Exercise 14.1 Practice
Theoretical Probability Concepts
Q1: Fill in the Blanks
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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) 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. $P(E) + P(\text{not } E) = 1$.
(ii) 0. Such an event is called an impossible event.
(iii) 1. Such an event is called a sure or certain event.
(iv) 1.
(v) 0 and 1. ($0 \le P(E) \le 1$).
Q2: Equally Likely Outcomes
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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. He/she 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) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. He/she 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. Depends on the car's condition.
(ii) Not equally likely. Depends on the player's skill.
(iii) Equally likely. Only two possibilities (True/False) with no bias assumed.
(iv) Equally likely. Boy or Girl are equally probable biologically in this context.
Q3: Fair Decision
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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?
When a coin is tossed, there are only two possible outcomes: Head or Tail.
These outcomes are equally likely, meaning each has a probability of 1/2.
The result of an individual toss is completely unpredictable. Therefore, it is a fair method.
Q4: Valid Probability
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Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$ (B) -1.5 (C) 15% (D) 0.7
(A) $\frac{2}{3}$ (B) -1.5 (C) 15% (D) 0.7
Probability of an event always lies between 0 and 1 (inclusive).
(A) 2/3 = 0.66 (Valid)
(B) -1.5 (Invalid, probability cannot be negative)
(C) 15% = 0.15 (Valid)
(D) 0.7 (Valid)
(B) -1.5 (Invalid, probability cannot be negative)
(C) 15% = 0.15 (Valid)
(D) 0.7 (Valid)
Option (B) is the answer.
Q5: Complementary Event
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If $P(E) = 0.08$, what is the probability of 'not E'?
$P(\text{not } E) = 1 - P(E)$.
$P(\text{not } E) = 1 - 0.08 = 0.92$.
Probability is 0.92.
Q6: Lemon Candies
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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) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) The bag has only lemon candies. Getting an orange candy is an impossible event. Probability = 0.
(ii) Getting a lemon candy is a sure event. Probability = 1.
Q7: Birthday Probability
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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?
Let E be the event "same birthday". $P(\text{not } E) = 0.992$.
$P(E) = 1 - P(\text{not } E) = 1 - 0.992 = 0.008$.
Probability is 0.008.
Q8: Red & Black Balls
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A bag contains 4 red balls and 6 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 = $4 + 6 = 10$.
(i) $P(\text{Red}) = \frac{\text{No. of red balls}}{\text{Total}} = \frac{4}{10} = \frac{2}{5} = 0.4$.
(ii) $P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - 0.4 = 0.6$. (Or $P(\text{Black}) = 6/10 = 0.6$).
Q9: Marbles
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A box contains 6 red marbles, 7 white marbles and 5 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 = $6 + 7 + 5 = 18$.
(i) $P(\text{Red}) = \frac{6}{18} = \frac{1}{3}$.
(ii) $P(\text{White}) = \frac{7}{18}$.
(iii) $P(\text{Not Green}) = \frac{\text{Red + White}}{18} = \frac{6+7}{18} = \frac{13}{18}$.
Q10: Piggy Bank
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A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹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 ₹5 coin?
(i) will be a 50p coin?
(ii) will not be a ₹5 coin?
Total coins = $100 + 50 + 20 + 10 = 180$.
(i) $P(\text{50p}) = \frac{100}{180} = \frac{5}{9}$.
(ii) No. of ₹5 coins = 10. No. of 'not ₹5' coins = $180 - 10 = 170$.
$P(\text{not } ₹5) = \frac{170}{180} = \frac{17}{18}$.
$P(\text{not } ₹5) = \frac{170}{180} = \frac{17}{18}$.
Q11: Fish Tank
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Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 6 male fish and 9 female fish. What is the probability that the fish taken out is a male fish?
Total fish = $6 + 9 = 15$.
$P(\text{Male}) = \frac{6}{15} = \frac{2}{5} = 0.4$.
Probability is 0.4.
Q12: Spinning Arrow
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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. What is the probability that it will point at:
(i) 5?
(ii) an even number?
(iii) a number greater than 3?
(iv) a number less than 9?
(i) 5?
(ii) an even number?
(iii) a number greater than 3?
(iv) a number less than 9?
Total outcomes = 8.
(i) P(5) = $\frac{1}{8}$.
(ii) Even: 2, 4, 6, 8 (4 numbers). $P(\text{Even}) = \frac{4}{8} = \frac{1}{2}$.
(iii) > 3: 4, 5, 6, 7, 8 (5 numbers). $P(>3) = \frac{5}{8}$.
(iv) < 9: All 8 numbers. $P(<9) = \frac{8}{8} = 1$.
Q13: Die Throw
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A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number lying between 2 and 5
(iii) an odd number
(i) a prime number
(ii) a number lying between 2 and 5
(iii) an odd number
Outcomes: 1, 2, 3, 4, 5, 6. Total = 6.
(i) Prime: 2, 3, 5 (3 numbers). $P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$.
(ii) Between 2 and 5: 3, 4 (2 numbers). $P(\text{Btwn 2-5}) = \frac{2}{6} = \frac{1}{3}$.
(iii) Odd: 1, 3, 5 (3 numbers). $P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$.
Q14: Playing Cards
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One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of black colour
(ii) a face card
(iii) a red face card
(iv) the jack of spades
(i) a queen of black colour
(ii) a face card
(iii) a red face card
(iv) the jack of spades
(i) Black Queens: Spade, Club (2). $P = \frac{2}{52} = \frac{1}{26}$.
(ii) Face cards: 4 Jacks, 4 Queens, 4 Kings (12). $P = \frac{12}{52} = \frac{3}{13}$.
(iii) Red Face cards: 6 (Heart/Diamond J, Q, K). $P = \frac{6}{52} = \frac{3}{26}$.
(iv) Jack of Spades: 1 card. $P = \frac{1}{52}$.
Q15: 5 Cards
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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?
(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?
(i) Total = 5. Queen = 1. $P(\text{Queen}) = \frac{1}{5}$.
(ii) Queen is gone. Total remaining = 4 (Ten, Jack, King, Ace).
(a) Ace = 1. $P(\text{Ace}) = \frac{1}{4}$.
(b) Queen = 0 (since it was put aside). $P(\text{Queen}) = 0$.
Q16: Defective Pens
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10 defective pens are accidentally mixed with 140 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.
Total pens = $10 + 140 = 150$.
Good pens = 140.
$P(\text{Good}) = \frac{140}{150} = \frac{14}{15}$.
Q17: Defective Bulbs
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(i) A lot of 50 bulbs contain 5 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?
(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?
Part (i): Total = 50. Defective = 5.
$P(\text{Defective}) = \frac{5}{50} = \frac{1}{10} = 0.1$.
$P(\text{Defective}) = \frac{5}{50} = \frac{1}{10} = 0.1$.
Part (ii): The bulb drawn was non-defective and not replaced.
Remaining Total = $50 - 1 = 49$.
Remaining Non-Defective = $(50 - 5) - 1 = 44$.
Remaining Total = $50 - 1 = 49$.
Remaining Non-Defective = $(50 - 5) - 1 = 44$.
$P(\text{Not Defective}) = \frac{44}{49}$.
Q18: Numbered Cards
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A box contains 100 cards which are numbered from 1 to 100. If one card 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 7
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 7
Total outcomes = 100.
(i) Two-digit numbers (10 to 100): Total = 91 numbers.
$P = \frac{91}{100} = 0.91$.
$P = \frac{91}{100} = 0.91$.
(ii) Perfect squares (1, 4, 9, ..., 100): There are 10 such numbers ($1^2$ to $10^2$).
$P = \frac{10}{100} = \frac{1}{10} = 0.1$.
$P = \frac{10}{100} = \frac{1}{10} = 0.1$.
(iii) Divisible by 7: 7, 14, ..., 98. $98 = 7 \times 14$, so 14 numbers.
$P = \frac{14}{100} = \frac{7}{50}$.
$P = \frac{14}{100} = \frac{7}{50}$.
Q19: Letter Die
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A child has a die whose six faces show the letters as given below:
P, Q, R, S, T, P
The die is thrown once. What is the probability of getting (i) P? (ii) T?
P, Q, R, S, T, P
The die is thrown once. What is the probability of getting (i) P? (ii) T?
Total faces = 6.
(i) P appears 2 times. $P(\text{P}) = \frac{2}{6} = \frac{1}{3}$.
(ii) T appears 1 time. $P(\text{T}) = \frac{1}{6}$.
Q20: Geometric Probability
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Suppose you drop a coin at random on a rectangular region of length 4 m and width 3 m. What is the probability that it will land inside a circle with diameter 1.4 m drawn inside the rectangle?
Total Area = $4 \times 3 = 12 \text{ m}^2$.
Circle Radius $r = 1.4/2 = 0.7$ m.
Circle Area = $\pi r^2 = \frac{22}{7} \times (0.7)^2 = \frac{22}{7} \times 0.49 = 22 \times 0.07 = 1.54 \text{ m}^2$.
Circle Area = $\pi r^2 = \frac{22}{7} \times (0.7)^2 = \frac{22}{7} \times 0.49 = 22 \times 0.07 = 1.54 \text{ m}^2$.
$P = \frac{\text{Circle Area}}{\text{Total Area}} = \frac{1.54}{12} = \frac{154}{1200} = \frac{77}{600}$.
Probability is approx 0.128.
Q21: Good vs Defective Pens
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A lot consists of 200 ball pens of which 25 are defective and the others are good. Rohit 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 him. What is the probability that:
(i) He will buy it?
(ii) He will not buy it?
(i) He will buy it?
(ii) He will not buy it?
Total pens = 200. Defective = 25. Good = 175.
(i) Buy (Good): $P = \frac{175}{200} = \frac{7}{8} = 0.875$.
(ii) Not Buy (Defective): $P = \frac{25}{200} = \frac{1}{8} = 0.125$.
Q22: Two Dice Sum
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Two dice are thrown at the same time.
(i) Find the probability that the sum of the two numbers appearing on the top is 10.
(ii) Find the probability that the sum is less than or equal to 5.
(i) Find the probability that the sum of the two numbers appearing on the top is 10.
(ii) Find the probability that the sum is less than or equal to 5.
Total outcomes = $6 \times 6 = 36$.
(i) Sum = 10: (4,6), (5,5), (6,4). Total 3 outcomes.
$P = \frac{3}{36} = \frac{1}{12}$.
$P = \frac{3}{36} = \frac{1}{12}$.
(ii) Sum $\le$ 5: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). Total 10 outcomes.
$P = \frac{10}{36} = \frac{5}{18}$.
$P = \frac{10}{36} = \frac{5}{18}$.
Q23: Coin Game
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A game consists of tossing a coin 3 times. A player wins if all the tosses give the same result (i.e., three heads or three tails), and loses otherwise. Calculate the probability that the player will lose the game.
Total outcomes ($2^3$): HHH, HHT, HTH, HTT, THH, THT, TTH, TTT (8 outcomes).
Winning outcomes (Same result): HHH, TTT (2 outcomes).
Losing outcomes: $8 - 2 = 6$.
$P(\text{Lose}) = \frac{6}{8} = \frac{3}{4} = 0.75$.
Q24: Die Thrown Twice
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A die is thrown twice. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
Total outcomes = 36.
Outcomes with 4: (4,1)...(4,6) [6 outcomes] and (1,4)...(6,4) [6 outcomes]. (4,4) is common.
Total outcomes with at least one 4 = $6 + 6 - 1 = 11$.
Total outcomes with at least one 4 = $6 + 6 - 1 = 11$.
(ii) $P(\text{4 at least once}) = \frac{11}{36}$.
(i) $P(\text{4 not come up}) = 1 - P(\text{4 at least once}) = 1 - \frac{11}{36} = \frac{25}{36}$.
Q25: Argument Check
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Which of the following arguments are correct? Give reasons.
(i) If two coins are tossed, there are three possible outcomes: two heads, two tails, or one of each. Therefore, probability of each 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) If two coins are tossed, there are three possible outcomes: two heads, two tails, or one of each. Therefore, probability of each 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. The possible outcomes are HH, HT, TH, TT. "One of each" corresponds to {HT, TH}, which has probability 2/4 = 1/2, not 1/3.
(ii) Correct. Odd numbers {1, 3, 5} and Even numbers {2, 4, 6} are equal in count (3 each). So $P(\text{Odd}) = 3/6 = 1/2$.