RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Probability is the chapter where careful counting matters more than algebra. Every answer is the same fraction — favourable outcomes divided by total equally likely outcomes — so the real work is listing the sample space without missing or repeating anything. Get into the habit of writing down the total number of outcomes first, then counting the favourable ones, and finally reducing the fraction to lowest terms.
A fair die is rolled once. The probability of getting a prime number is:
Answer
The probability is 1/2. The total number of outcomes is 6, namely 1, 2, 3, 4, 5 and 6. The prime numbers among them are 2, 3 and 5, so there are 3 favourable outcomes. P(prime) = 3/6 = 1/2. Remember that 1 is not a prime number, a slip that would give the wrong answer 2/3.
One card is drawn at random from a well-shuffled pack of 52 playing cards. The probability that it is a face card is:
Answer
The probability is 3/13. Each of the four suits contains 3 face cards — the jack, the queen and the king. So the number of face cards = 4 × 3 = 12. P(face card) = 12/52. Dividing numerator and denominator by 4 gives 3/13.
If the probability of an event E happening is 0.37, then the probability that E does not happen is:
Answer
The probability is 0.63. For any event, P(E) + P(not E) = 1. So P(not E) = 1 - P(E) = 1 - 0.37 = 0.63. The option 1.37 can be rejected immediately, since no probability can exceed 1.
Two fair coins are tossed together. The probability of getting at least one head is:
Answer
The probability is 3/4. The sample space is HH, HT, TH, TT, so there are 4 equally likely outcomes. 'At least one head' means one head or two heads, which happens in HH, HT and TH — that is 3 outcomes. P(at least one head) = 3/4. Using the complement: P(no head) = P(TT) = 1/4, so P(at least one head) = 1 - 1/4 = 3/4.
Assertion (A): The probability of an event can never be 1.5. Reason (R): For any event E of a random experiment, 0 ≤ P(E) ≤ 1.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is a basic property: P(E) is the ratio (favourable outcomes)/(total outcomes), and the number of favourable outcomes can never be less than 0 nor greater than the total. So the ratio must lie between 0 and 1 inclusive. Since 1.5 is greater than 1, it lies outside this range and cannot be a probability, which is exactly Assertion A. Hence R is the correct explanation of A.
A bag contains 5 red marbles, 8 green marbles and 7 yellow marbles, all identical except for colour. One marble is drawn at random. Find the probability that it is (i) green, (ii) not yellow.
Answer
Total number of marbles = 5 + 8 + 7 = 20, so there are 20 equally likely outcomes. (i) There are 8 green marbles. P(green) = 8/20 = 2/5. (ii) There are 7 yellow marbles, so P(yellow) = 7/20. P(not yellow) = 1 - 7/20 = 13/20. Check: the non-yellow marbles are 5 + 8 = 13, which agrees. ✔
Identical cards bearing the numbers 3, 4, 5, …, 22 are placed in a box and mixed thoroughly. One card is drawn at random. Find the probability that the number on it is (i) a multiple of 4, (ii) a prime number, (iii) a perfect square.
Answer
The numbers run from 3 to 22, so the total number of cards = 22 - 3 + 1 = 20. (i) The multiples of 4 in this range are 4, 8, 12, 16 and 20 — that is 5 numbers. P(multiple of 4) = 5/20 = 1/4. (ii) The primes in this range are 3, 5, 7, 11, 13, 17 and 19 — that is 7 numbers. P(prime) = 7/20. (iii) The perfect squares in this range are 4, 9 and 16 — that is 3 numbers. P(perfect square) = 3/20.
A fair die is thrown once. Find the probability that the number obtained is not a multiple of 3.
Answer
The total number of outcomes is 6. The multiples of 3 on a die are 3 and 6, so there are 2 such outcomes. P(multiple of 3) = 2/6 = 1/3. P(not a multiple of 3) = 1 - 1/3 = 2/3. Check by direct counting: the favourable numbers are 1, 2, 4 and 5, giving 4/6 = 2/3. ✔
Two fair dice, one red and one blue, are thrown together and the numbers on their tops are noted. Find the probability that (i) the sum of the numbers is 9, (ii) the sum is a prime number, (iii) both dice show the same number, (iv) the product of the numbers is 12.
Answer
Since each die has 6 faces and the two dice are distinguishable, the total number of equally likely outcomes = 6 × 6 = 36. (i) Sum equal to 9. The favourable pairs (red, blue) are (3, 6), (4, 5), (5, 4) and (6, 3) — that is 4 outcomes. P(sum is 9) = 4/36 = 1/9. (ii) Sum is a prime number. The possible sums range from 2 to 12, and the primes among them are 2, 3, 5, 7 and 11. Sum 2: (1, 1) → 1 outcome. Sum 3: (1, 2), (2, 1) → 2 outcomes. Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) → 4 outcomes. Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 outcomes. Sum 11: (5, 6), (6, 5) → 2 outcomes. Total favourable = 1 + 2 + 4 + 6 + 2 = 15. P(sum is prime) = 15/36 = 5/12. (iii) Both dice show the same number. The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) — that is 6 outcomes. P(same number) = 6/36 = 1/6. (iv) Product equal to 12. The favourable pairs are (2, 6), (6, 2), (3, 4) and (4, 3) — that is 4 outcomes. Note that (12, 1) is impossible because a die shows at most 6. P(product is 12) = 4/36 = 1/9.
A drawer contains 80 identical plastic tokens numbered 1, 2, 3, …, 80. One token is taken out at random. Find the probability that the number on it is (i) a two-digit number, (ii) a perfect square, (iii) divisible by 7, (iv) divisible by both 4 and 6.
Answer
The total number of tokens is 80, so there are 80 equally likely outcomes. (i) Two-digit numbers. The one-digit numbers are 1 to 9, which is 9 tokens. So the two-digit numbers are 80 - 9 = 71 in count (they run from 10 to 80). P(two-digit number) = 71/80. (ii) Perfect squares. They are 1, 4, 9, 16, 25, 36, 49 and 64. The next square, 81, is beyond 80. So there are 8 favourable tokens. P(perfect square) = 8/80 = 1/10. (iii) Numbers divisible by 7. They are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 and 77 — that is 11 numbers, since 7 × 11 = 77 and 7 × 12 = 84 is too large. P(divisible by 7) = 11/80. (iv) Numbers divisible by both 4 and 6. A number divisible by both must be divisible by their LCM. LCM of 4 and 6 = 12. The multiples of 12 up to 80 are 12, 24, 36, 48, 60 and 72 — that is 6 numbers. P(divisible by both 4 and 6) = 6/80 = 3/40.
Read the following and answer the questions that follow: At a school fair, a 'lucky draw' stall keeps a cloth bag containing tokens that are identical in shape and size but differ in colour: 6 blue tokens, 9 white tokens and 5 red tokens. A player shakes the bag and draws one token at random without looking. (a) Find the probability that the token drawn is white. (b) Find the probability that the token drawn is not blue. (c) Find the probability that the token drawn is either red or blue. (d) Later the organiser adds some extra red tokens to the same bag so that the probability of drawing a red token becomes 1/2. How many red tokens were added?
Answer
Total number of tokens at the start = 6 + 9 + 5 = 20. (a) There are 9 white tokens. P(white) = 9/20. (b) There are 6 blue tokens, so P(blue) = 6/20 = 3/10. P(not blue) = 1 - 3/10 = 7/10. Check: the non-blue tokens number 9 + 5 = 14, and 14/20 = 7/10. ✔ (c) The red and blue tokens together number 5 + 6 = 11. P(red or blue) = 11/20. (d) Let k red tokens be added. The number of red tokens becomes 5 + k and the total becomes 20 + k. (5 + k)/(20 + k) = 1/2 Cross-multiplying: 2(5 + k) = 20 + k 10 + 2k = 20 + k 2k - k = 20 - 10 k = 10. So 10 red tokens were added. (The bag then holds 15 red out of 30 tokens, and 15/30 = 1/2. ✔)
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