OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Chapter Test

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Chapter Test

Question 1. In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?
Answer: When rolling two fair dice, the total number of possible outcomes is \( n(S) = 6^2 = 36 \). Let A be the event of obtaining a sum greater than 3 but not exceeding 6. This means the sum can be 4, 5, or 6. The outcomes for these sums are: Sum 4: (1,3), (2,2), (3,1) Sum 5: (1,4), (2,3), (3,2), (4,1) Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) So, the set A is \(\{(1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1), (1,5), (2,4), (3,3), (4,2), (5,1)\}\). The number of favorable outcomes is \( n(A) = 3 + 4 + 5 = 12 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} \).
\( \implies P(A) = \frac{12}{36} \)
\( \implies P(A) = \frac{1}{3} \) Thus, the probability of obtaining a sum greater than 3 but not exceeding 6 is \( \frac{1}{3} \).In simple words: When you roll two dice, there are 36 possible results. We want the sum to be 4, 5, or 6. If you count all the ways to get these sums, there are 12 of them. So, the chance is 12 out of 36, which simplifies to 1 out of 3.

🎯 Exam Tip: Always list all possible outcomes for each sum systematically when dealing with dice problems to avoid missing any combination.

Question 2. Find the probability of obtaining a sum 8 in a single throw of two dice.
Answer: When throwing two dice, the total number of possible outcomes is \( n(S) = 6^2 = 36 \). Let A be the event of obtaining a sum of 8. The outcomes that sum to 8 are: \(\{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}\). The number of favorable outcomes is \( n(A) = 5 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} \).
\( \implies P(A) = \frac{5}{36} \) So, the probability of getting a sum of 8 when throwing two dice is \( \frac{5}{36} \).In simple words: When you roll two dice, there are 36 possible outcomes. To get a sum of 8, there are 5 specific pairs of numbers you can roll. So, the chance of getting a sum of 8 is 5 out of 36.

🎯 Exam Tip: When listing outcomes for a specific sum, start with the smallest possible number on the first die (e.g., 2 for a sum of 8) and increase it systematically to ensure all pairs are included.

Question 3. A bag contains 4 red, 6 white and 5 black balls. 2 balls are drawn at random. Find the probability of getting one red and one white ball.
Answer: Given number of red balls = 4, number of white balls = 6, and number of black balls = 5. The total number of balls in the bag is \( 4 + 6 + 5 = 15 \). When 2 balls are drawn at random from 15, the total number of possible outcomes (sample space) is \( n(S) = ^{15}C_2 \).
\( \implies n(S) = \frac{15 \times 14}{2 \times 1} = 15 \times 7 = 105 \) Let A be the event of getting one red ball and one white ball. To get one red ball out of 4, the number of ways is \( ^4C_1 = 4 \). To get one white ball out of 6, the number of ways is \( ^6C_1 = 6 \). Since we need one red AND one white ball, the number of favorable outcomes is \( n(A) = ^4C_1 \times ^6C_1 \).
\( \implies n(A) = 4 \times 6 = 24 \) The probability of event A is \( P(A) = \frac{n(A)}{n(S)} \).
\( \implies P(A) = \frac{24}{105} \) We can simplify this fraction by dividing both numerator and denominator by 3.
\( \implies P(A) = \frac{24 \div 3}{105 \div 3} = \frac{8}{35} \) Therefore, the probability of drawing one red and one white ball is \( \frac{8}{35} \).In simple words: There are 15 balls in total. We want to pick 2 balls. There are 105 ways to do this. We want one red ball (from 4 red ones) and one white ball (from 6 white ones). This can happen in 24 ways. So, the chance is 24 out of 105, which simplifies to 8 out of 35.

🎯 Exam Tip: Remember to use combinations (nCr) when the order of selection does not matter. Also, "and" usually implies multiplication for probabilities of independent selections.

Question 4. Out of 26 cards numbered from 1 to 26, one card is chosen. Find the probability that it is not divisible by 4.
Answer: The total number of cards is 26, so the total number of outcomes is \( n(S) = 26 \). Let A be the event that the chosen card has a number divisible by 4. The numbers between 1 and 26 that are divisible by 4 are \(\{4, 8, 12, 16, 20, 24\}\). The number of outcomes favorable to event A is \( n(A) = 6 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} \).
\( \implies P(A) = \frac{6}{26} = \frac{3}{13} \) We need to find the probability that the chosen card is NOT divisible by 4. This is the complement of event A, denoted as \( P(\overline{A}) \). Using the complement rule, \( P(\overline{A}) = 1 - P(A) \).
\( \implies P(\overline{A}) = 1 - \frac{3}{13} \) To subtract, we find a common denominator:
\( \implies P(\overline{A}) = \frac{13}{13} - \frac{3}{13} = \frac{10}{13} \) Thus, the probability that the chosen card is not divisible by 4 is \( \frac{10}{13} \).In simple words: We have 26 cards. First, we find how many cards can be divided by 4 without a remainder. There are 6 such cards. The chance of picking one of these is 6 out of 26. The chance of NOT picking one of these is 1 minus that probability, which comes out to 10 out of 13.

🎯 Exam Tip: It is often easier to calculate the probability of an event happening and then subtract it from 1 to find the probability of it not happening, especially when "not" is in the question.

Question 5. If A and B are mutually exclusive events with \( P(B) = \frac{1}{2} \) and \( A \cup B = S \), the sample space. Find P (A).
Answer: We are given that A and B are mutually exclusive events. This means that they cannot happen at the same time, so the probability of their intersection is zero: \( P(A \cap B) = 0 \). We are given \( P(B) = \frac{1}{2} \). We are also given that \( A \cup B = S \), where S is the sample space. This means that the union of A and B covers all possible outcomes. Therefore, the probability of their union is 1: \( P(A \cup B) = P(S) = 1 \). For any two events A and B, the general formula for the probability of their union is: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) Since A and B are mutually exclusive, \( P(A \cap B) = 0 \). The formula simplifies to: \( P(A \cup B) = P(A) + P(B) \) Now, we substitute the given values into the simplified formula:
\( \implies 1 = P(A) + \frac{1}{2} \) To find \( P(A) \), we subtract \( \frac{1}{2} \) from both sides:
\( \implies P(A) = 1 - \frac{1}{2} \)
\( \implies P(A) = \frac{1}{2} \) Thus, the probability of event A is \( \frac{1}{2} \).In simple words: If two events, A and B, cannot happen together and they cover all possibilities, then their probabilities must add up to 1. Since event B has a chance of 1/2, event A must also have a chance of 1/2 for the total to be 1.

🎯 Exam Tip: For mutually exclusive events, if their union forms the entire sample space, then their probabilities sum to 1. This is a key property to remember.

Question 6. A and B are two events, such that P (A) = 0.42, P (B) = 0.48, P (A and B) = 0.16 Determine (i) P (not A), (ii) P (not B), (iii) P (A or B)
Answer: We are given the following probabilities: \( P(A) = 0.42 \) \( P(B) = 0.48 \) \( P(A \text{ and } B) = P(A \cap B) = 0.16 \) (i) To find \( P(\text{not A}) \), we use the complement rule: \( P(\text{not A}) = P(\overline{A}) = 1 - P(A) \)
\( \implies P(\overline{A}) = 1 - 0.42 = 0.58 \) (ii) To find \( P(\text{not B}) \), we use the complement rule: \( P(\text{not B}) = P(\overline{B}) = 1 - P(B) \)
\( \implies P(\overline{B}) = 1 - 0.48 = 0.52 \) (iii) To find \( P(A \text{ or } B) \), we use the general addition rule for probabilities: \( P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = 0.42 + 0.48 - 0.16 \) First, add \( P(A) \) and \( P(B) \):
\( \implies P(A \cup B) = 0.90 - 0.16 \) Then, subtract \( P(A \cap B) \):
\( \implies P(A \cup B) = 0.74 \) Thus, the probabilities are: (i) \( P(\text{not A}) = 0.58 \), (ii) \( P(\text{not B}) = 0.52 \), and (iii) \( P(A \text{ or } B) = 0.74 \).In simple words: We know the chances of A, B, and both A and B happening. To find the chance of "not A" or "not B", we subtract their individual chances from 1. To find the chance of "A or B" happening, we add the chances of A and B, and then subtract the chance of both happening so we don't count it twice.

🎯 Exam Tip: Clearly understand the difference between "and" (intersection, \( \cap \)) and "or" (union, \( \cup \)) in probability questions. The complement rule \( P(\text{not } E) = 1 - P(E) \) is crucial for many problems.