Chapter 15: Probability (NCERT Solutions)

Exercise 15.1: Theoretical Probability

Q1. Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ___.
(ii) The probability of an event that cannot happen is ___. Such an event is called ___.
(iii) The probability of an event that is certain to happen is ___. Such an event is called ___.
(iv) The sum of the probabilities of all the elementary events of an experiment is ___.
(v) The probability of an event is greater than or equal to ___ and less than or equal to ___.

(i) 1
(ii) 0, Impossible event
(iii) 1, Sure event or Certain event
(iv) 1
(v) 0, 1

Q2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

(i) Not equally likely. It depends on various factors (mechanical condition, fuel, etc.).
(ii) Not equally likely. Depends on the player's skill.
(iii) Equally likely. Only two possibilities, chances are equal if guessing.
(iv) Equally likely. Chances are 50-50.

Q3. 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?

Because the result of tossing a coin is completely unpredictable and there are only two outcomes (Head or Tail) which are equally likely. Hence, it is considered fair.

Q4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7

(B) -1.5. Probability cannot be negative. It always lies between 0 and 1.

Q5. If P(E) = 0.05, what is the probability of 'not E'?

P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.

Q6. 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?

(i) 0 (Impossible event, as bag contains only lemon candies).
(ii) 1 (Sure event).

Q7. 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?

P(same birthday) = 1 - P(not same birthday)
= 1 - 0.992 = 0.008.

Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Total balls = 3 + 5 = 8.
(i) P(Red) = 3/8 = 0.375.
(ii) P(Not Red) = 1 - 3/8 = 5/8 = 0.625 (which is P(Black)).

Q9. 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?

Total marbles = 5 + 8 + 4 = 17.
(i) P(Red) = 5/17.
(ii) P(White) = 8/17.
(iii) P(Not Green) = (Total - Green)/Total = (17-4)/17 = 13/17.

Q10. A piggy bank contains hundred 50p 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 50p coin? (ii) will not be a Rs 5 coin?

Total coins = 100 + 50 + 20 + 10 = 180.
(i) P(50p) = 100/180 = 10/18 = 5/9.
(ii) P(Not Rs 5) = (180 - 10)/180 = 170/180 = 17/18.

Q11. 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?

Total fish = 5 + 8 = 13.
P(Male) = 5/13.

Q12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at:
(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Total outcomes = 8.
(i) P(8) = 1/8.
(ii) Odd (1,3,5,7): P(Odd) = 4/8 = 1/2.
(iii) >2 (3,4,5,6,7,8): P(>2) = 6/8 = 3/4.
(iv) <9 (all): P(<9)=8/8=1.

Q13. 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.

Outcomes: 1, 2, 3, 4, 5, 6. Total = 6.
(i) Prime (2, 3, 5): 3 outcomes. P(Prime) = 3/6 = 1/2.
(ii) Between 2 and 6 (3, 4, 5): 3 outcomes. P = 3/6 = 1/2.
(iii) Odd (1, 3, 5): 3 outcomes. P(Odd) = 3/6 = 1/2.

Q14. 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

Total cards = 52.
(i) Red Kings (Heart, Diamond) = 2. P = 2/52 = 1/26.
(ii) Face cards (J,Q,K in 4 suits) = 12. P = 12/52 = 3/13.
(iii) Red face cards (6). P = 6/52 = 3/26.
(iv) Jack of Hearts (1). P = 1/52.
(v) Spades (13). P = 13/52 = 1/4.
(vi) Queen of Diamonds (1). P = 1/52.

Q15. 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?

Total = 5 cards.
(i) Queen (1). P(Queen) = 1/5.
(ii) Queen removed. Total = 4 cards.
(a) Ace (1). P(Ace) = 1/4.
(b) Queen (0). P(Queen) = 0.

Q16. 12 balls balls are mixed with 132 good ones. Note: 12 defective pens mixed with 132 good ones. One pen is taken out. Determine the probability that the pen taken out is a good one.

Total pens = 12 + 132 = 144.
Good pens = 132.
P(Good) = 132/144 = 11/12.

Q17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

(i) Defective = 4, Total = 20. P(Defective) = 4/20 = 1/5.
(ii) Remaining total = 19. Remaining good = 16 - 1 = 15.
P(Not Defective) = 15/19 = 0.789.

Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Total = 90.
(i) Two-digit (10 to 90): 90 - 9 = 81. P = 81/90 = 9/10.
(ii) Perfect squares (1,4,9,16,25,36,49,64,81): 9. P = 9/90 = 1/10.
(iii) Divisible by 5 (5,10,...90): 18 numbers. P = 18/90 = 1/5.

Q19. A child has a die whose six faces show the letters: A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?

Total = 6.
(i) A appears 2 times. P(A) = 2/6 = 1/3.
(ii) D appears 1 time. P(D) = 1/6.

Q20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m? (Rectangle: 3m x 2m).

Area of Rectangle = 3 × 2 = 6 m².
Area of Circle = π(0.5)² = 0.25π m².
P = Area of Circle / Area of Rectangle = 0.25π/6 = π/24.
Answer: π/24.

Q21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?

Total = 144. Defective = 20. Good = 124.
(i) P(Buy) = P(Good) = 124/144 = 31/36.
(ii) P(Not Buy) = P(Defective) = 20/144 = 5/36.

Q22. Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12?

Total outcomes = 36.
(i) Sum = 8 (2,6), (3,5), (4,4), (5,3), (6,2): 5 outcomes. P = 5/36.
(ii) Sum = 13: 0 outcomes (Max sum is 12). P = 0.
(iii) Sum ≤ 12: All 36 outcomes. P = 1.

Q23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Total outcomes (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT) = 8.
Win cases (HHH, TTT) = 2.
Lose cases = 8 - 2 = 6.
P(Lose) = 6/8 = 3/4.

Q24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?

Total = 36.
(ii) 5 comes up at least once: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6) = 11 outcomes.
P(at least once) = 11/36.
(i) P(not come up) = 1 - 11/36 = 25/36.

Q25. Which of the following arguments are correct and which are not correct? Give reasons.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

(i) Incorrect. Possible outcomes are HH, HT, TH, TT. P(one of each) is 2/4 = 1/2, not 1/3. HH is 1/4, TT is 1/4.
(ii) Correct. Odd (1,3,5) and Even (2,4,6) are equally likely. P(odd) = 3/6 = 1/2.

Probability - RD Sharma Important Questions

Q1. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.

Let green = G, blue = B. G + B = 24.
P(Green) = G/24 = 2/3.
G = 24 × (2/3) = 16.
B = 24 - 16 = 8 blue marbles.

Q2. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Let blue balls = x. Total = 5 + x.
P(Blue) = x / (5+x), P(Red) = 5 / (5+x).
x / (5+x) = 2 × [5 / (5+x)].
x = 10.
Answer: 10 blue balls.

Q3. Two dice are thrown at the same time. Find the probability of getting:
(i) same number on both dice.
(ii) different numbers on both dice.

Total = 36.
(i) Doublets: (1,1), (2,2), ... (6,6). Total 6.
P(Same) = 6/36 = 1/6.
(ii) P(Different) = 1 - P(Same) = 1 - 1/6 = 5/6.

Q4. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is:
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king.
(iv) either a king or a queen.

Total = 52.
(i) Spades (13) + Aces (3 others) = 16. P = 16/52 = 4/13.
(ii) Black Kings (2). P = 2/52 = 1/26.
(iii) Jacks (4) + Kings (4) = 8. Neither = 52 - 8 = 44. P = 44/52 = 11/13.
(iv) Kings (4) + Queens (4) = 8. P = 8/52 = 2/13.

Q5. A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is:
(i) a square number.
(ii) a multiple of 7.

Total cards = 123 - 11 + 1 = 113.
(i) Squares (16, 25, 36, 49, 64, 81, 100, 121) = 8. P = 8/113.
(ii) Multiples of 7 (14, 21, ..., 119): 14 = 7×2, 119 = 7×17. Total = 17-2+1=16. P = 16/113.

Q6. Find the probability that a leap year selected at random will contain 53 Sundays and 53 Mondays.

Leap year = 366 days = 52 weeks + 2 days.
Outcomes for 2 days: (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun). Total 7.
For 53 Sun AND 53 Mon, we need (Sun, Mon). 1 outcome.
Probability = 1/7.

Q7. Three distinct coins are tossed together. Find the probability of getting:
(i) at least 2 heads.
(ii) at most 2 heads.

Total = 8 (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT).
(i) At least 2H (HHH, HHT, HTH, THH): 4 cases. P = 4/8 = 1/2.
(ii) At most 2H (All except HHH): 7 cases. P = 7/8.

Q8-Q15. [Additional problems on complex card probabilities, geometrical probability, and unknown variables]

Key concepts:
• Leap year problems (53 Sundays).
• Complex conditions on deck of cards.
• Probability from areas.

Probability - Formulas & PYQs

Key Concepts

1. Probability Formula

P(E) = (Number of outcomes favourable to E) / (Total number of possible outcomes)

2. Properties

0 ≤ P(E) ≤ 1

P(Sure Event) = 1

P(Impossible Event) = 0

P(E) + P(not E) = 1

3. Cards (Deck of 52)

Suits: Spades (♠), Hearts (♥), Diamonds (♦), Clubs (♣) - 13 each.

Red cards: 26 (Hearts, Diamonds). Black: 26 (Spades, Clubs).

Face cards: K, Q, J (3 in each suit, total 12).

Previous Year Questions (CBSE/JKBOSE)

Q1. A die is thrown once. Find the probability of getting a number divisible by 3. (CBSE 2020)

Outcomes: 1, 2, 3, 4, 5, 6.
Divisible by 3: 3, 6 (2 outcomes).
Probability = 2/6 = 1/3.

Q2. Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6. (CBSE 2019)

Total = 36.
Product 6: (1,6), (2,3), (3,2), (6,1). Total 4.
Probability = 4/36 = 1/9.

Q3. A card is drawn at random from a well shuffled deck of 52 playing cards. Find the probability of getting neither a red card nor a queen. (CBSE 2016)

Red cards = 26.
Black Queens = 2.
Total cards to AVOID = 26 + 2 = 28.
Cards remaining = 52 - 28 = 24.
Probability = 24/52 = 6/13.

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