CBSE Class 10 Mathematics Probability MCQs Set 10

Answer: (c) \( \frac{5}{18} \)

Explanation:
1. Total outcomes = 6 × 6 = 36.
2. Favourable outcomes: (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5), which are 10.
3. P = \( \frac{10}{36} \) = \( \frac{5}{18} \).

Teacher's Note:
1. List the pairs with difference 1 in both orders.
2. Reduce the fraction at the end.

Answer: (d) \( \frac{9}{100} \)

Explanation:
1. The pages are numbered 1 to 100, so there are 100 outcomes.
2. Doublet pages: 11, 22, 33, 44, 55, 66, 77, 88, 99, which are 9.
3. P = \( \frac{9}{100} \).

Teacher's Note:
1. Single-digit pages are not doublets.
2. Count only the numbers with two equal digits.

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

Explanation:
1. Total outcomes = 36.
2. The second die must show 5 or 6, and the first die can show any of 6 numbers, so there are 6 × 2 = 12 favourable outcomes.
3. P = \( \frac{12}{36} \) = \( \frac{1}{3} \).

Teacher's Note:
1. Only the second die is restricted.
2. Multiply the choices for the two dice.

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

Explanation:
1. The ways of choosing two different numbers from 1 to 5 are (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5), which are 10.
2. Consecutive pairs: (1, 2), (2, 3), (3, 4), (4, 5), which are 4.
3. P = \( \frac{4}{10} \) = \( \frac{2}{5} \).

Teacher's Note:
1. Treat (a, b) and (b, a) as the same choice here.
2. Count the consecutive pairs separately.

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

Explanation:
1. The sum lies between 2 and 12, so the possible primes are 2, 3, 5, 7 and 11.
2. Number of ways: sum 2 = 1, sum 3 = 2, sum 5 = 4, sum 7 = 6, sum 11 = 2, a total of 15.
3. P = \( \frac{15}{36} \) = \( \frac{5}{12} \).

Teacher's Note:
1. List every prime between 2 and 12.
2. Count the ways for each sum separately.

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

Explanation:
1. Total outcomes = 36, and the outcomes with the same number are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), which are 6.
2. Different numbers: 36 − 6 = 30.
3. P = \( \frac{30}{36} \) = \( \frac{5}{6} \).

Teacher's Note:
1. Use the complement: subtract the doublets from the total.
2. P(different) = 1 − P(same).

Answer: (c) 10

Explanation:
1. Number of Rs. 2 coins = \( \frac{5}{8} \) × 40 = 25.
2. After removing x coins: \( \frac{25 − x}{40 − x} \) = \( \frac{1}{2} \).
3. 2(25 − x) = 40 − x, so 50 − 2x = 40 − x and x = 10.

Teacher's Note:
1. First find the number of Rs. 2 coins from the first probability.
2. Both the favourable and the total number decrease by x.

Answer: (c) \( \frac{2}{3} \)

Explanation:
1. Workers in February = 120 + 15 = 135 and the number of female workers is still 90.
2. P(female) = \( \frac{90}{135} \) = \( \frac{2}{3} \).

Teacher's Note:
1. Only the total changes, because the new workers are male.
2. Reduce the fraction by dividing by 45.

Answer: (d) \( \frac{1}{3} \)

Explanation:
1. Number of marbles = 38 − 18 + 1 = 21.
2. Multiples of 3 from 18 to 38: 18, 21, 24, 27, 30, 33, 36, which are 7.
3. P = \( \frac{7}{21} \) = \( \frac{1}{3} \).

Teacher's Note:
1. Count the end numbers both: 38 − 18 + 1.
2. Divide by 3 to check that 18 is the first multiple.

Answer: (b) II and III

Explanation:
1. There are 13 diamonds, so P(diamond) = 13/52 = 1/4, and I is false.
2. There is 1 ace of hearts, so P = 1/52 and II is true.
3. P(not a heart) = 1 − 13/52 = 3/4, so III is true. P(king or queen) = 8/52 = 2/13, not 1/26, so IV is false.
4. Statements II and III are true.

Teacher's Note:
1. Check each statement separately.
2. Remember a deck has 4 kings and 4 queens.

Answer: (c) \( \frac{1}{6} \)

Explanation:
1. The numbers less than 5 are 1, 2, 3 and 4. Of these, only 4 is composite (1 is neither prime nor composite).
2. P = \( \frac{1}{6} \).

Teacher's Note:
1. 1 is neither prime nor composite.
2. List the numbers below 5 before choosing.

Answer: (c) \( \frac{5}{26} \)

Explanation:
1. The English alphabet has 26 letters, and the word DELHI has 5 different letters.
2. P = \( \frac{5}{26} \).

Teacher's Note:
1. Count distinct letters of the word.
2. The sample space is the 26 letters of the alphabet.

Answer: (c) 0.06

Explanation:
1. The two days are independent events, so P(both) = P(day 1) × P(day 2).
2. = 0.2 × 0.3 = 0.06.

Teacher's Note:
1. For independent events multiply the probabilities.
2. Check that the answer is smaller than both.

Answer: (b) −1/5

Explanation:
1. A probability always lies between 0 and 1 (inclusive).
2. 2/3, 15% (= 0.15) and 0.7 all lie in this range.
3. −1/5 is negative, so it cannot be a probability.

Teacher's Note:
1. Probability can never be negative or more than 1.
2. Convert percentages to decimals.

Answer: (d) \( \frac{4}{15} \)

Explanation:
1. There are 90 two digit numbers (10 to 99).
2. Multiples of 3: 30; multiples of 15 (both 3 and 5): 15, 30, 45, 60, 75, 90, which are 6.
3. Favourable = 30 − 6 = 24, so P = \( \frac{24}{90} \) = \( \frac{4}{15} \).

Teacher's Note:
1. Subtract the common multiples of 3 and 5, that is multiples of 15.
2. Count two digit numbers as 99 − 9 = 90.

Answer: (a) 2/3

Explanation:
1. Numbers greater than 2 are 3, 4, 5 and 6, which are 4 numbers out of 6.
2. P = 4/6 = 2/3.

Teacher's Note:
1. Greater than 2 does not include 2.
2. A probability can never exceed 1, so 4/3 is impossible.

Answer: (b) \( \frac{1}{4} \)

Explanation:
1. The one digit prime numbers are 2, 3, 5 and 7, which are 4 numbers.
2. Only 2 is even, so P = \( \frac{1}{4} \).

Teacher's Note:
1. List the one digit primes first.
2. 2 is the only even prime.

Answer: (c) \( \frac{4}{11} \)

Explanation:
1. PROBABILITY has 11 letters.
2. The vowels are O, A, I, I, which are 4 letters.
3. P = \( \frac{4}{11} \).

Teacher's Note:
1. Count repeated letters each time they appear.
2. The sample space is all 11 letters.

Answer: (a) \( \frac{2}{5} \)

Explanation:
1. A number is odd when its units digit is odd. Among the five digits only 3 and 7 are odd.
2. The units digit can be chosen from the 5 digits with 2 favourable choices, so P = \( \frac{2}{5} \).

Teacher's Note:
1. Only the units digit decides whether the number is odd.
2. Every digit is equally likely to be in the units place.

Answer: (d) 12

Explanation:
1. Total balls = 6 + a, so P(green) = \( \frac{a}{6 + a} \) and P(blue) = \( \frac{6}{6 + a} \).
2. Given \( \frac{a}{6 + a} \) = 2 × \( \frac{6}{6 + a} \), so a = 12.
3. Check: P(green) = 12/18 and P(blue) = 6/18, and 12/18 = 2 × 6/18.

Teacher's Note:
1. Both probabilities have the same denominator.
2. So the number of green balls is double the number of blue balls.

Answer: (c) \( \frac{3}{4} \)

Explanation:
1. The outcomes are HH, HT, TH, TT, which are 4.
2. At most one head means HT, TH, TT, which are 3.
3. P = \( \frac{3}{4} \).

Teacher's Note:
1. At most one means 0 or 1 head.
2. Use the complement: 1 − P(HH).

Answer: (c) \( \frac{1}{5} \)

Explanation:
1. Total tickets = 5 + 20 = 25.
2. P(prize) = \( \frac{5}{25} \) = \( \frac{1}{5} \).

Teacher's Note:
1. The total is prizes plus blanks.
2. Reduce 5/25.

Answer: (a) P(E) + P(not E) = 1

Explanation:
1. An event and its complement together cover the whole sample space.
2. So P(E) + P(not E) = 1.

Teacher's Note:
1. P(not E) = 1 − P(E).
2. The other options would give a probability above 1.

Answer: (c) 480

Explanation:
1. Probability = (tickets bought) ÷ (total tickets).
2. Tickets bought = 0.08 × 6000 = 480.

Teacher's Note:
1. Each ticket has an equal chance of winning.
2. Multiply the probability by the total.

Answer: (a) \( \frac{5}{9} \)

Explanation:
1. Total three-digit numbers = 3 × 3 × 2 = 18 (the first digit cannot be 0).
2. Divisible by 5 means the last digit is 0 or 5. Ending in 0: 3 × 2 = 6 numbers. Ending in 5: 2 × 2 = 4 numbers (first digit 1 or 2). Total = 10.
3. P = \( \frac{10}{18} \) = \( \frac{5}{9} \).

Teacher's Note:
1. A number is divisible by 5 when it ends in 0 or 5.
2. Fix the last digit first, then fill the others.

Answer: (c) \( \frac{3}{8} \)

Explanation:
1. Total pairs = 4 × 4 = 16.
2. Products above 16: 2 × 9 = 18, 3 × 9 = 27, 4 × 9 = 36, 2 × 16 = 32, 3 × 16 = 48, 4 × 16 = 64, which are 6 (1 × 16 = 16 is not more than 16).
3. P = \( \frac{6}{16} \) = \( \frac{3}{8} \).

Teacher's Note:
1. Check 1 × 16 = 16 carefully: it is not greater than 16.
2. List the products row by row.

Answer: (c) \( \frac{1}{5} \)

Explanation:
1. After removing 29 the data are 1, 4, 7, 16, 27, which are 5 numbers.
2. Only 7 is prime, so P = \( \frac{1}{5} \).

Teacher's Note:
1. 29 would have been prime, but it is removed.
2. 1 is not prime.

Answer: (c) \( \frac{8}{25} \)

Explanation:
1. Number of cards = 27 − 3 + 1 = 25.
2. Prime numbers: 3, 5, 7, 11, 13, 17, 19, 23, which are 8.
3. P = \( \frac{8}{25} \).

Teacher's Note:
1. Count the cards from 3 to 27 inclusive.
2. List the primes in order.

Answer: (d) \( \frac{1}{9} \)

Explanation:
1. Total outcomes = 36.
2. Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), which are 4.
3. P = \( \frac{4}{36} \) = \( \frac{1}{9} \).

Teacher's Note:
1. List the pairs in both orders.
2. Reduce 4/36.

Answer: (b) 0.99

Explanation:
1. For x > 0, \( x + \frac{1}{x} - 2 = \frac{(x-1)^2}{x} \ge 0 \), with equality only when x = 1.
2. So \( x + \frac{1}{x} \gt 2 \) for x = 2, 3, …, 100, which are 99 values.
3. P = \( \frac{99}{100} \) = 0.99.

Teacher's Note:
1. Rewrite the inequality as (x − 1)² > 0.
2. Only x = 1 fails.

Answer: (a) 1

Explanation:
1. A coin must land either head or tail, so the event is certain.
2. P = 2/2 = 1.

Teacher's Note:
1. A sure event has probability 1.
2. Both outcomes are favourable.

Answer: (b) The probability of getting a multiple of 3 is 1/3.

Explanation:
1. P(odd) = 3/6 = 1/2, so (a) is wrong.
2. P(multiple of 3) = 2/6 = 1/3 (numbers 3 and 6), so (b) is correct.
3. P(prime) = 3/6 = 1/2 (numbers 2, 3, 5), so (c) is wrong. P(greater than 5) = 1/6, so (d) is wrong.

Teacher's Note:
1. Compute each probability instead of guessing.
2. Prime numbers on a die are 2, 3 and 5.

Answer: (b) negative

Explanation:
1. Probability lies between 0 and 1, both included.
2. It can be positive, zero or one, but it can never be negative.

Teacher's Note:
1. Remember 0 ≤ P(E) ≤ 1.
2. Zero is for an impossible event and one for a sure event.

Answer: (c) impossible event

Explanation:
1. An event that can never happen has probability 0.
2. Such an event is called an impossible event.

Teacher's Note:
1. Probability 1 gives a sure or certain event.
2. Probability 0 gives an impossible event.

Answer: (d) None of these

Explanation:
1. (i) is the definition of an impossible event.
2. (ii) A compound event has more than one favourable outcome, so it is correct.
3. (iii) is the definition of the sample space.
4. All the three statements are correct, so none of them is incorrect.

Teacher's Note:
1. Read the question: it asks for the statement which is not correct.
2. If all three are correct, the answer is none of these.

Answer: (a) \( \frac{1}{26} \)

Explanation:
1. There are 2 red kings (hearts and diamonds) in 52 cards.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \).

Teacher's Note:
1. Count only the red suits for this question.
2. Reduce 2/52.

Answer: (d) \( \frac{3}{13} \)

Explanation:
1. Face cards are the jack, queen and king of each of the 4 suits, which are 3 × 4 = 12.
2. P = \( \frac{12}{52} \) = \( \frac{3}{13} \).

Teacher's Note:
1. Aces are not face cards.
2. Reduce 12/52 by 4.

Answer: (b) \( \frac{1}{52} \)

Explanation:
1. There is exactly one jack of hearts in the deck.
2. P = \( \frac{1}{52} \).

Teacher's Note:
1. Name the card and suit: only one such card exists.
2. P = 1/52.

Answer: (a) \( \frac{3}{26} \)

Explanation:
1. Red face cards: jack, queen, king of hearts and of diamonds, which are 6.
2. P = \( \frac{6}{52} \) = \( \frac{3}{26} \).

Teacher's Note:
1. Only hearts and diamonds are red.
2. 3 face cards × 2 red suits.

Answer: (d) \( \frac{1}{4} \)

Explanation:
1. There are 13 spades in the deck.
2. P = \( \frac{13}{52} \) = \( \frac{1}{4} \).

Teacher's Note:
1. Each suit has 13 cards.
2. Reduce 13/52.