Probability
Section A: Multiple Choice Questions
1. A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is:
Solution Total cards = 52. Red face cards (Jack, Queen, King of Hearts & Diamonds) = \(2 \times 3 = 6\).
Probability = \(6/52 = 3/26\).
Probability = \(6/52 = 3/26\).
2. A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a spade (or) King?
Solution Total cards = 52. Spades = 13 (including King of Spades). Other Kings = 3 (Clubs, Diamonds, Hearts).
Total favorable = \(13 + 3 = 16\). Probability = \(16/52 = 4/13\).
Total favorable = \(13 + 3 = 16\). Probability = \(16/52 = 4/13\).
3. The probability that cannot exist among the following:
Solution Probability cannot be negative. Therefore, -1.6 is invalid.
4. The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4 is:
Solution Multiples of 4 are {4, 8, 12}. Total outcomes = 15.
Probability = \(3/15 = 1/5\).
Probability = \(3/15 = 1/5\).
5. For an event E, if \(P(E) + P(\bar{E}) = q\), then the value of \(q^2 - 4\) is:
Solution \(P(E) + P(\bar{E}) = 1\), so \(q = 1\).
\(q^2 - 4 = 1^2 - 4 = 1 - 4 = -3\).
\(q^2 - 4 = 1^2 - 4 = 1 - 4 = -3\).
6. From the letters of the word “MOBILE”, a letter is selected at random. The probability that the selected letter is a vowel is:
Solution Total letters = 6 (M, O, B, I, L, E). Vowels = 3 (O, I, E).
Probability = \(3/6 = 1/2\).
Probability = \(3/6 = 1/2\).
7. Two dice are tossed simultaneously. The probability of getting odd numbers on both dice is:
Solution Total outcomes = 36. Favorable outcomes (Odd, Odd): (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5). Total = 9.
Probability = \(9/36\).
Probability = \(9/36\).
8. Two coins are tossed simultaneously. The probability of having exactly one head is:
Solution Outcomes = {HH, HT, TH, TT}. Exactly one head = {HT, TH} (2 outcomes).
Probability = \(2/4 = 1/2\).
Probability = \(2/4 = 1/2\).
9. Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. The probability of choosing a rotten egg is:
Solution Total eggs = \(3 + 12 = 15\). Rotten = 3.
Probability = \(3/15 = 1/5\).
Probability = \(3/15 = 1/5\).
10. 17 cards are numbered 1, 2, 3, ..., 17. The probability of picking a card divisible by 3 and 2 both is:
Solution Divisible by 3 and 2 means divisible by 6. Numbers are {6, 12}. Total = 2.
Probability = \(2/17\).
Probability = \(2/17\).
Section B: Assertion-Reason Questions
Directions:
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): In a simultaneous throw of a pair of dice, probability of getting a doublet is \(1/6\).
Reason (R): Probability of an event may be negative.
Reason (R): Probability of an event may be negative.
Solution Assertion: Doublets are (1,1) to (6,6), total 6. Prob = 6/36 = 1/6 (True).
Reason: Probability cannot be negative (False).
Reason: Probability cannot be negative (False).
12. Assertion (A): The probability of winning a game is 0.4, then the probability of losing it is 0.6.
Reason (R): \(P(E) + P(\text{not } E) = 1\).
Reason (R): \(P(E) + P(\text{not } E) = 1\).
Solution \(P(\text{Losing}) = 1 - P(\text{Winning}) = 1 - 0.4 = 0.6\). Assertion is true. Reason explains it correctly.
13. Assertion (A): The probability of getting exactly one head in tossing a pair of coins is \(1/2\).
Reason (R): Sample space = {HH, TT, HT, TH}, n=4.
Reason (R): Sample space = {HH, TT, HT, TH}, n=4.
Solution A is true (HT, TH gives 2/4). R is true and explains the sample space.
14. Assertion (A): The probability that a leap year has 53 Sundays is \(2/7\).
Reason (R): The probability that a non-leap year has 53 Sundays is \(5/7\).
Reason (R): The probability that a non-leap year has 53 Sundays is \(5/7\).
Solution Leap year has 2 extra days, so prob of 53 Sundays is 2/7 (True).
Non-leap year has 1 extra day, so prob of 53 Sundays is 1/7. Reason says 5/7 (False).
Non-leap year has 1 extra day, so prob of 53 Sundays is 1/7. Reason says 5/7 (False).
15. Assertion (A): In a lot of 400 eggs with prob of bad egg 0.035, good eggs are 386.
Reason (R): \(P(E) + P(\bar{E}) = 1\).
Reason (R): \(P(E) + P(\bar{E}) = 1\).
Solution \(P(\text{Bad}) = 0.035\). Bad eggs = \(400 \times 0.035 = 14\). Good eggs = \(400 - 14 = 386\). True. Reason correctly explains complement probability logic.
16. Assertion (A): Batsman hits boundary 9 times out of 45 balls. Prob of NOT hitting boundary is \(4/5\).
Reason (R): \(P(E) + P(\text{not } E) = 1\).
Reason (R): \(P(E) + P(\text{not } E) = 1\).
Solution \(P(\text{Hit}) = 9/45 = 1/5\). \(P(\text{Not Hit}) = 1 - 1/5 = 4/5\). Assertion True. Reason explains it.
17. Assertion (A): Tanvi and Manvi born in year 2000. Prob of same birthday is \(1/366\).
Reason (R): Leap year has 366 days.
Reason (R): Leap year has 366 days.
Solution Year 2000 was a leap year (366 days). Same birthday can be any one day. \(P = 1/366\). Reason explains why denominator is 366.
18. Assertion (A): 4-digit number formed using 1,2,5,6,8 without repetition. Prob it is even is \(3/5\).
Reason (R): The units digit of an even number is also an even number.
Reason (R): The units digit of an even number is also an even number.
Solution Total digits = 5. Even digits = 2, 6, 8 (3 digits).
\(P(\text{Even}) = 3/5\). Assertion True. Reason is true but does not explain the calculation directly.
\(P(\text{Even}) = 3/5\). Assertion True. Reason is true but does not explain the calculation directly.
19. Assertion (A): A dice is rolled. Probability of composite number is \(1/3\).
Reason (R): Probability of prime number is \(2/3\).
Reason (R): Probability of prime number is \(2/3\).
Solution Composite numbers on die: {4, 6} (Count=2). \(P(\text{Composite}) = 2/6 = 1/3\) (True).
Primes: {2, 3, 5} (Count=3). \(P(\text{Prime}) = 3/6 = 1/2\). Reason says \(2/3\), which is False.
Primes: {2, 3, 5} (Count=3). \(P(\text{Prime}) = 3/6 = 1/2\). Reason says \(2/3\), which is False.
20. Assertion (A): When dice is rolled, prob of getting number which is multiple of 3 and 5 both is 0.
Reason (R): Probability of impossible event is 0.
Reason (R): Probability of impossible event is 0.
Solution Multiples of 3 and 5 is multiple of 15. No number on a die (1-6) is divisible by 15. Impossible event. Assertion True. Reason True and explains it.