.

Teacher's Note:
1. Exhaustive and mutually exclusive events add to 1.