GSEB Class 12 Statistics Solutions Chapter 1 Probability Exercise 1.3

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.3

Question 1.2 cards are drawn from a pack of 52 cards. Find the probability that both the cards drawn are
(1) of the same suit
(2) of the same colour.
Answer:To pick 2 cards from a standard deck of 52 cards, the total number of possible ways is calculated as \( n = {^{52}C_2} \). \[ n = \frac{52 \times 51}{2 \times 1} = 1326 \] (1) Let A be the event where the two cards drawn are of the same suit. This means both cards are spades, or both are clubs, or both are hearts, or both are diamonds. Each suit in a pack of 52 cards has 13 cards. The number of favorable outcomes for event A is found by choosing 2 cards from any of the 4 suits. \[ m = {^{13}C_2} + {^{13}C_2} + {^{13}C_2} + {^{13}C_2} \] \[ m = 4 \times {^{13}C_2} = 4 \times \frac{13 \times 12}{2 \times 1} \] \[ m = 4 \times 78 = 312 \] The probability of event A is \( P(A) = \frac{m}{n} \). \[ P(A) = \frac{312}{1326} = \frac{4}{17} \] (2) Let B be the event where the two cards drawn are of the same colour. This means both cards are black, or both cards are red. In a pack of 52 cards, there are 26 black cards and 26 red cards. The number of favorable outcomes for event B is found by choosing 2 cards from the black cards or 2 cards from the red cards. \[ m = {^{26}C_2} + {^{26}C_2} = 2 \times {^{26}C_2} \] \[ m = 2 \times \frac{26 \times 25}{2 \times 1} = 2 \times 325 \] \[ m = 650 \] The probability of event B is \( P(B) = \frac{m}{n} \). \[ P(B) = \frac{650}{1326} = \frac{25}{51} \]In simple words: First, we find all the ways to pick 2 cards from 52. Then, for cards of the same suit, we find ways to pick 2 from spades, 2 from clubs, 2 from hearts, or 2 from diamonds, and add them up. For cards of the same color, we find ways to pick 2 from black cards or 2 from red cards and add them up. We divide these by the total ways to get the probability.

🎯 Exam Tip: Remember that "same suit" means cards must be from one of the four specific suits (spades, clubs, hearts, diamonds), while "same colour" means cards must be either both black or both red. Clearly identifying `n` (total outcomes) and `m` (favorable outcomes) is crucial for accurate probability calculation.

Question 2.3 books of Statistics and 4 of Mathematics are arranged on a shelf. Two books are randomly selected from these books. Find the probability that both the books selected are of the same subject.
Answer:There are 3 books on Statistics and 4 books on Mathematics, making a total of \( 3 + 4 = 7 \) books on the shelf. When 2 books are selected randomly from these 7 books, the total number of possible outcomes is \( n \). \[ n = {^7C_2} = \frac{7 \times 6}{2 \times 1} = 21 \] Let A be the event that both selected books are of the same subject. This means either both books are from Statistics, or both books are from Mathematics. The number of favorable outcomes for event A is calculated by adding the ways to choose 2 Statistics books and the ways to choose 2 Mathematics books. \[ m = {^3C_2} + {^4C_2} \] \[ m = \frac{3 \times 2}{2 \times 1} + \frac{4 \times 3}{2 \times 1} = 3 + 6 = 9 \] The probability of event A is \( P(A) = \frac{m}{n} \). \[ P(A) = \frac{9}{21} = \frac{3}{7} \]In simple words: We have 7 books in total. We want to pick 2 books that are either both Statistics or both Mathematics. We count how many ways this can happen and divide by the total ways to pick any 2 books from the 7.

🎯 Exam Tip: When dealing with selections, remember to use combinations (`nCr`). For "same subject," consider separate cases for each subject and add their outcomes to find the total favorable outcomes before calculating probability.

Question 3.One card is randomly drawn from a pack of 52 cards. Find the probability that it is
(1) spade card or ace
(2) neither spade nor ace.
Answer:When one card is drawn randomly from a pack of 52 cards, the total number of possible outcomes is \( n \). \[ n = {^{52}C_1} = 52 \] (1) To find the probability that the card is a spade or an ace: Let A be the event that the card drawn is a spade. There are 13 spades in a pack. \[ P(A) = \frac{13}{52} \] Let B be the event that the card drawn is an ace. There are 4 aces in a pack. \[ P(B) = \frac{4}{52} \] Let A \(\cap\) B be the event that the card drawn is both a spade and an ace, which is the Ace of Spades. There is only 1 such card. \[ P(A \cap B) = \frac{1}{52} \] The event that the card is a spade or an ace is A \(\cup\) B. Using the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). \[ P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \] (2) To find the probability that the card is neither a spade nor an ace: This is the complement of the event "spade or ace", which is \( P(A \cup B)' \). Using the complement rule: \( P(A \cup B)' = 1 - P(A \cup B) \). \[ P(A \cup B)' = 1 - \frac{4}{13} = \frac{13 - 4}{13} = \frac{9}{13} \]In simple words: First, we find the chance of getting a spade, and the chance of getting an ace. We also find the chance of getting both (Ace of Spades). We add the first two chances and subtract the "both" chance to get "spade or ace". Then, to find "neither spade nor ace", we subtract the "spade or ace" chance from 1.

🎯 Exam Tip: Clearly define your events A and B. Remember the addition rule \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) for "or" probabilities and the complement rule \( P(E') = 1 - P(E) \) for "neither/nor" scenarios.

Question 4.A number is selected from the natural numbers 1 to 100. Find the probability of the event that the selected number is a multiple of 3 or 5.
Answer:The universal set U contains natural numbers from 1 to 100, so \( U = \{1, 2, 3, \dots, 100\} \). When one number is selected from U, the total number of possible outcomes is \( n = {^{100}C_1} = 100 \). Let A be the event that the selected number is a multiple of 3. The multiples of 3 are \( \{3, 6, 9, \dots, 99\} \). The number of favorable outcomes for A is \( m_A = 33 \). The probability of event A is \( P(A) = \frac{m_A}{n} = \frac{33}{100} \). Let B be the event that the selected number is a multiple of 5. The multiples of 5 are \( \{5, 10, 15, \dots, 100\} \). The number of favorable outcomes for B is \( m_B = 20 \). The probability of event B is \( P(B) = \frac{m_B}{n} = \frac{20}{100} \). Let A \(\cap\) B be the event that the selected number is a multiple of both 3 and 5, which means it is a multiple of 15. The multiples of 15 are \( \{15, 30, 45, 60, 75, 90\} \). The number of favorable outcomes for A \(\cap\) B is \( m_{A \cap B} = 6 \). The probability of event A \(\cap\) B is \( P(A \cap B) = \frac{m_{A \cap B}}{n} = \frac{6}{100} \). The event that the number selected is a multiple of 3 or 5 is A \(\cup\) B. Using the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). \[ P(A \cup B) = \frac{33}{100} + \frac{20}{100} - \frac{6}{100} \] \[ P(A \cup B) = \frac{53 - 6}{100} = \frac{47}{100} \]In simple words: We pick one number from 1 to 100. We want to find the chance it's a multiple of 3 or 5. We find how many numbers are multiples of 3 and how many are multiples of 5. Then, we find how many are multiples of both (multiples of 15). We add the multiples of 3 and 5, then subtract the multiples of 15 because they were counted twice. Finally, we divide by 100.

🎯 Exam Tip: For "multiple of X or Y", remember to account for numbers that are multiples of both X and Y (i.e., multiples of their Least Common Multiple) to avoid double-counting in your favorable outcomes.

Question 5.Two balanced dice are thrown simultaneously. Find the probability that the sum of numbers on two dice is a multiple of 2 or 3.
Answer:When two balanced dice are thrown simultaneously, the total number of possible outcomes is \( n = 6^2 = 36 \). Let A be the event that the sum of the numbers on the two dice is a multiple of 2 (an even number). The possible sums are 2, 4, 6, 8, 10, or 12. The outcomes for event A are: \( \{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} \) The number of favorable outcomes for A is \( m_A = 18 \). The probability of event A is \( P(A) = \frac{18}{36} \). Let B be the event that the sum of the numbers on the dice is a multiple of 3. The possible sums are 3, 6, 9, or 12. The outcomes for event B are: \( \{(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\} \) The number of favorable outcomes for B is \( m_B = 12 \). The probability of event B is \( P(B) = \frac{12}{36} \). Let A \(\cap\) B be the event that the sum of the numbers on the dice is a multiple of both 2 and 3, meaning it is a multiple of 6. The possible sums are 6 or 12. The outcomes for A \(\cap\) B are: \( \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)\} \) The number of favorable outcomes for A \(\cap\) B is \( m_{A \cap B} = 6 \). The probability of event A \(\cap\) B is \( P(A \cap B) = \frac{6}{36} \). The event that the sum is a multiple of 2 or 3 is A \(\cup\) B. Using the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). \[ P(A \cup B) = \frac{18}{36} + \frac{12}{36} - \frac{6}{36} \] \[ P(A \cup B) = \frac{30 - 6}{36} = \frac{24}{36} = \frac{2}{3} \]In simple words: When rolling two dice, there are 36 possible results. We want the sum to be a multiple of 2 (even) or a multiple of 3. We count how many ways the sum is even, and how many ways the sum is a multiple of 3. Then we count how many ways the sum is a multiple of both (multiple of 6). We add the first two counts, subtract the "both" count, and divide by 36.

🎯 Exam Tip: Listing all possible outcomes (sample space) for two dice helps in correctly identifying favorable outcomes for sums that are multiples of 2, 3, and 6. Remember to apply the inclusion-exclusion principle for "or" events.

Question 6.The probability that the price of potato rises in the vegetable market during festive days is 0.8. The probability that the price of onion rises is 0.7. The probability of rise in price of both potato and onion is 0.6. Find the probability of rise in price of at least one of the two, potato and onion.
Answer:Let A be the event that the price of potato rises. Let B be the event that the price of onion rises. The event that the prices of both potato and onion rise is A \(\cap\) B. We are given the following probabilities: \( P(A) = 0.8 \) \( P(B) = 0.7 \) \( P(A \cap B) = 0.6 \) The event that the price of at least one of the two (potato or onion) rises is A \(\cup\) B. Using the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). \[ P(A \cup B) = 0.8 + 0.7 - 0.6 \] \[ P(A \cup B) = 1.5 - 0.6 = 0.9 \]In simple words: We know the chance potato prices go up, onion prices go up, and both go up. To find the chance that at least one goes up, we add the potato chance and the onion chance, then subtract the "both" chance.

🎯 Exam Tip: This question directly applies the addition rule for probabilities. Ensure you correctly identify `P(A)`, `P(B)`, and `P(A \cap B)` from the problem statement to accurately calculate `P(A \cup B)`. The phrase "at least one" strongly suggests the union of events.

Question 7.Two aircrafts drop bombs to destroy a bridge. The probability that a bomb dropped from the first aircraft hits the target is 0.9 and the probability that a bomb from the second aircraft hits the target is 0.7. The probability of bombs from both aircraft hitting the target is 0.63. The bridge is destroyed even if one bomb drops on it. Find the probability that the bridge is destroyed.
Answer:Let A be the event that a bomb from the first aircraft hits the target. Let B be the event that a bomb from the second aircraft hits the target. The event that bombs from both aircraft hit the target is A \(\cap\) B. We are given the following probabilities: \( P(A) = 0.9 \) \( P(B) = 0.7 \) \( P(A \cap B) = 0.63 \) The bridge is destroyed if one or more bombs hit it, which means at least one bomb hits the target. This represents the event A \(\cup\) B. Using the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). \[ P(A \cup B) = 0.9 + 0.7 - 0.63 \] \[ P(A \cup B) = 1.6 - 0.63 = 0.97 \]In simple words: We know the chance of the first plane hitting, the second plane hitting, and both hitting. The bridge is destroyed if either plane hits. So, we add the chances of each plane hitting and then subtract the chance of both hitting to find the total chance the bridge is destroyed.

🎯 Exam Tip: The phrase "bridge is destroyed even if one bomb drops on it" translates to "at least one bomb hits," which means you need to find the probability of the union of the two events (P(A \(\cup\) B)).

Question 8.The probability that a teenager coming to a restaurant for dinner orders pizza is 0.63. The probability of ordering cold drink is 0.54. The probability that the teenager orders at least one out of pizza and cold drink is 0.88. Find the probability that the teenager coming for dinner on a certain day orders only one of the two items from pizza and cold drink.
Answer:Let A be the event that a teenager orders pizza. Let B be the event that a teenager orders a cold drink. The event that a teenager orders at least one of the two (pizza or cold drink) is A \(\cup\) B. We are given: \( P(A) = 0.63 \) \( P(B) = 0.54 \) \( P(A \cup B) = 0.88 \) First, find the probability that a teenager orders both pizza and a cold drink, which is A \(\cap\) B. Using the addition rule rearranged: \( P(A \cap B) = P(A) + P(B) - P(A \cup B) \). \[ P(A \cap B) = 0.63 + 0.54 - 0.88 \] \[ P(A \cap B) = 1.17 - 0.88 = 0.29 \] Let C be the event that a teenager orders only one of the two items. This can happen in two mutually exclusive ways: 1. Orders pizza but not a cold drink (A \(\cap\) B'). 2. Orders a cold drink but not pizza (A' \(\cap\) B). The probability of ordering only pizza is \( P(A \cap B') = P(A) - P(A \cap B) \). The probability of ordering only a cold drink is \( P(A' \cap B) = P(B) - P(A \cap B) \). Since A \(\cap\) B' and A' \(\cap\) B are mutually exclusive events (they cannot happen at the same time), the probability of event C is their sum: \[ P(C) = P(A \cap B') + P(A' \cap B) \] \[ P(C) = [P(A) - P(A \cap B)] + [P(B) - P(A \cap B)] \] Substituting the values: \[ P(C) = [0.63 - 0.29] + [0.54 - 0.29] \] \[ P(C) = 0.34 + 0.25 = 0.59 \]In simple words: We know the chance of ordering pizza, cold drink, and at least one. We first find the chance of ordering both. Then, to find the chance of ordering *only one* item, we add the chance of ordering pizza but not cold drink, and the chance of ordering cold drink but not pizza.

🎯 Exam Tip: The phrase "only one of the two" implies finding \( P(A \cap B') + P(A' \cap B) \). This can be calculated as \( P(A) + P(B) - 2P(A \cap B) \) or \( P(A \cup B) - P(A \cap B) \) for a quicker method if \( P(A \cup B) \) is given. Ensure clear understanding of set operations.

Question 9.If A and B are mutually exclusive and exhaustive events in a sample space U and P(A) = 2P(B), then find P(A).
Answer:Events A and B are given to be mutually exclusive and exhaustive. "Mutually exclusive" means they cannot happen at the same time, so \( P(A \cap B) = 0 \). "Exhaustive" means together they cover all possible outcomes in the sample space U, so \( A \cup B = U \). Therefore, \( P(A \cup B) = P(U) = 1 \). Since A and B are mutually exclusive, the probability of their union is the sum of their individual probabilities: \( P(A \cup B) = P(A) + P(B) \) Substituting \( P(A \cup B) = 1 \), we get: \[ P(A) + P(B) = 1 \quad \text{(Equation 1)} \] We are also given the relationship between their probabilities: \[ P(A) = 2P(B) \] From this, we can write \( P(B) = \frac{P(A)}{2} \). Substitute this expression for \( P(B) \) into Equation 1: \[ P(A) + \frac{P(A)}{2} = 1 \] To solve for \( P(A) \), find a common denominator: \[ \frac{2P(A)}{2} + \frac{P(A)}{2} = 1 \] \[ \frac{3P(A)}{2} = 1 \] Multiply both sides by 2: \[ 3P(A) = 2 \] Divide by 3: \[ P(A) = \frac{2}{3} \]In simple words: A and B are events that cannot happen together but cover all possibilities. This means their probabilities add up to 1. We also know that the chance of A is twice the chance of B. By using these two facts, we can figure out the exact chance of A.

🎯 Exam Tip: When events are both mutually exclusive and exhaustive, their probabilities sum to 1. Use algebraic substitution with the given relationships between probabilities to solve for the unknown probability.

Question 10.Three events A, B and C in a sample space are mutually exclusive and exhaustive. If 4P (A) = 5P (B) = 3P (C), then find P (A \(\cup\) C) and P(B \(\cup\) C).
Answer:Given that A, B, and C are mutually exclusive and exhaustive events, their probabilities sum to 1: \( P(A) + P(B) + P(C) = 1 \) We are given the relationship: \( 4P(A) = 5P(B) = 3P(C) \). Let \( 4P(A) = 5P(B) = 3P(C) = x \). From these equalities, we can express each probability in terms of \( x \): 1. \( 4P(A) = x \implies P(A) = \frac{x}{4} \) 2. \( 5P(B) = x \implies P(B) = \frac{x}{5} \) 3. \( 3P(C) = x \implies P(C) = \frac{x}{3} \) Substitute these expressions into the sum of probabilities equation: \[ \frac{x}{4} + \frac{x}{5} + \frac{x}{3} = 1 \] To solve for \( x \), find a common denominator, which is 60: \[ \frac{15x}{60} + \frac{12x}{60} + \frac{20x}{60} = 1 \] \[ \frac{15x + 12x + 20x}{60} = 1 \] \[ \frac{47x}{60} = 1 \] \[ 47x = 60 \implies x = \frac{60}{47} \] Now, substitute the value of \( x \) back into the expressions for \( P(A) \), \( P(B) \), and \( P(C) \): \[ P(A) = \frac{x}{4} = \frac{60/47}{4} = \frac{60}{4 \times 47} = \frac{15}{47} \] \[ P(B) = \frac{x}{5} = \frac{60/47}{5} = \frac{60}{5 \times 47} = \frac{12}{47} \] \[ P(C) = \frac{x}{3} = \frac{60/47}{3} = \frac{60}{3 \times 47} = \frac{20}{47} \] Next, find \( P(A \cup C) \). Since A and C are mutually exclusive, \( P(A \cup C) = P(A) + P(C) \). \[ P(A \cup C) = \frac{15}{47} + \frac{20}{47} = \frac{35}{47} \] Finally, find \( P(B \cup C) \). Since B and C are mutually exclusive, \( P(B \cup C) = P(B) + P(C) \). \[ P(B \cup C) = \frac{12}{47} + \frac{20}{47} = \frac{32}{47} \]In simple words: We have three events that cover all possibilities and cannot happen together. We are given how their chances relate to each other. We use a placeholder 'x' to find the individual chances of A, B, and C. Then, to find the chance of "A or C", we just add the chances of A and C. We do the same to find the chance of "B or C".

🎯 Exam Tip: For mutually exclusive and exhaustive events, remember that \( \sum P(Event_i) = 1 \). When probabilities are given as ratios (e.g., \( k_1 P(A) = k_2 P(B) \)), setting them equal to a common variable \( x \) simplifies finding individual probabilities. For unions of mutually exclusive events, simply add their probabilities.

Question 11.Given the following information about three events A, B and C in a sample space: P(A) = 0.65, P(B) = 0.45, P(C) = 0.25, P(A \(\cap\) B) = 0.25, P(A \(\cap\) C) = 0.15, P(B \(\cap\) C) = 0.2, P(A \(\cap\) B \(\cap\) C) = 0.05. Find P(A \(\cup\) B \(\cup\) C).
Answer:We are given the following probabilities for events A, B, and C: \( P(A) = 0.65 \) \( P(B) = 0.45 \) \( P(C) = 0.25 \) We are also given the probabilities of their intersections: \( P(A \cap B) = 0.25 \) \( P(A \cap C) = 0.15 \) \( P(B \cap C) = 0.2 \) And the probability of their triple intersection: \( P(A \cap B \cap C) = 0.05 \) To find the probability of the union of three events, \( P(A \cup B \cup C) \), we use the Principle of Inclusion-Exclusion: \( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \) Substitute the given values into the formula: \[ P(A \cup B \cup C) = 0.65 + 0.45 + 0.25 - 0.25 - 0.15 - 0.2 + 0.05 \] \[ P(A \cup B \cup C) = 1.35 - 0.6 + 0.05 \] \[ P(A \cup B \cup C) = 0.75 + 0.05 = 0.8 \]In simple words: We know the chances of A, B, C happening alone, the chances of any two happening together, and the chance of all three happening together. To find the chance of "A or B or C" happening, we add the individual chances, then subtract the chances of two events happening at once, and finally add back the chance of all three happening at once.

🎯 Exam Tip: Memorize the Principle of Inclusion-Exclusion formula for three events. Carefully substitute the given values, paying attention to positive and negative signs for each term in the formula.

Question 12.Three events A, B and C in a sample space are mutually exclusive and exhaustive. If P(C') = 0.8 and 3P(B) = 2P(A'), then find P(A) and P(B).
Answer:Given that A, B, and C are mutually exclusive and exhaustive events, their probabilities sum to 1: \( P(A) + P(B) + P(C) = 1 \) (Equation 1) We are given \( P(C') = 0.8 \). Using the complement rule, \( P(C) = 1 - P(C') \). \( P(C) = 1 - 0.8 = 0.2 \) We are also given \( 3P(B) = 2P(A') \). Using the complement rule, \( P(A') = 1 - P(A) \). Substitute \( P(A') \) into the given relationship: \( 3P(B) = 2(1 - P(A)) \) \( 3P(B) = 2 - 2P(A) \) Now, express \( P(B) \) in terms of \( P(A) \): \[ P(B) = \frac{2 - 2P(A)}{3} \] Substitute the expressions for \( P(B) \) and the value of \( P(C) \) into Equation 1: \[ P(A) + \frac{2 - 2P(A)}{3} + 0.2 = 1 \] First, move the constant term to the right side: \[ P(A) + \frac{2 - 2P(A)}{3} = 1 - 0.2 \] \[ P(A) + \frac{2 - 2P(A)}{3} = 0.8 \] To combine the terms on the left, find a common denominator (3): \[ \frac{3P(A)}{3} + \frac{2 - 2P(A)}{3} = 0.8 \] \[ \frac{3P(A) + 2 - 2P(A)}{3} = 0.8 \] Simplify the numerator: \[ \frac{P(A) + 2}{3} = 0.8 \] Multiply both sides by 3: \[ P(A) + 2 = 0.8 \times 3 \] \[ P(A) + 2 = 2.4 \] Subtract 2 from both sides to find \( P(A) \): \[ P(A) = 2.4 - 2 = 0.4 \] Now that we have \( P(A) = 0.4 \), substitute it back into the expression for \( P(B) \): \[ P(B) = \frac{2 - 2P(A)}{3} = \frac{2 - 2(0.4)}{3} \] \[ P(B) = \frac{2 - 0.8}{3} = \frac{1.2}{3} \] \[ P(B) = 0.4 \] So, \( P(A) = 0.4 \) and \( P(B) = 0.4 \).In simple words: We know that A, B, and C cover all outcomes and cannot happen together, so their probabilities add up to 1. We use the given information about C to find its chance. Then, we use the given rule about A and B to write a connection between their chances. We put all this into the first rule to find the chance of A, and then use that to find the chance of B.

🎯 Exam Tip: The terms "mutually exclusive" and "exhaustive" are key starting points, implying \( \sum P(E_i) = 1 \). Use complement rules to convert probabilities of complements (e.g., \( P(C') \) to \( P(C) \), \( P(A') \) to \( P(A) \)). Set up and solve algebraic equations carefully.