GSEB Class 12 Statistics Solutions Chapter 1 Probability Exercise 1.4

Gujarat Board Textbook Solutions Class 12 Part 2 Chapter 1 Probability Ex 1.4

 

Question 1. There are two children in a family. If the first child is a girl, find the probability that both children in the family are girls.

Answer: Let B represent a Boy and G represent a Girl. The possible outcomes for a family with two children are: U = {BB, BG, GB, GG} The total number of outcomes, \(n = 4\). Let A be the event that the first child is a girl. A = {GB, GG} The number of favorable outcomes for event A is \(m = 2\). So, the probability of event A is \(P(A) = \frac{m}{n} = \frac{2}{4} = \frac{1}{2}\). Let B be the event that both children are girls. B = {GG} The number of favorable outcomes for event B is \(m = 1\). The probability of event B is \(P(B) = \frac{m}{n} = \frac{1}{4}\). The intersection of A and B, denoted as A \( \cap \) B, is the event that the first child is a girl AND both children are girls. This means both children must be girls. A \( \cap \) B = {GG} So, \(P(A \cap B) = P(B) = \frac{1}{4}\). Now, we need to find the probability that both children are girls, given that the first child is a girl. This is a conditional probability, P(B|A). The formula for conditional probability is: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\)
\( = \frac{\frac{1}{4}}{\frac{1}{2}}\)
\( = \frac{1}{4} \times \frac{2}{1}\)
\( = \frac{1}{2}\)In simple words: We are given that the first child is a girl. We want to find the chance that the second child is also a girl, making both children girls. Out of the cases where the first child is a girl (GB, GG), only one case (GG) has both children as girls.

🎯 Exam Tip: Clearly define your events and sample space. Remember that conditional probability \(P(B|A)\) calculates the likelihood of event B happening, given that event A has already occurred.

 

Question 2. Two six-faced balanced dice are thrown simultaneously. If the sum of numbers on both the dice is more than 7, then find the probability that both the dice show same numbers.

Answer: When two six-faced balanced dice are thrown at the same time, the total number of possible outcomes is \(n = 6^2 = 36\). Let A be the event that the sum of the numbers on both dice is more than 7. This means the sum can be 8, 9, 10, 11, or 12. A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} The number of favorable outcomes for event A is \(m = 15\). So, the probability of event A is \(P(A) = \frac{m}{n} = \frac{15}{36}\). Let B be the event that both dice show the same numbers. B = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} The number of favorable outcomes for event B is \(m = 6\). Now, let's find the intersection of A and B (A \( \cap \) B), which means the sum of numbers is more than 7 AND both dice show the same numbers. A \( \cap \) B = {(4, 4), (5, 5), (6, 6)} The number of favorable outcomes for event A \( \cap \) B is \(m = 3\). So, the probability of A \( \cap \) B is \(P(A \cap B) = \frac{m}{n} = \frac{3}{36}\). We need to find the conditional probability P(B|A), which is the probability that both dice show the same numbers, given that the sum of the numbers is more than 7. Using the law of conditional probability: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\)
\( = \frac{\frac{3}{36}}{\frac{15}{36}}\)
\( = \frac{3}{36} \times \frac{36}{15}\)
\( = \frac{3}{15}\)
\( = \frac{1}{5}\)In simple words: We know the dice sum to more than 7. We want to find how likely it is that both dice show the same number within those cases. We list outcomes where the sum is greater than 7. Then, from those outcomes, we count how many also have matching numbers on both dice.

🎯 Exam Tip: When dealing with two dice, carefully list all possible outcomes for events. Conditional probability reduces the sample space to only the outcomes where the given event (A) has occurred.

 

Question 3. Among the various vehicle owners visiting a petrol pump, 80% vehicle owners visit to fill petrol in their vehicle and 60% vehicle owners visit to fill air in their vehicles. 50 % vehicle owners visit to fill air and petrol in their vehicle. Find the probability for the following events:
(1) If a vehicle owner has come to fill petrol in his vehicle, then that vehicle owner will fill air.
(2) If a vehicle owner has come to fill air in his vehicle, then that vehicle owner will fill petrol in his vehicle.

Answer: Let A be the event that vehicle owners fill petrol in their vehicles. The probability of event A is \(P(A) = \frac{80}{100} = \frac{4}{5}\). Let B be the event that vehicle owners fill air in their vehicles. The probability of event B is \(P(B) = \frac{60}{100} = \frac{3}{5}\). The intersection of A and B (A \( \cap \) B) is the event that vehicle owners fill both petrol and air in their vehicles. The probability of A \( \cap \) B is \(P(A \cap B) = \frac{50}{100} = \frac{1}{2}\). (1) We need to find the probability that a vehicle owner will fill air, given that they have come to fill petrol. This is \(P(B|A)\). According to the law of conditional probability: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\)
\( = \frac{\frac{1}{2}}{\frac{4}{5}}\)
\( = \frac{1}{2} \times \frac{5}{4}\)
\( = \frac{5}{8}\) (2) We need to find the probability that a vehicle owner will fill petrol, given that they have come to fill air. This is \(P(A|B)\). According to the law of conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\( = \frac{\frac{1}{2}}{\frac{3}{5}}\)
\( = \frac{1}{2} \times \frac{5}{3}\)
\( = \frac{5}{6}\)In simple words: This problem asks us to find probabilities based on one event already happening. For example, if we know a person is getting petrol, what's the chance they also get air? We use a special formula for this, dividing the chance of both things happening by the chance of the thing we already know.

🎯 Exam Tip: Clearly distinguish between \(P(A|B)\) and \(P(B|A)\). The event after the vertical bar is the condition that is known to have occurred, and it forms the denominator in the conditional probability formula.

 

Question 4. 80% customers hold saving account and 50% customers hold current account of a nationalised bank. 90% of the customers hold at least one of the saving account and the current account. If one of the account holders randomly selected from this bank holds a current account, find the probability that he holds a saving account.

Answer: Let A be the event that customers hold a saving account. The probability of event A is \(P(A) = \frac{80}{100} = \frac{4}{5}\). Let B be the event that customers hold a current account. The probability of event B is \(P(B) = \frac{50}{100} = \frac{1}{2}\). Let A \( \cup \) B be the event that customers hold at least one of the saving or current accounts. The probability of A \( \cup \) B is \(P(A \cup B) = \frac{90}{100} = \frac{9}{10}\). Now, let's find the probability that customers hold both saving and current accounts, denoted as P(A \( \cap \) B). Using the addition rule of probability: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\( \frac{9}{10} = \frac{4}{5} + \frac{1}{2} - P(A \cap B)\)
\( P(A \cap B) = \frac{4}{5} + \frac{1}{2} - \frac{9}{10}\) To sum the fractions, find a common denominator (10):
\( P(A \cap B) = \frac{8}{10} + \frac{5}{10} - \frac{9}{10}\)
\( = \frac{8 + 5 - 9}{10}\)
\( = \frac{4}{10}\)
\( = \frac{2}{5}\) We need to find the probability that a selected account holder holds a saving account, given that they hold a current account. This is \(P(A|B)\). According to the law of conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\( = \frac{\frac{2}{5}}{\frac{1}{2}}\)
\( = \frac{2}{5} \times \frac{2}{1}\)
\( = \frac{4}{5}\)In simple words: We know the percentages of people with different bank accounts. We are told that someone has a current account. We want to find the chance that this same person also has a saving account. We use a formula that connects the chance of having both types of accounts to the chance of having just a current account.

🎯 Exam Tip: Remember the addition rule for probabilities: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). It's crucial for finding the intersection when union and individual probabilities are given, especially before calculating conditional probabilities.

 

Question 5. If \(P(A) = \frac{2}{3}\), \(P(B) = \frac{3}{5}\) and \(P(B|A) = \frac{3}{4}\) for two events in the sample space of a random experiment, then find \(P(A|B)\).

Answer: We are given the following probabilities: \(P(A) = \frac{2}{3}\) \(P(B) = \frac{3}{5}\) \(P(B|A) = \frac{3}{4}\) First, we use the definition of conditional probability to find \(P(A \cap B)\): \(P(B|A) = \frac{P(A \cap B)}{P(A)}\) Rearranging the formula to solve for \(P(A \cap B)\): \(P(A \cap B) = P(B|A) \times P(A)\)
\( = \frac{3}{4} \times \frac{2}{3}\)
\( = \frac{6}{12}\)
\( = \frac{1}{2}\) Now that we have \(P(A \cap B)\), we can find \(P(A|B)\) using its definition: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\( = \frac{\frac{1}{2}}{\frac{3}{5}}\)
\( = \frac{1}{2} \times \frac{5}{3}\)
\( = \frac{5}{6}\)In simple words: We are given some probabilities, including the chance of B happening if A has already happened. We first use this information to find the chance that both A and B happen together. Then, we use that result to find the chance of A happening if B has already happened.

🎯 Exam Tip: This question tests your ability to manipulate the conditional probability formula. It's often a two-step process: first, find the probability of the intersection, and then use it to find the other conditional probability.

 

Question 6. If \(P(M) = P(F) = \frac{1}{2}\), \(P(A | M) = \frac{1}{10}\) and \(P(A|F) = \frac{1}{2}\) for events A, M and F, then find \(P(A \cap M)\) and \(P (A \cap F)\).

Answer: We are given the following probabilities: \(P(M) = \frac{1}{2}\) \(P(F) = \frac{1}{2}\) \(P(A|M) = \frac{1}{10}\) \(P(A|F) = \frac{1}{2}\) To find \(P(A \cap M)\): Using the conditional probability formula: \(P(A|M) = \frac{P(A \cap M)}{P(M)}\) Rearranging to solve for \(P(A \cap M)\): \(P(A \cap M) = P(A|M) \times P(M)\)
\( = \frac{1}{10} \times \frac{1}{2}\)
\( = \frac{1}{20}\) To find \(P(A \cap F)\): Using the conditional probability formula: \(P(A|F) = \frac{P(A \cap F)}{P(F)}\) Rearranging to solve for \(P(A \cap F)\): \(P(A \cap F) = P(A|F) \times P(F)\)
\( = \frac{1}{2} \times \frac{1}{2}\)
\( = \frac{1}{4}\)In simple words: We are given the probability of some events and also the probability of one event happening if another has already happened. We use these values to find the probability that two specific events happen at the same time.

🎯 Exam Tip: This question is a direct application of the conditional probability formula. Make sure you correctly rearrange the formula to find the probability of the intersection of events.

 

Question 7. There are 2 gold coins and 4 silver coins in a box. The other box contains 3 gold and 5 silver coins. One coin is selected from each box. Find the probability that one of the selected coins is a gold coin and the other is a silver coin.

Answer: In the first box, there are a total of (2 gold + 4 silver) = 6 coins. In the second box, there are a total of (3 gold + 5 silver) = 8 coins. One coin is selected from each box. Let A be the event that one selected coin is gold and the other is silver. There are two ways for event A to occur, which are mutually exclusive (they cannot happen at the same time): 1. A gold coin (G₁) is selected from the first box AND a silver coin (S₂) is selected from the second box. 2. A silver coin (S₁) is selected from the first box AND a gold coin (G₂) is selected from the second box. The probability of event A is the sum of the probabilities of these two scenarios: \(P(A) = P(G_1 \cap S_2) + P(S_1 \cap G_2)\) Since the selections from the two boxes are independent events: \(P(A) = P(G_1) \cdot P(S_2) + P(S_1) \cdot P(G_2)\) Let's calculate the individual probabilities: Probability of selecting a gold coin from the first box: \(P(G_1) = \frac{2}{6} = \frac{1}{3}\) Probability of selecting a silver coin from the first box: \(P(S_1) = \frac{4}{6} = \frac{2}{3}\) Probability of selecting a gold coin from the second box: \(P(G_2) = \frac{3}{8}\) Probability of selecting a silver coin from the second box: \(P(S_2) = \frac{5}{8}\) Now, substitute these probabilities into the formula for P(A): \(P(A) = \left(\frac{1}{3} \times \frac{5}{8}\right) + \left(\frac{2}{3} \times \frac{3}{8}\right)\)
\( = \frac{5}{24} + \frac{6}{24}\)
\( = \frac{5+6}{24}\)
\( = \frac{11}{24}\)In simple words: We have two boxes of coins. We pick one coin from each. We want to find the chance that we end up with one gold and one silver coin. This can happen in two ways: either gold from the first box and silver from the second, OR silver from the first and gold from the second. We add the chances of these two different ways.

🎯 Exam Tip: For "one of each" type problems with multiple sources (like two boxes), remember to consider all possible combinations that satisfy the condition and sum their probabilities, especially if the events are mutually exclusive.

 

Question 8. One joint family has 3 sons and 2 daughters whereas the other joint family has 2 sons and 4 daughters. One joint family is selected from two joint families and a child is randomly selected from that family. Find the probability that the selected child is a girl.

Answer: In the first joint family, there are (3 sons + 2 daughters) = 5 children. In the second joint family, there are (2 sons + 4 daughters) = 6 children. Let \(F_1\) be the event that the first joint family is selected. Let \(F_2\) be the event that the second joint family is selected. Since one family is selected randomly from the two, the probability of selecting each family is: \(P(F_1) = P(F_2) = \frac{1}{2}\) Let G be the event that the selected child is a girl. The event G can occur in two ways: 1. The first family (\(F_1\)) is selected AND a girl is chosen from it. 2. The second family (\(F_2\)) is selected AND a girl is chosen from it. We use the law of total probability to find P(G): \(P(G) = P(G|F_1) \cdot P(F_1) + P(G|F_2) \cdot P(F_2)\) Now, let's calculate the conditional probabilities: Probability of selecting a girl, given the first family is selected (\(P(G|F_1)\)): There are 2 daughters out of 5 children in the first family. \(P(G|F_1) = \frac{2}{5}\) Probability of selecting a girl, given the second family is selected (\(P(G|F_2)\)): There are 4 daughters out of 6 children in the second family. \(P(G|F_2) = \frac{4}{6} = \frac{2}{3}\) Substitute these values into the total probability formula: \(P(G) = \left(\frac{2}{5} \times \frac{1}{2}\right) + \left(\frac{2}{3} \times \frac{1}{2}\right)\)
\( = \frac{1}{5} + \frac{1}{3}\) To sum the fractions, find a common denominator (15):
\( = \frac{3}{15} + \frac{5}{15}\)
\( = \frac{3+5}{15}\)
\( = \frac{8}{15}\)In simple words: We have two families, each with a different number of sons and daughters. We first pick a family by chance, then pick a child from that chosen family. We want to find the overall chance that the child we pick is a girl. We calculate the chance of picking a girl from each family and then combine these chances based on how likely it is to pick each family.

🎯 Exam Tip: This problem is a classic example of the "Law of Total Probability." Remember to break down the main event (selecting a girl) into mutually exclusive paths (girl from family 1, girl from family 2) and sum their probabilities.

 

Question 9. There are 10 ice cream cones in a box of which 3 cones weigh less than the specification and the rest of the 7 cones have the specified weight. Two cones are randomly selected one by one with replacement. Find the probability that both the cones selected weigh less than the specified weight.

Answer: In a box, there are a total of 10 ice cream cones. Number of cones weighing less than the specified weight = 3. Number of cones with specified weight = 7. Two cones are selected one by one with replacement. This means that after the first cone is selected and its weight is checked, it is put back into the box, so the total number of cones remains the same for the second selection. This makes the two selections independent events. Let A be the event that the first cone selected weighs less than the specified weight. Number of favorable outcomes for A = 3. Total number of outcomes = 10. So, the probability of event A is \(P(A) = \frac{3}{10}\). Let B be the event that the second cone selected weighs less than the specified weight. Since the first cone is replaced, the conditions for the second selection are identical to the first. Number of favorable outcomes for B = 3. Total number of outcomes = 10. So, the probability of event B is \(P(B) = \frac{3}{10}\). We want to find the probability that both cones selected weigh less than the specified weight, which is \(P(A \cap B)\). Since events A and B are independent (because of replacement), the probability of both happening is the product of their individual probabilities: \(P(A \cap B) = P(A) \times P(B)\)
\( = \frac{3}{10} \times \frac{3}{10}\)
\( = \frac{9}{100}\)In simple words: We pick two ice cream cones from a box, putting the first one back before picking the second. We want to know the chance that both cones we picked are underweight. Since we put the first cone back, the chance of picking an underweight cone stays the same for both picks, so we just multiply the chances together.

🎯 Exam Tip: The phrase "with replacement" is key for identifying independent events. If selections are independent, the probability of their intersection is simply the product of their individual probabilities.

 

Question 10. There are 10 CDs in a CD rack in which 6 are action film CDs and 4 are drama film CDs. Two CDs are randomly selected one by one without replacement from this box. Find the probability that the first selected CD is of action film and the second CD is of drama film.

Answer: There are a total of (6 action film + 4 drama film) = 10 CDs in the rack. Two CDs are selected one by one without replacement. This means that the first selected CD is not put back, making the second selection dependent on the first. Let A be the event that the first selected CD is an action film CD. Number of action film CDs = 6. Total number of CDs = 10. So, the probability of event A is \(P(A) = \frac{6}{10}\). After the first CD (an action film) is selected and not replaced, the total number of CDs remaining in the rack is \(10 - 1 = 9\). The number of action film CDs remaining is \(6 - 1 = 5\). The number of drama film CDs remaining is 4. Let B be the event that the second selected CD is a drama film CD. This event is conditional on A having occurred. So, we are looking for \(P(B|A)\). Number of drama film CDs remaining = 4. Total number of CDs remaining = 9. So, the conditional probability of B given A is \(P(B|A) = \frac{4}{9}\). We want to find the probability that the first CD is an action film AND the second CD is a drama film, which is \(P(A \cap B)\). Using the multiplication rule for dependent events: \(P(A \cap B) = P(A) \times P(B|A)\)
\( = \frac{6}{10} \times \frac{4}{9}\)
\( = \frac{24}{90}\)
\( = \frac{4}{15}\)In simple words: We pick two CDs from a rack, but we don't put the first one back. We want the chance that the first CD is an action film and the second is a drama film. First, we find the chance of picking an action CD. Then, because we didn't replace it, the total number of CDs and the number of action CDs change for the second pick. We then find the chance of picking a drama CD from the remaining ones and multiply these two chances.

🎯 Exam Tip: The phrase "without replacement" signifies dependent events. In such cases, the probability of the second event is conditional on the outcome of the first event, and the sample space (and sometimes favorable outcomes) changes.

 

Question 11. If two balanced dice are thrown, then find the probability that
(1) at least one die shows number 5
(2) the first die shows the number 5 or 6 and the other die shows an even number.

Answer: When two balanced dice are thrown, the total number of possible outcomes is \(6 \times 6 = 36\). (1) Probability that at least one die shows number 5: Let A be the event that the first die shows a 5. \(P(A) = \frac{1}{6}\) Let B be the event that the second die shows a 5. \(P(B) = \frac{1}{6}\) The event that both dice show a 5 is A \( \cap \) B. Since the dice throws are independent: \(P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) The probability that at least one die shows a 5 is \(P(A \cup B)\). Using the addition rule for probabilities: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\( = \frac{1}{6} + \frac{1}{6} - \frac{1}{36}\) To sum the fractions, find a common denominator (36):
\( = \frac{6}{36} + \frac{6}{36} - \frac{1}{36}\)
\( = \frac{6+6-1}{36}\)
\( = \frac{11}{36}\) (2) Probability that the first die shows 5 or 6 AND the other die shows an even number: Let A be the event that the first die shows a 5 or a 6. The favorable outcomes for A are {5, 6}. So, the number of favorable outcomes is 2. The probability of event A is \(P(A) = \frac{2}{6} = \frac{1}{3}\). Let B be the event that the second die shows an even number. The favorable outcomes for B are {2, 4, 6}. So, the number of favorable outcomes is 3. The probability of event B is \(P(B) = \frac{3}{6} = \frac{1}{2}\). We want to find the probability that the first die shows 5 or 6 AND the second die shows an even number. This is \(P(A \cap B)\). Since the two dice throws are independent events: \(P(A \cap B) = P(A) \times P(B)\)
\( = \frac{1}{3} \times \frac{1}{2}\)
\( = \frac{1}{6}\)In simple words: For two dice, we find two different chances. For the first part, we want at least one '5'. We add the chance of the first die being a '5' and the second die being a '5', then subtract the chance of both being '5' so we don't count it twice. For the second part, we want the first die to be '5' or '6' AND the second die to be an even number. Since these are separate dice rolls, we just multiply their individual chances.

🎯 Exam Tip: Distinguish carefully between "at least one" (which often uses the addition rule) and "both/and" (which uses the multiplication rule). For independent events, the multiplication rule is straightforward.

 

Question 12. A problem in Mathematics is given to Tania, Kathan and Kirti to solve. The probabilities of them solving the problem correctly are \(\frac{2}{3}\), \(\frac{3}{4}\) and \(\frac{1}{2}\) respectively. Find the probability that the problem is solved correctly.

Answer: Let A be the event that Tania solves the problem correctly. \(P(A) = \frac{2}{3}\). The probability that Tania does NOT solve the problem is \(P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}\). Let B be the event that Kathan solves the problem correctly. \(P(B) = \frac{3}{4}\). The probability that Kathan does NOT solve the problem is \(P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}\). Let C be the event that Kirti solves the problem correctly. \(P(C) = \frac{1}{2}\). The probability that Kirti does NOT solve the problem is \(P(C') = 1 - P(C) = 1 - \frac{1}{2} = \frac{1}{2}\). The problem is solved correctly if at least one of them solves it. It's easier to find the probability that the problem is *not* solved at all, and then subtract that from 1. The event that the problem is NOT solved means that Tania does NOT solve it AND Kathan does NOT solve it AND Kirti does NOT solve it. This is \(P(A' \cap B' \cap C')\). Assuming their attempts are independent events: \(P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')\)
\( = \frac{1}{3} \times \frac{1}{4} \times \frac{1}{2}\)
\( = \frac{1}{24}\) The probability that the problem is solved correctly is 1 minus the probability that it is not solved: \(P(\text{problem is solved}) = 1 - P(A' \cap B' \cap C')\)
\( = 1 - \frac{1}{24}\)
\( = \frac{24 - 1}{24}\)
\( = \frac{23}{24}\)In simple words: Three friends try to solve a math problem, and we know how likely each one is to solve it. We want to find the chance that at least one of them solves it. It's simpler to first find the chance that *none* of them solve it, and then subtract that from 1 to get the answer.

🎯 Exam Tip: For "at least one" scenarios involving independent events, it's often more efficient to calculate the probability of the complementary event ("none of them") and subtract it from 1. This avoids complex calculations with unions of multiple events.

 

Question 13. Person A can hit the target in 3 out of 5 attempts whereas person B can hit the target in 5 out of 6 attempts. If both of them attempt simultaneously, find the probability that the target is hit.

Answer: Let A be the event that person A hits the target. The probability of event A is \(P(A) = \frac{3}{5}\). Let B be the event that person B hits the target. The probability of event B is \(P(B) = \frac{5}{6}\). The target is hit if A hits it, or B hits it, or both hit it. This is represented by the union \(P(A \cup B)\). Assuming A and B attempt independently, the probability that both hit the target (A \( \cap \) B) is: \(P(A \cap B) = P(A) \times P(B)\)
\( = \frac{3}{5} \times \frac{5}{6}\)
\( = \frac{15}{30}\)
\( = \frac{1}{2}\) Using the addition rule for probabilities to find \(P(A \cup B)\): \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\( = \frac{3}{5} + \frac{5}{6} - \frac{1}{2}\) To sum the fractions, find a common denominator (30):
\( = \frac{18}{30} + \frac{25}{30} - \frac{15}{30}\)
\( = \frac{18+25-15}{30}\)
\( = \frac{28}{30}\)
\( = \frac{14}{15}\)In simple words: Two people try to hit a target. We know how likely each person is to hit it. We want to find the chance that the target gets hit by at least one of them. We add their individual chances of hitting, then subtract the chance of both hitting to avoid counting that situation twice.

🎯 Exam Tip: This problem also uses the "at least one" concept, for which the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) is generally applied. If events are independent, \(P(A \cap B)\) can be found by multiplying \(P(A)\) and \(P(B)\).

 

Question 14. Person A speaks truth in 90 % cases whereas person B speaks truth in 80 % cases. Find the probability that persons A and B differ in stating the same fact.

Answer: Let A be the event that person A speaks the truth. The probability of event A is \(P(A) = \frac{90}{100} = 0.9\). The probability that person A does NOT speak the truth is \(P(A') = 1 - P(A) = 1 - 0.9 = 0.1\). Let B be the event that person B speaks the truth. The probability of event B is \(P(B) = \frac{80}{100} = 0.8\). The probability that person B does NOT speak the truth is \(P(B') = 1 - P(B) = 1 - 0.8 = 0.2\). Persons A and B differ in stating the same fact if one speaks the truth and the other does not. There are two scenarios for this: 1. A speaks the truth AND B does not speak the truth (event A \( \cap \) B'). 2. A does not speak the truth AND B speaks the truth (event A' \( \cap \) B). Since A and B' are independent events, and A' and B are also independent events: \(P(A \cap B') = P(A) \times P(B') = 0.9 \times 0.2 = 0.18\) \(P(A' \cap B) = P(A') \times P(B) = 0.1 \times 0.8 = 0.08\) The two scenarios (A \( \cap \) B') and (A' \( \cap \) B) are mutually exclusive. Therefore, the total probability that A and B differ is the sum of their individual probabilities: \(P(\text{A and B differ}) = P(A \cap B') + P(A' \cap B)\)
\( = 0.18 + 0.08\)
\( = 0.26\)In simple words: We know how often two people tell the truth. We want to find the chance that when they talk about the same thing, one tells the truth and the other lies. This can happen in two ways: either the first person tells the truth and the second lies, or the first person lies and the second tells the truth. We calculate the chance for each way and add them up.

🎯 Exam Tip: For problems involving two independent events and their complements, remember to define all four probabilities (P(A), P(A'), P(B), P(B')). "Differ" implies one event occurs while the other's complement occurs, which are mutually exclusive scenarios.

 

Question 15. If three events A, B and C of a random experiment are independent events and \(P(A) = 0.2\), \(P(B) = 0.3\) and \(P(C) = 0.5\), then find \(P(A \cup B \cup C)\).

Answer: We are given the following probabilities for independent events A, B, and C: \(P(A) = 0.2\) \(P(B) = 0.3\) \(P(C) = 0.5\) Since A, B, and C are independent, the probabilities of their intersections are the products of their individual probabilities: \(P(A \cap B) = P(A) \times P(B) = 0.2 \times 0.3 = 0.06\) \(P(A \cap C) = P(A) \times P(C) = 0.2 \times 0.5 = 0.10\) \(P(B \cap C) = P(B) \times P(C) = 0.3 \times 0.5 = 0.15\) \(P(A \cap B \cap C) = P(A) \times P(B) \times P(C) = 0.2 \times 0.3 \times 0.5 = 0.03\) To find \(P(A \cup B \cup C)\), we use the Principle of Inclusion-Exclusion for three events: \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\) Substitute the calculated values: \(P(A \cup B \cup C) = 0.2 + 0.3 + 0.5 - 0.06 - 0.10 - 0.15 + 0.03\)
\( = (0.2 + 0.3 + 0.5) - (0.06 + 0.10 + 0.15) + 0.03\)
\( = 1.0 - 0.31 + 0.03\)
\( = 0.72\)In simple words: We have three independent events, and we know the chance of each happening. We want to find the chance that at least one of these three events happens. We use a special formula that adds the individual chances, subtracts the chances of any two happening together, and then adds back the chance of all three happening, to make sure nothing is counted wrong.

🎯 Exam Tip: The Principle of Inclusion-Exclusion for three events is a fundamental formula. Remember its structure carefully, especially the alternating signs, and correctly calculate all pairwise and triple intersections when events are independent.