CBSE Class 10 Maths Probability Worksheet Set 04

Question. A girl has a cube one letter written on each face, as shown below: M, N, P, M, N, M The cube is thrown once. The probability of getting M is
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: (c) \( \frac{1}{2} \)
Explanation: Number of possible outcomes (getting M) = 3. Number of total outcomes = 6. Required Probability = \( \frac{3}{6} = \frac{1}{2} \).

Question. If P(E) = 0.05, what will be the probability of 'not E'?
(a) 0.55
(b) 0.59
(c) 0.95
(d) 0.095
Answer: (c) 0.95
Explanation: We know that P(E) + P(not E) = 1. \( \therefore \) P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.

Question. A bag contains 50 balls of which 2x are red, 3x are white and 5x are blue. A ball is selected at random. The probability that it is not white is
(a) \( \frac{7}{10} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{7}{45} \)
(d) \( \frac{3}{5} \)
Answer: (a) \( \frac{7}{10} \)
Explanation: Here, \( 2x + 3x + 5x = 50 \Rightarrow 10x = 50 \Rightarrow x = 5 \).
Number of red balls = \( 2 \times 5 = 10 \)
Number of white balls = \( 3 \times 5 = 15 \)
Number of blue balls = \( 5 \times 5 = 25 \)
Now, Number of possible outcomes (not white) = 25 + 10 = 35. And Number of total outcomes = 50. Required Probability = \( \frac{35}{50} = \frac{7}{10} \).

Question. From a well-shuffled pack of 52 cards, one card is drawn at random. The probability of getting a face card is
(a) \( \frac{4}{13} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{6}{13} \)
Answer: (b) \( \frac{3}{13} \)
Explanation: Face Cards are = 4 kings + 4 queens + 4 jacks = 12. Number of possible outcomes = 12. Number of Total outcomes = 52. Required Probability = \( \frac{12}{52} = \frac{3}{13} \).

Question. A card is drawn from a pack of 52 cards at random. The probability of getting either an ace or a king card is
(a) \( \frac{3}{13} \)
(b) \( \frac{8}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{4}{13} \)
Answer: (c) \( \frac{2}{13} \)
Explanation: Number of Total outcomes = 52. Number of aces and Number of kings = 4 + 4 = 8. Required Probability = \( \frac{8}{52} = \frac{2}{13} \).

Question. If the probability of winning a game is 0.7, what is the probability of losing it?
Answer: Let E be the event of winning the game. Then, P(E) = 0.7. Probability of losing the game = 1 - P(E) = (1 - 0.7) = 0.3.

Question. A bag contains lemon flavoured candies only. Shalini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy?
Answer: Bag contains only lemon flavoured candies. So, getting an orange flavoured candy is impossible. probability = \( \frac{\text{Number of favorable outcome}}{\text{Total number of outcome}} \). \( \therefore \) P(orange flavoured candies) = \( \frac{0}{1} = 0 \).

Question. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is
i. a king,
ii. of red colour,
iii. a face card,
iv. a queen.
Answer: After removing 2 red queens and 2 black jacks, the number of remaining cards = 52 - (2 + 2) = 48.
i. Out of 48 cards, there are 4 kings. \( \therefore \) P(getting a king) = \( \frac{4}{48} = \frac{1}{12} \).
ii. Number of cards of red colour = 26 - 2 = 24. Total number of cards = 48. \( \therefore \) P(getting a card of red colour) = \( \frac{24}{48} = \frac{1}{2} \).
iii. Number of face cards = 12 - (2 + 2) = 8. Total number of cards = 48. \( \therefore \) P(getting a face card) = \( \frac{8}{48} = \frac{1}{6} \).
iv. Number of queens in 48 cards = 4 - 2 = 2. \( \therefore \) P(getting a queen) = \( \frac{2}{48} = \frac{1}{24} \).

Question. Cards marked with number 3, 4, 5, ......., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Answer: Total number of cases = 48. Possible outcomes are {4, 9, 16, 25, 36, 49} = 6. No. of favourable outcomes = 6. Probability of event happen P(E) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \). \( \Rightarrow \) P(perfect square number) = \( \frac{6}{48} = \frac{1}{8} \).

Question. A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability that the numbers obtained have a product less than 16?
Answer: Clearly, there are 36 elementary events. \( \therefore n(\text{Total number of throws}) = 36 \). Number of pairs such that the numbers obtained have a product less than 16 can be selected as listed below: {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2)}. Therefore, \( n(\text{Favourable events}) = 25 \). P(the number obtained appearing have a product less than 16) = \( \frac{25}{36} \).

Question. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:(i) red? (ii) white? (iii) not green?
Answer: Total number of marbles in the box = 5 + 8 + 4 = 17. \( \therefore \) Total number of elementary events = 17. Probability of the event = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \).
i. There are 5 red marbles in the box. Favourable number of elementary events = 5. P (getting a red marble) = \( \frac{5}{17} \).
ii. There are 8 white marbles in the box. Favourable number of elementary events = 8. P (getting a white marble) = \( \frac{8}{17} \).
iii. There are 5 + 8 = 13 marbles in the box, which are not green. Favourable number of elementary events = 13. P (not getting a green marble) = \( \frac{13}{17} \).

Question. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 2, 2, 3, 3, 4 respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting (i) sum 7 (ii) sum is a perfect square.
Answer: Total possible outcomes = 36.
i. Sum of number = 7. Favourable outcomes are (3, 4), (4, 3), (4, 3), (5, 2), (5, 2), (6, 1). Favourable ways = 6. Probability that sum of number is 7 = \( \frac{6}{36} = \frac{1}{6} \).
ii. Sum is a perfect square i.e., sum is 4 or 9. Favourable outcomes are (1, 3), (1, 3), (2, 2), (2, 2), (5, 4), (6, 3), (6, 3) = 7. Probability = \( \frac{7}{36} \).

Question. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble?
Answer: There 5 marbles in the bag. Out of these 5 marbles one can be chosen in 5 ways. Total number of elementary events = 5. Since the bag contains 2 blue marbles. Therefore, one blue marble can be drawn in 2 ways. Favourable number of elementary events = 2. Hence, P (Getting a blue marble) = \( \frac{2}{5} \).

Question. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is:
i. an even number
ii. a number less than 14
iii. a number which is a perfect square
iv. a prime number less than 20.
Answer: There are 100 cards in the box out of which one card can be drawn in 100 ways. Total number of elementary events = 100.
i. From numbers 2 to 101, there are 50 even numbers, namely, 2, 4, 6, 8, ..., 100. Out of these 50 even numbered cards, one card can be chosen in 50 ways. Favourable number of elementary events = 50. Hence, P (Getting an even numbered card) = \( \frac{50}{100} = \frac{1}{2} \).
ii. There are 12 cards bearing numbers less than 14 i.e. numbers 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. Favourable number of elementary events = 12. Hence, required probability = \( \frac{12}{100} = \frac{3}{25} \).
iii. Those numbers from 2 to 101 which are perfect squares are 4, 9, 16, 25, 36, 49, 64, 81, 100 i.e. squares of 2, 3, 4, 5, ..., and 10 respectively. Therefore, there are 9 cards marked with the numbers which are perfect squares. Favourable number of elementary events = 9. Hence, P (Getting a card marked with a number which is a perfect square) = \( \frac{9}{100} \).
iv. Prime numbers less than 20 in the numbers from 2 to 101 are 2, 3, 5, 7, 11, 13, 17 and 19. Thus, there are 8 cards marked with prime numbers less than 20. Out of these 8 cards one card can be chose in 8 ways. Favourable number of elementary events = 8. Hence, P (Getting a card marked with a prime number less than 20) = \( \frac{8}{100} = \frac{2}{25} \).

Question. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card
i. is divisible by 9 and is a perfect square.
ii. is a prime number greater than 80.
Answer: Total no. of possibilities are {1, 2, 3 ... 99, 100}. So \( n = 100 \).
i. Number divisible by 9 and perfect square are {9, 36, 81}. So \( m = 3 \). Required probability P = \( \frac{3}{100} \).
ii. Now the prime number more than 80 upto 100 are 83, 89, 97. So \( m = 3 \). Hence, the probability P = \( \frac{3}{100} \).

Question. Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25?
Answer: Peter throws two dice together, \( \therefore \text{Total number of possible outcomes} = 6^2 = 36 \). Possible outcome where product of two numbers is 25 is (5, 5). No. of favourable outcomes = 1. P(product is 25) = \( \frac{1}{36} \).
Rina throws one dice, \( \therefore \text{Total number of all possible outcomes} = 6 \). The number where square is 25 is 5. No. of favourable outcomes = 1. P(a number whose square is 25) = \( \frac{1}{6} \).
On comparison \( \frac{1}{6} > \frac{1}{36} \). Hence, Rina has better chances to get the number 25.

Question. The king, queen and jack of club are removed from a deck of 52 cards. Then the cards are well-shuffled. One card is drawn at random from the remaining cards. Find the probability of getting
i. a heart
ii. a king
iii. a club
iv. a '10' of hearts.
Answer: According to the question, Cards removed = king, queen and jack of clubs = 3. \( \therefore \) Cards left = 52 - 3 = 49. Probability = \( \frac{\text{favourable outcomes}}{\text{Total outcomes}} \).
i. Number of hearts = 13. \( \therefore \) Probability of drawing a heart = \( \frac{13}{49} \).
ii. Total number of kings = 4. Number of kings left = 4 - 1 = 3. \( \therefore \) Probability of drawing a king = \( \frac{3}{49} \).
iii. Number of clubs left = 13 - 3 = 10. Probability of drawing a club = \( \frac{10}{49} \).
iv. There is only one '10' of hearts. \( \therefore \) Probability of drawing one '10' of hearts = \( \frac{1}{49} \).

Question. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting
i. a queen
ii. a diamond
iii. a king or an ace
iv. a red ace.
Answer: A standard deck has 52 cards. 13 ranks of cards are available in 4 different suits namely ♠ Spades - Black in colour, ♥ Hearts - Red in colour, ♦ Diamonds - Red in colour, ♣ Clubs - Black in colour. A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K are the cards available in all the suits. Jack, Queen, King are referred as face cards. 2, 3, 4, 5, 6, 7, 8, 9, 10 are referred as number cards. Total number of all possible outcomes = 52.
i. Total number of queens = 4. P(getting a queen) = \( \frac{4}{52} = \frac{1}{13} \).
ii. Number of diamond suits = 13. P(getting a diamond) = \( \frac{13}{52} = \frac{1}{4} \).
iii. Total number of kings = 4, Total number of aces = 4. Let E be the event of getting a king or an ace card. Then, the favorable outcomes = 4 + 4 = 8. P(getting a king or an ace) = P(E) = \( \frac{8}{52} = \frac{2}{13} \).
iv. Number of red aces = 2. P(getting a red ace) = \( \frac{2}{52} = \frac{1}{26} \).

Question. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Answer: Total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ekta are: (T, T), (T, W), (T, TH), (T, F), (T, S), (W, T), (W, W), (W, TH), (W, F), (W, S), (TH, T), (TH, W), (TH, TH), (TH, F), (TH, S), (F, T), (F, W), (F, TH), (F, F), (F, S), (S, T), (S, W), (S, TH), (S, F), (S, S). Total number of outcomes = 25.
i. The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S). Number of favourable outcomes = 5. Hence required probability = \( \frac{5}{25} = \frac{1}{5} \).
ii. The favourable outcomes of visiting on consecutive days are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S). Number of favourable outcomes = 8. Hence required probability = \( \frac{8}{25} \).
iii. Number of favourable outcomes of visiting on different days are 25 - 5 = 20. Number of favourable outcomes = 20. Hence required probability = \( \frac{20}{25} = \frac{4}{5} \).

Question. A dice is thrown twice. Find the probability that
i. 5 may not come either time.
ii. same number may not come on the dice thrown two times.
Answer: No. of total outcome = 36.
i. Number of outcomes when 5 may not come either time = Outcome except (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5, 5), (5, 6), (6, 5) = 36 - 11 = 25. Probability that 5 may not come either time = \( \frac{25}{36} \).
ii. Number of outcomes when same number may not come on the dice thrown two times = Outcome except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) = 36 - 6 = 30. Probability that same number may not come on the dice thrown two times = \( \frac{30}{36} = \frac{5}{6} \).

Selected NCERT Questions

Question. If \( P(E) = 0.05 \), what is the probability of ‘not \( E \)’?
Answer: As we know that, \( P(E) + P(\text{not } E) = 1 \)
\( P(\text{not } E) = 1 - P(E) = 1 - 0.05 = 0.95 \)

Question. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Answer: (i) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out an orange flavoured candy is an impossible event. So, its probability is 0.
(ii) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out lemon flavoured candies is sure event. So, its probability is 1.

Question. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer: Here, total number of pens \( = 132 + 12 = 144 \)
\( \therefore \) Total number of elementary outcomes \( = 144 \)
Now, favourable number of elementary events \( = 132 \)
\( \therefore \) Probability that a pen taken out is good one \( = \frac{132}{144} = \frac{11}{12} \).

Question. A die is thrown once. Find the probability of getting:
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number. 
Answer: We have, the total number of possible outcomes associated with the random experiment of throwing a die is 6 (i.e., 1, 2, 3, 4, 5, 6).
(i) Let \( E \) denotes the event of getting a prime number.
So, favourable number of outcomes \( = 3 \) (i.e., 2, 3, 5)
\( \therefore P(E) = \frac{3}{6} = \frac{1}{2} \)
(ii) Let \( E \) be the event of getting a number lying between 2 and 6.
\( \therefore \) Favourable number of elementary events (outcomes) \( = 3 \) (i.e., 3, 4, 5)
\( \therefore P(E) = \frac{3}{6} = \frac{1}{2} \)
(iii) Let \( E \) be the event of getting an odd number.
\( \therefore \) Favourable number of elementary events \( = 3 \) (i.e., 1, 3, 5)
\( \therefore P(E) = \frac{3}{6} = \frac{1}{2} \)

Question. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Answer: Here, the total number of possible outcomes \( = 5 \).
(i) Since, there is only one queen.
\( \therefore \) Favourable number of elementary events \( = 1 \)
\( \therefore \) Probability of getting the card of queen \( = \frac{1}{5} \)
(ii) Now, the total number of possible outcomes \( = 4 \).
(a) Since, there is only one ace.
\( \therefore \) Favourable number of elementary events \( = 1 \)
\( \therefore \) Probability of getting an ace card \( = \frac{1}{4} \)
(b) Since, there is no queen (as queen is put aside).
\( \therefore \) Favourable number of elementary events \( = 0 \)
\( \therefore \) Probability of getting a queen \( = \frac{0}{4} = 0 \).

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour.
(ii) a face card.
(iii) a red face card.
(iv) the jack of hearts.
(v) a spade.
(vi) the queen of diamonds.
Answer: Here, total number of possible outcomes \( = 52 \)
(i) As we know that there are two suits of red cards, i.e., diamond and heart and each suit contains one king.
\( \therefore \) Favourable number of outcomes \( = 2 \)
\( \therefore \) Probability of getting a king of red colour \( = \frac{2}{52} = \frac{1}{26} \).
(ii) As we know that kings, queens and jacks are called face cards i.e., there are 12 face cards.
\( \therefore \) Favourable number of elementary events \( = 12 \)
\( \therefore \) Probability of getting a face card \( = \frac{12}{52} = \frac{3}{13} \)
(iii) As we know there are two suits of red cards, i.e., diamond and heart and each suit contains 3 face cards.
\( \therefore \) Favourable number of elementary events \( = 2 \times 3 = 6 \)
\( \therefore \) Probability of getting red face card \( = \frac{6}{52} = \frac{3}{26} \)
(iv) Since, there is only one jack of hearts.
\( \therefore \) Favourable number of elementary events \( = 1 \)
\( \therefore \) Probability of getting the jack of heart \( = \frac{1}{52} \).
(v) Since, there are 13 cards of spade.
\( \therefore \) Favourable number of elementary events \( = 13 \)
\( \therefore \) Probability of getting a spade \( = \frac{13}{52} = \frac{1}{4} \).
(vi) Since, there is only one queen of diamonds.
\( \therefore \) Favourable number of outcomes (elementary events) \( = 1 \)
\( \therefore \) Probability of getting a queen of diamonds \( = \frac{1}{52} \).

Question. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer: The experiment "tossing a coin" have equally likely outcomes head and tail. So, the result of a coin toss is completely unpredictable.

Question. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Answer: Let \( S \) be sample space and \( E \), getting male fish, be event corresponding to given experiment.
From equation \( n(S) = 5 + 8 = 13 \) (number of fish)
\( n(E) = 5 \) (number of male fish)
\( \therefore P(E) = \frac{n(E)}{n(S)} = \frac{5}{13} \)

Question. A piggy bank contains hundred 50 P coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 P coin? (ii) will not be a Rs. 5 coin?
Answer: Let \( S \) be the sample space of corresponding experiment.
Here, \( n(S) = \) Total number of coins \( = 100 + 50 + 20 + 10 = 180 \)
(i) Let \( E \) be the event of falling a coin of 50 P
Here, \( n(E) = \) Number of 50 P coins \( = 100 \)
\( \therefore P(E) = \frac{n(E)}{n(S)} = \frac{100}{180} = \frac{5}{9} \)
(ii) Let \( E \) be the event that falling coin is not a Rs. 5 coin
Here, \( n(E) = \) Number of coins except Rs. 5 coin \( = 100 + 50 + 20 \)
\( \therefore P(E) = \frac{n(E)}{n(S)} = \frac{170}{180} = \frac{17}{18} \)

Question. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday? 
Answer: Let \( E \) be the event of having the same birthday.
Therefore, \( \overline{E} \) is the event of not having the same birthday.
i.e., \( P(\overline{E}) = 0.992 \) (Given)
Now, we have
\( P(E) + P(\overline{E}) = 1
\implies P(E) = 1 - P(\overline{E}) = 1 - 0.992 = 0.008 \)

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. A bag contains 3 red, 5 black and 7 white balls. A ball is drawn from the bag at random. The probability that the ball drawn is not black, is 
(a) \( \frac{1}{3} \)
(b) \( \frac{9}{15} \)
(c) \( \frac{5}{10} \)
(d) \( \frac{2}{3} \)
Answer: (d) \( \frac{2}{3} \)

Question. If \( P(A) \) denotes the probability of an event \( A \), then 
(a) \( P(A) < 0 \)
(b) \( P(A) > 1 \)
(c) \( 0 \leq P(A) \leq 1 \)
(d) \( -1 \leq P(A) \leq 1 \)
Answer: (c) \( 0 \leq P(A) \leq 1 \)

Question. Which of the following cannot be the probability of an event? 
(a) \( \frac{1}{3} \)
(b) 0.1
(c) 3%
(d) \( \frac{17}{16} \)
Answer: (d) \( \frac{17}{16} \)

Question. Kirti has a box containing four cards labelled \( A, B, C \) and \( D \). She randomly picks a card from the box, records the label on the card and put it back in the box. She repeats this experiment 80 times and records her observation in the table shown below.
Card A: 11
Card B: 16
Card C: 25
Card D: 28
Which of the following shows the empirical probability and theoretical probability of picking Card \( C \) the next time?
(a) Empirical probability \( = \frac{5}{11} \), Theoretical probability \( = \frac{1}{2} \)
(b) Empirical probability \( = \frac{5}{11} \), Theoretical probability \( = \frac{1}{4} \)
(c) Empirical probability \( = \frac{5}{16} \), Theoretical probability \( = \frac{1}{2} \)
(d) Empirical probability \( = \frac{5}{16} \), Theoretical probability \( = \frac{1}{4} \)
Answer: (d) Empirical probability \( = \frac{5}{16} \), Theoretical probability \( = \frac{1}{4} \)

Question. Romy is blindfolded and asked to pick one ball from each of the jars. (Jar 1, Jar 2, Jar 3). The chance of Romy picking a red ball is same in
(a) jars 2 and 3
(b) jars 1 and 3
(c) jars 1 and 2
(d) All of the options
Answer: (b) jars 1 and 3

Question. If the probability of an event is \( p \), the probability of its complementary event will be 
(a) \( p - 1 \)
(b) \( 1 - p \)
(c) \( 1 - \frac{1}{p} \)
(d) \( p \)
Answer: (b) \( 1 - p \)

Question. A card is selected from a deck of 52 cards. The probability of being a red face card is 
(a) \( \frac{3}{26} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{1}{2} \)
Answer: (a) \( \frac{3}{26} \)

Question. If a card is drawn from a deck of cards, what is the probability of a card drawn to be a red or a black card and what can we say about that event?
(a) 0 and it is a sure event.
(b) 1 and it is a sure event.
(c) 0 and it is an impossible event.
(d) 1 and it is an impossible event.
Answer: (b) 1 and it is a sure event.

Question. To win a prize in a game, you need to first choose one of the 4 doors, 1, 2, 3, 4 and then need to choose one of the three boxes \( A, B, C \) and then need to choose between two colours red and green. How many of the possible outcomes of this game include selecting box \( A \) and red colour?
(a) 2
(b) 4
(c) 8
(d) 12
Answer: (b) 4

Question. A card is drawn from a deck of 52 cards. The event \( E \) is that card is not an ace of hearts. The number of outcomes favourable to \( E \) is 
(a) 4
(b) 13
(c) 48
(d) 51
Answer: (d) 51

Question. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6,000 tickets are sold, how many tickets has she bought? 
(a) 40
(b) 240
(c) 480
(d) 750
Answer: (c) 480

Question. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(a) 7
(b) 14
(c) 21
(d) 28
Answer: (b) 14

Question. A bag contains three green marbles, four blue marbles and two orange marbles. If a marble is picked at random, then the probability that it is not an orange marble is
(a) \( \frac{7}{9} \)
(b) \( \frac{2}{9} \)
(c) \( \frac{4}{9} \)
(d) none of the options
Answer: (a) \( \frac{7}{9} \)

Question. If a number \( x \) is chosen from the numbers 1, 2, 3 and a number \( y \) is selected from the numbers 1, 4, 9. Then \( P(xy < 9) \) is
(a) \( \frac{3}{9} \)
(b) \( \frac{4}{9} \)
(c) \( \frac{1}{9} \)
(d) \( \frac{5}{9} \)
Answer: (d) \( \frac{5}{9} \)

Question. A box has 10 equal size cards. Of the 10 cards, 4 are blue, 3 are green, 2 are yellow and 1 is red. If a card is randomly drawn from the box, which is the colour that the card is most likely to have?
(a) red
(b) green
(c) blue
(d) yellow
Answer: (c) blue

Question. Look at the numbers shown below.
(i) -0.5
(ii) 0.00001
(iii) \( \frac{1}{2} \)
(iv) 1
(v) 1.00001
(vi) 99%
Which of the above numbers represent probabilities of events? 
(a) only (i) and (iii)
(b) only (i), (ii), (iii) and (iv)
(c) only (ii), (iii), (iv) and (v)
(d) only (ii), (iii), (iv) and (vi)
Answer: (d) only (ii), (iii), (iv) and (vi)

Question. Two identical fair dice have numbers 1 to 6 written on their faces. Both are tossed simultaneously. What is the probability that the product of the numbers that turn up is 12?  
(a) \( \frac{1}{36} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{3} \)
Answer: (b) \( \frac{1}{9} \)