GSEB Class 11 Maths Solutions Chapter 16 Probability Exercise 16.3

 

Question 1. Which of the following cannot be valid assignments of probability for outcomes of sample space \( S = \{W_1, W_2, W_3, W_4, W_5, W_6, W_7\} \).
Answer:
(a) For this assignment, the sum of probabilities is:
\( 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00 \)
Since the sum equals 1.00 and all probabilities are non-negative, this assignment of probabilities is valid.
(b) For this assignment, the sum of probabilities is:
\( \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{7}{7} = 1 \)
Since the sum equals 1 and all probabilities are non-negative, this assignment of probabilities is valid.
(c) For this assignment, the sum of probabilities is:
\( 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 \)
Since the sum of probabilities is greater than 1, this assignment is not valid.
(d) For this assignment, two probabilities are negative: -0.1 and -0.2.
Probabilities for any event must always be non-negative. Therefore, this assignment is not valid.
(e) For this assignment, the probability of \( \omega_7 \) is \( \frac{15}{14} \).
Since \( \frac{15}{14} \) is greater than 1, this assignment is not valid.
In simple words: A valid probability assignment must have all probabilities between 0 and 1 (inclusive), and their total sum must be exactly 1. Options (c), (d), and (e) break these rules, so they are not valid.

Exam Tip: Remember the two fundamental rules of probability: (1) The probability of any event must be between 0 and 1, inclusive. (2) The sum of probabilities of all possible outcomes in a sample space must equal 1.

 

Question 2. A coin is tossed twice, what is the probability that at least one tail occurs?
Answer: When a coin is tossed twice, the sample space (S) consists of all possible outcomes.
\( S = \{HH, HT, TH, TT\} \)
The total number of outcomes, \( n(S) \), is 4.
Let A be the event of getting at least one tail. This means getting one tail or two tails.
\( A = \{HT, TH, TT\} \)
The number of favorable outcomes for event A, \( n(A) \), is 3.
The probability of event A, \( P(A) \), is calculated as:
\( P(A) = \frac{n(A)}{n(S)} = \frac{3}{4} \)
In simple words: When a coin is flipped two times, there are four possible results. We want results where we get one or more tails. Three of the four results have at least one tail. So, the chance is three out of four.

Exam Tip: Always list the full sample space first, then identify the favorable outcomes for the specific event, and finally apply the probability formula \( P(A) = \frac{n(A)}{n(S)} \).

 

Question 3. A die is thrown. Find the probability of the following events:
(i) A prime number will appear.
(ii) A number greater than or equal to 3 will appear.
(iii) A number less than or equal to 1 will appear.
(iv) A number more than 6 will appear.
(v) A number less than 6 will appear.
Answer: When a die is thrown, the sample space (S) consists of:
\( S = \{1, 2, 3, 4, 5, 6\} \)
The total number of possible outcomes, \( n(S) \), is 6.
(i) Let A be the event that a prime number will appear.
Prime numbers in the sample space are \( \{2, 3, 5\} \).
So, \( n(A) = 3 \).
The probability is \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \).
(ii) Let B be the event that a number greater than or equal to 3 will appear.
Numbers greater than or equal to 3 are \( \{3, 4, 5, 6\} \).
So, \( n(B) = 4 \).
The probability is \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{6} = \frac{2}{3} \).
(iii) Let C be the event that a number less than or equal to 1 will appear.
Numbers less than or equal to 1 are \( \{1\} \).
So, \( n(C) = 1 \).
The probability is \( P(C) = \frac{n(C)}{n(S)} = \frac{1}{6} \).
(iv) Let D be the event that a number more than 6 will appear.
Numbers more than 6 in the sample space are none, so \( D = \{\} \).
So, \( n(D) = 0 \).
The probability is \( P(D) = \frac{n(D)}{n(S)} = \frac{0}{6} = 0 \).
(v) Let E be the event that a number less than 6 will appear.
Numbers less than 6 are \( \{1, 2, 3, 4, 5\} \).
So, \( n(E) = 5 \).
The probability is \( P(E) = \frac{n(E)}{n(S)} = \frac{5}{6} \).
In simple words: First, list all possible results when rolling a die (1 to 6). Then, for each part, count how many of those results fit the description and divide that by the total number of results (which is 6). Remember, if no number fits, the probability is 0.

Exam Tip: Be very careful when interpreting phrases like "at least," "at most," "greater than or equal to," and "less than or equal to" to correctly define your event set.

 

Question 4. A card is selected from a pack of 52 cards.
(a) How many points are there in this sample space?
(b) Calculate the probability that card is an ace of spade,
(c) Calculate the probability that the card is (i) an ace (ii) black card.
Answer: The total number of cards in a pack is 52.
(a) The sample space represents all possible outcomes. Since there are 52 cards, there are 52 points in this sample space.
(b) We need to calculate the probability that the card drawn is an ace of spade.
There is only 1 ace of spade in a standard deck.
Let A be the event of drawing an ace of spade.
So, \( n(A) = 1 \) and \( n(S) = 52 \).
The probability \( P(A) = \frac{n(A)}{n(S)} = \frac{1}{52} \).
(c) We need to calculate the probability that the card is (i) an ace (ii) a black card.
(i) Let B be the event of drawing an ace card.
There are 4 aces in a standard deck (Ace of Hearts, Diamonds, Clubs, Spades).
So, \( n(B) = 4 \) and \( n(S) = 52 \).
The probability \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13} \).
(ii) Let C be the event of drawing a black card.
There are 26 black cards in a standard deck (13 Clubs and 13 Spades).
So, \( n(C) = 26 \) and \( n(S) = 52 \).
The probability \( P(C) = \frac{n(C)}{n(S)} = \frac{26}{52} = \frac{1}{2} \).
In simple words: When picking a card from a deck of 52, the sample space is 52. For specific events, count how many cards fit that event (like 1 ace of spades, 4 aces total, or 26 black cards) and divide by 52 to get the probability.

Exam Tip: Always clearly define your event and sample space. Knowing the composition of a standard deck of cards (52 total, 4 suits, 13 ranks, 2 colors) is essential for probability questions involving cards.

 

Question 5. A fair coin marked 1 on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.
Answer: A fair coin has two possible outcomes: 1 or 6. A fair die has six possible outcomes: 1, 2, 3, 4, 5, or 6.
When both are tossed, the total number of possible outcomes in the sample space is \( 2 \times 6 = 12 \).
The sample space (S) can be listed as:
\( S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \)
(i) Find the probability that the sum of numbers that turn up is 3.
Let A be the event where the sum is 3.
Only one outcome results in a sum of 3: (1,2) (1 from coin, 2 from die).
So, \( n(A) = 1 \).
The probability \( P(A) = \frac{n(A)}{n(S)} = \frac{1}{12} \).
(ii) Find the probability that the sum of numbers that turn up is 12.
Let B be the event where the sum is 12.
Only one outcome results in a sum of 12: (6,6) (6 from coin, 6 from die).
So, \( n(B) = 1 \).
The probability \( P(B) = \frac{n(B)}{n(S)} = \frac{1}{12} \).
In simple words: List all possible pairs when you flip the special coin and roll the die. Then, for each part, count how many pairs add up to the target number and divide by the total number of pairs.

Exam Tip: Always construct the full sample space when dealing with multiple independent events to ensure you capture all possible outcomes accurately. A tree diagram or a table can help organize outcomes.

 

Question 6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer: The total number of members on the city council is the sum of men and women.
Total members = \( 4 \text{ men} + 6 \text{ women} = 10 \text{ members} \).
We are selecting one council member at random.
The number of favorable outcomes for selecting a woman is the number of women on the council.
Number of women = 6.
The probability of selecting a woman is given by:
\( P(\text{Woman}) = \frac{\text{Number of women}}{\text{Total number of members}} = \frac{6}{10} = 0.6 \)
In simple words: To find the chance of picking a woman, divide the number of women by the total number of people available to pick from.

Exam Tip: For simple probability questions, identify the total number of items (sample space) and the number of desired items (favorable outcomes), then express the probability as a fraction or decimal.

 

Question 7. A fair coin is tossed four times and a person wins Rs 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Answer: When a fair coin is tossed four times, the total number of possible outcomes is \( 2^4 = 16 \).
Let H denote a head and T denote a tail.
The possible combinations of heads and tails, and their corresponding money amounts and probabilities, are:
(i) **0 Heads and 4 Tails (TTTT):**
There is only 1 way to get 0 heads and 4 tails: TTTT.
Amount = \( (0 \times \text{Rs } 1) + (4 \times \text{-Rs } 1.50) = 0 - 6.00 = \text{-Rs } 6.00 \). (A loss of Rs 6.00)
Probability \( P(\text{0H, 4T}) = \frac{1}{16} \).
(ii) **1 Head and 3 Tails:**
There are 4 ways to get 1 head and 3 tails: HTTT, THTT, TTHT, TTTH.
Amount = \( (1 \times \text{Rs } 1) + (3 \times \text{-Rs } 1.50) = 1 - 4.50 = \text{-Rs } 3.50 \). (A loss of Rs 3.50)
Probability \( P(\text{1H, 3T}) = \frac{4}{16} = \frac{1}{4} \).
(iii) **2 Heads and 2 Tails:**
There are 6 ways to get 2 heads and 2 tails: HHTT, HTHT, HTTH, THHT, THTH, TTHH.
Amount = \( (2 \times \text{Rs } 1) + (2 \times \text{-Rs } 1.50) = 2 - 3.00 = \text{-Rs } 1.00 \). (A loss of Rs 1.00)
Probability \( P(\text{2H, 2T}) = \frac{6}{16} = \frac{3}{8} \).
(iv) **3 Heads and 1 Tail:**
There are 4 ways to get 3 heads and 1 tail: HHHT, HHTH, HTHH, THHH.
Amount = \( (3 \times \text{Rs } 1) + (1 \times \text{-Rs } 1.50) = 3 - 1.50 = \text{Rs } 1.50 \). (A gain of Rs 1.50)
Probability \( P(\text{3H, 1T}) = \frac{4}{16} = \frac{1}{4} \).
(v) **4 Heads and 0 Tails (HHHH):**
There is only 1 way to get 4 heads and 0 tails: HHHH.
Amount = \( (4 \times \text{Rs } 1) + (0 \times \text{-Rs } 1.50) = 4 - 0 = \text{Rs } 4.00 \). (A gain of Rs 4.00)
Probability \( P(\text{4H, 0T}) = \frac{1}{16} \).
The different amounts of money you can have are: Rs -6.00, Rs -3.50, Rs -1.00, Rs 1.50, and Rs 4.00.
In simple words: When you flip a coin four times, you can get different numbers of heads and tails. For each possible mix (like all tails, one head, etc.), figure out how much money you win or lose based on the rules. Then, count how many ways each mix can happen and divide by the total possible outcomes (which is 16) to get the probability.

Exam Tip: For problems involving repeated trials (like coin tosses), listing all possible outcomes (sample space) systematically or using combinations can help determine the number of ways each event can occur. Carefully calculate the monetary outcome for each scenario.

 

Question 8. Three coins are tossed once. Find the probability of getting.
(i) 3 heads
(ii) exactly 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head
(vi) 3 tails
(vii) exactly 2 tails
(viii) no tail
(ix) at most two tails
Answer: When three coins are tossed once, the sample space (S) is given by:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
The total number of exhaustive cases, \( n(S) \), is 8.
(i) **3 heads:**
The only favorable case is HHH.
So, \( n(A) = 1 \).
Probability \( P(\text{3 heads}) = \frac{1}{8} \).
(ii) **Exactly 2 heads:**
The favorable cases are HHT, HTH, THH.
So, \( n(A) = 3 \).
Probability \( P(\text{exactly 2 heads}) = \frac{3}{8} \).
(iii) **At least 2 heads (meaning 2 or 3 heads):**
The favorable cases are HHT, HTH, THH, HHH.
So, \( n(A) = 4 \).
Probability \( P(\text{at least 2 heads}) = \frac{4}{8} = \frac{1}{2} \).
(iv) **At most 2 heads (meaning not 3 heads):**
This is the complement of getting 3 heads.
Probability \( P(\text{at most 2 heads}) = 1 - P(\text{3 heads}) = 1 - \frac{1}{8} = \frac{7}{8} \).
(v) **No head (meaning all tails):**
The only favorable case is TTT.
So, \( n(A) = 1 \).
Probability \( P(\text{no head}) = \frac{1}{8} \).
(vi) **3 tails:**
The only favorable case is TTT.
So, \( n(A) = 1 \).
Probability \( P(\text{3 tails}) = \frac{1}{8} \).
(vii) **Exactly 2 tails:**
The favorable cases are HTT, THT, TTH.
So, \( n(A) = 3 \).
Probability \( P(\text{exactly 2 tails}) = \frac{3}{8} \).
(viii) **No tail (meaning all heads):**
The only favorable case is HHH.
So, \( n(A) = 1 \).
Probability \( P(\text{no tail}) = \frac{1}{8} \).
(ix) **At most two tails (meaning not 3 tails):**
This is the complement of getting 3 tails.
Probability \( P(\text{at most two tails}) = 1 - P(\text{3 tails}) = 1 - \frac{1}{8} = \frac{7}{8} \).
In simple words: When you flip three coins, there are eight total possibilities. For each event listed, count how many of those eight possibilities match. For example, '3 heads' only happens one way. 'At least 2 heads' means 2 or 3 heads. Then, divide the number of matches by 8 to get the probability.

Exam Tip: Be precise with "at least," "at most," "exactly," "no," and "all." They define different subsets of your sample space. Complementary events (like "at most 2 heads" being "not 3 heads") can simplify calculations.

 

Question 9. If \( \frac{2}{11} \) is the probability of an event, what is the probability of the event 'not A'.
Answer: Let P(A) be the probability of event A. We are given \( P(A) = \frac{2}{11} \).
The probability of the event 'not A' (also denoted as \( P(A') \) or \( P(A^c) \)) is found using the complement rule:
\( P(\text{not } A) = 1 - P(A) \)
Substituting the given value:
\( P(\text{not } A) = 1 - \frac{2}{11} \)
To subtract, find a common denominator:
\( P(\text{not } A) = \frac{11}{11} - \frac{2}{11} = \frac{11 - 2}{11} = \frac{9}{11} \)
In simple words: If the chance of something happening is 2 out of 11, then the chance of it *not* happening is simply 1 minus that chance. You subtract the given probability from 1.

Exam Tip: Always remember the complement rule: \( P(A') = 1 - P(A) \). This is a quick way to find the probability of an event not occurring if you know the probability of it occurring.

 

Question 10. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that the letter is
(i) a vowel
(ii) a consonant.
Answer: First, let's count the total number of letters in the word "ASSASSINATION".
Total letters = 13. So, \( n(S) = 13 \).
Next, let's identify the vowels and consonants in the word.
Vowels: A, A, A, I, I, O (6 vowels)
Consonants: S, S, S, S, N, N, T (7 consonants)
(i) Probability that the chosen letter is a vowel.
Number of vowels = 6. Let E be the event of choosing a vowel. So, \( n(E) = 6 \).
\( P(\text{vowel}) = \frac{\text{Number of vowels}}{\text{Total number of letters}} = \frac{6}{13} \)
(ii) Probability that the chosen letter is a consonant.
Number of consonants = 7. Let F be the event of choosing a consonant. So, \( n(F) = 7 \).
\( P(\text{consonant}) = \frac{\text{Number of consonants}}{\text{Total number of letters}} = \frac{7}{13} \)
In simple words: First, count all the letters in the word. Then, count how many of them are vowels and how many are consonants. The probability for each is that count divided by the total number of letters.

Exam Tip: For problems involving letters in a word, ensure you correctly identify and count all unique letters, vowels, and consonants, including repetitions, as each instance is a distinct outcome.

 

Question 11. A person chooses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability, of winning the prize in the game? [Hint : Order of the number is not important.]
Answer: There are 20 natural numbers available to choose from (1 to 20).
The person chooses 6 different numbers, and the order of selection does not matter. This is a combination problem.
The total number of ways to choose 6 numbers from 20 is given by \( ^{20}C_6 \).
\[ ^{20}C_6 = \frac{20!}{6!(20-6)!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \]
\[ ^{20}C_6 = 20 \times 19 \times 3 \times 17 \times 2 = 38760 \]
So, the total number of possible outcomes, \( n(S) \), is 38760.
There is only 1 favorable case: the chosen six numbers perfectly match the six numbers fixed by the lottery committee.
So, \( n(A) = 1 \).
The probability of winning the lottery is:
\[ P(\text{Winning}) = \frac{n(A)}{n(S)} = \frac{1}{38760} \]
In simple words: To win, you must pick exactly the right six numbers out of twenty, and the order doesn't matter. First, calculate all the different ways you can pick six numbers. Since only one set of numbers will win, the chance of winning is 1 divided by that large total number of ways.

Exam Tip: Recognize when a problem involves combinations (order doesn't matter) versus permutations (order matters). The phrase "order is not important" is a strong hint for using combinations. The formula for combinations is \( ^nC_r = \frac{n!}{r!(n-r)!} \).

 

Question 12. Check whether the following probabilities P(A) and P(B) are consistent by definition:
(i) P(A) = 0.5, P(B) = 0.7, P(A \( \cap \) B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A \( \cup \) B) = 0.8
Answer: For probabilities to be consistent, they must satisfy basic probability axioms. One important rule is that the probability of an intersection of two events cannot be greater than the probability of either individual event. Also, the union formula must hold true.
(i) Given: \( P(A) = 0.5 \), \( P(B) = 0.7 \), \( P(A \cap B) = 0.6 \)
For consistency, \( P(A \cap B) \) must be less than or equal to both \( P(A) \) and \( P(B) \).
Here, \( P(A \cap B) = 0.6 \).
We observe that \( 0.6 > P(A) = 0.5 \).
Since the probability of the intersection is greater than the probability of event A, the given probabilities are not consistent.
(ii) Given: \( P(A) = 0.5 \), \( P(B) = 0.4 \), \( P(A \cup B) = 0.8 \)
We use the formula for the probability of the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Let's find \( P(A \cap B) \) using the given values:
\( 0.8 = 0.5 + 0.4 - P(A \cap B) \)
\( 0.8 = 0.9 - P(A \cap B) \)
\( P(A \cap B) = 0.9 - 0.8 \)
\( P(A \cap B) = 0.1 \)
Now we check for consistency:
1. All probabilities \( P(A), P(B), P(A \cap B), P(A \cup B) \) are between 0 and 1. (True: 0.5, 0.4, 0.1, 0.8)
2. \( P(A \cap B) \le P(A) \): \( 0.1 \le 0.5 \) (True)
3. \( P(A \cap B) \le P(B) \): \( 0.1 \le 0.4 \) (True)
Since all conditions are met, the given probabilities are consistently defined.
In simple words: To check if probabilities make sense together, first make sure the "AND" probability isn't bigger than either separate event. Then, use the main probability formula to see if the "OR" and "AND" values match up with the individual event probabilities. If they do, they are consistent.

Exam Tip: The two key consistency checks are: (1) \( P(A \cap B) \le \min(P(A), P(B)) \) and (2) \( P(A \cup B) \ge \max(P(A), P(B)) \). The formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is fundamental for solving such problems.

 

Question 13. Fill in the blanks in the following table:

Answer: We use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) to fill in the blanks.
(i) Given: \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{5} \), \( P(A \cap B) = \frac{1}{15} \). Find \( P(A \cup B) \).
\[ P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} \]
To sum these fractions, find a common denominator, which is 15.
\[ P(A \cup B) = \frac{5}{15} + \frac{3}{15} - \frac{1}{15} = \frac{5+3-1}{15} = \frac{7}{15} \]
So, \( P(A \cup B) = \frac{7}{15} \).
(ii) Given: \( P(A) = 0.35 \), \( P(A \cap B) = 0.25 \), \( P(A \cup B) = 0.6 \). Find \( P(B) \).
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the known values into the equation:
\[ 0.6 = 0.35 + P(B) - 0.25 \]
\[ 0.6 = (0.35 - 0.25) + P(B) \]
\[ 0.6 = 0.10 + P(B) \]
\[ P(B) = 0.6 - 0.10 \]
\[ P(B) = 0.5 \]
So, \( P(B) = 0.5 \).
(iii) Given: \( P(A) = 0.5 \), \( P(B) = 0.35 \), \( P(A \cup B) = 0.7 \). Find \( P(A \cap B) \).
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the known values into the equation:
\[ 0.7 = 0.5 + 0.35 - P(A \cap B) \]
\[ 0.7 = 0.85 - P(A \cap B) \]
\[ P(A \cap B) = 0.85 - 0.7 \]
\[ P(A \cap B) = 0.15 \]
So, \( P(A \cap B) = 0.15 \).
In simple words: Use the main formula for combined probability, \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \). For each row in the table, put the numbers you know into this formula and then solve for the missing one.

Exam Tip: This question tests your ability to apply the addition rule of probability. Make sure you can rearrange the formula to solve for any of the four variables: \( P(A \cup B) \), \( P(A) \), \( P(B) \), or \( P(A \cap B) \).

 

Question 14. Given \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{1}{5} \), find \( P(A \text{ or } B) \), if A and B are mutually exclusive events.
Answer: We are given the probabilities of two events, A and B: \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{1}{5} \).
The events A and B are stated to be mutually exclusive.
For mutually exclusive events, the probability of either A or B occurring (A union B) is simply the sum of their individual probabilities because they cannot happen at the same time.
The formula for mutually exclusive events is:
\( P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) \)
Substitute the given values:
\[ P(A \cup B) = \frac{3}{5} + \frac{1}{5} \]
\[ P(A \cup B) = \frac{3+1}{5} \]
\[ P(A \cup B) = \frac{4}{5} \]
In simple words: When two things can't happen together (mutually exclusive), the chance of one OR the other happening is just the sum of their individual chances.

Exam Tip: Remember the special rule for mutually exclusive events: \( P(A \cap B) = 0 \). This simplifies the general addition rule to \( P(A \cup B) = P(A) + P(B) \). Always check if events are mutually exclusive first.

 

Question 15. If E and F are events such that \( P(E) = \frac{1}{4} \), \( P(F) = \frac{1}{2} \) and \( P(E \text{ and } F) = \frac{1}{8} \). Find
(i) \( P(E \text{ or } F) \)
(ii) \( P(\text{not } E \text{ and not } F) \).
Answer: We are given the following probabilities for events E and F:
\( P(E) = \frac{1}{4} \)
\( P(F) = \frac{1}{2} \)
\( P(E \text{ and } F) = P(E \cap F) = \frac{1}{8} \)
(i) Find \( P(E \text{ or } F) \), which is \( P(E \cup F) \).
Using the general addition rule for probabilities:
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
Substitute the given values:
\[ P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} \]
To add and subtract these fractions, find a common denominator, which is 8.
\[ P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8} \]
So, \( P(E \text{ or } F) = \frac{5}{8} \).
(ii) Find \( P(\text{not } E \text{ and not } F) \), which is \( P(E' \cap F') \).
Using De Morgan's Law, the event "not E and not F" is equivalent to "not (E or F)".
So, \( P(E' \cap F') = P((E \cup F)') \).
Using the complement rule, \( P((E \cup F)') = 1 - P(E \cup F) \).
We found \( P(E \cup F) = \frac{5}{8} \) in part (i).
\[ P(E' \cap F') = 1 - \frac{5}{8} \]
\[ P(E' \cap F') = \frac{8}{8} - \frac{5}{8} = \frac{3}{8} \]
So, \( P(\text{not } E \text{ and not } F) = \frac{3}{8} \).
In simple words: For 'E or F', add their individual chances and subtract the chance of both happening. For 'neither E nor F', calculate the chance of 'E or F' happening, and then subtract that from 1.

Exam Tip: This question effectively combines the addition rule of probability with De Morgan's Law and the complement rule. Mastering these three concepts is crucial for solving many probability problems.

 

Question 16. Events E and F are such that P (not E or not F) = 0.25. State whether E and F are mutually exclusive.
Answer: We are given \( P(\text{not } E \text{ or not } F) = 0.25 \).
This can be written as \( P(E' \cup F') = 0.25 \).
Using De Morgan's Law, \( P(E' \cup F') \) is equivalent to \( P((E \cap F)') \).
So, \( P((E \cap F)') = 0.25 \).
The complement rule states that \( P(A') = 1 - P(A) \). Applying this:
\( 1 - P(E \cap F) = 0.25 \)
Now, we solve for \( P(E \cap F) \):
\( P(E \cap F) = 1 - 0.25 \)
\( P(E \cap F) = 0.75 \)
For events E and F to be mutually exclusive, their intersection must be an empty set, meaning \( P(E \cap F) \) must be 0.
Since \( P(E \cap F) = 0.75 \), which is not equal to 0, events E and F are not mutually exclusive. They have a common outcome.
In simple words: The chance that neither E nor F happens is given. We use a rule to turn this into the chance of E AND F happening. If that chance is not zero, then E and F are not mutually exclusive, meaning they can happen at the same time.

Exam Tip: The definition of mutually exclusive events is \( P(A \cap B) = 0 \). If you can calculate \( P(A \cap B) \) and it is not zero, then the events are not mutually exclusive. De Morgan's Law is a powerful tool for manipulating set operations in probability.

 

Question 17. A and B are events such that P(A) = 0.42, P(B) = 0.48 and 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) Determine \( P(\text{not } A) \). This is \( P(A') \).
Using the complement rule:
\( P(A') = 1 - P(A) \)
\( P(A') = 1 - 0.42 = 0.58 \)
(ii) Determine \( P(\text{not } B) \). This is \( P(B') \).
Using the complement rule:
\( P(B') = 1 - P(B) \)
\( P(B') = 1 - 0.48 = 0.52 \)
(iii) Determine \( P(A \text{ or } B) \). This is \( P(A \cup B) \).
Using the general addition rule for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the given values:
\[ P(A \cup B) = 0.42 + 0.48 - 0.16 \]
\[ P(A \cup B) = 0.90 - 0.16 \]
\[ P(A \cup B) = 0.74 \]
In simple words: For 'not A' or 'not B', just subtract their chances from 1. For 'A or B', add their individual chances and then subtract the chance of both happening at the same time.

Exam Tip: Be mindful of the terminology: "not A" means \( A' \), "A and B" means \( A \cap B \), and "A or B" means \( A \cup B \). Applying the correct formulas for each term is essential.

 

Question 18. In class XI of a school, 40% of the students study Mathematics and 30% study Biology and 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer: Let M be the event that a student studies Mathematics.
Let B be the event that a student studies Biology.
We are given the following probabilities:
Probability of studying Mathematics, \( P(M) = 40\% = \frac{40}{100} = 0.4 \).
Probability of studying Biology, \( P(B) = 30\% = \frac{30}{100} = 0.3 \).
Probability of studying both Mathematics and Biology, \( P(M \text{ and } B) = P(M \cap B) = 10\% = \frac{10}{100} = 0.1 \).
We need to find the probability that a student studies Mathematics or Biology, which is \( P(M \cup B) \).
Using the general addition rule for probabilities:
\[ P(M \cup B) = P(M) + P(B) - P(M \cap B) \]
Substitute the given values:
\[ P(M \cup B) = 0.4 + 0.3 - 0.1 \]
\[ P(M \cup B) = 0.7 - 0.1 \]
\[ P(M \cup B) = 0.6 \]
So, the probability that a randomly selected student studies Mathematics or Biology is 0.6.
In simple words: To find the chance that a student takes either Maths OR Biology, add the chance of taking Maths to the chance of taking Biology. Then, subtract the chance of taking BOTH, because those students were counted twice.

Exam Tip: This is a classic application of the addition rule for probabilities. Remember to convert percentages to decimal probabilities for calculations. A Venn diagram can be a helpful visual aid for such problems.

 

Question 19. In an entrance test, that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer: Let A be the event that a student passes the first examination.
Let B be the event that a student passes the second examination.
We are given the following probabilities:
Probability of passing the first examination, \( P(A) = 0.8 \).
Probability of passing the second examination, \( P(B) = 0.7 \).
Probability of passing at least one examination, \( P(A \text{ or } B) = P(A \cup B) = 0.95 \).
We need to find the probability of passing both examinations, which is \( P(A \text{ and } B) = P(A \cap B) \).
Using the general addition rule for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the known values into the equation:
\[ 0.95 = 0.8 + 0.7 - P(A \cap B) \]
\[ 0.95 = 1.5 - P(A \cap B) \]
Now, solve for \( P(A \cap B) \):
\[ P(A \cap B) = 1.5 - 0.95 \]
\[ P(A \cap B) = 0.55 \]
Therefore, the probability that the student will pass both examinations is 0.55.
In simple words: We know the chances of passing the first test, the second test, and at least one test. To find the chance of passing *both* tests, we add the individual test chances, then subtract the chance of passing at least one, which gives us the overlap.

Exam Tip: This problem requires rearranging the addition rule of probability to solve for the intersection. Clearly define your events and carefully plug in the values to avoid calculation errors.

 

Question 20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing Hindi examination?
Answer: Let E denote the event that a student passes the English examination.
Let H denote the event that a student passes the Hindi examination.
We are given:
Probability of passing both English and Hindi: \( P(E \text{ and } H) = P(E \cap H) = 0.5 \).
Probability of passing neither subject: \( P(\text{neither } E \text{ nor } H) = P(E' \cap H') = 0.1 \).
Probability of passing English examination: \( P(E) = 0.75 \).
We need to find the probability of passing the Hindi examination, \( P(H) \).
First, let's use the information about passing neither subject. According to De Morgan's Law:
\( P(E' \cap H') = P((E \cup H)') \)
Using the complement rule, \( P((E \cup H)') = 1 - P(E \cup H) \).
So, \( 0.1 = 1 - P(E \cup H) \).
Solving for \( P(E \cup H) \):
\( P(E \cup H) = 1 - 0.1 = 0.9 \).
Now, use the general addition rule for probabilities:
\[ P(E \cup H) = P(E) + P(H) - P(E \cap H) \]
Substitute the known values:
\[ 0.9 = 0.75 + P(H) - 0.5 \]
Combine the known probabilities on the right side:
\[ 0.9 = (0.75 - 0.5) + P(H) \]
\[ 0.9 = 0.25 + P(H) \]
Now, solve for \( P(H) \):
\[ P(H) = 0.9 - 0.25 \]
\[ P(H) = 0.65 \]
Therefore, the probability that the student will pass the Hindi examination is 0.65.
In simple words: We know the chance of passing both English and Hindi, and the chance of passing neither. We use a rule to find the chance of passing at least one. Then, using the general probability formula, we can figure out the chance of passing Hindi, given the other known values.

Exam Tip: This problem cleverly uses both De Morgan's Law and the addition rule. Always start by translating the given statements into probability notation and then identify which formula or law applies best to find the missing value.

 

Question 21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 23 opted for both NCC and NSS. If one of these students selected at random, find the probability that
(i) The student opted for NCC or NSS,
(ii) The student has opted neither NCC nor NSS,
(iii) The student has opted NSS but not NCC.
Answer: Total number of students in the class, \( n(S) = 60 \).
Let A be the event that a student opted for NCC.
Number of students who opted for NCC, \( n(A) = 30 \).
So, \( P(A) = \frac{30}{60} = 0.5 \).
Let B be the event that a student opted for NSS.
Number of students who opted for NSS, \( n(B) = 32 \).
So, \( P(B) = \frac{32}{60} \).
Number of students who opted for both NCC and NSS, \( n(A \cap B) = 23 \).
So, \( P(A \cap B) = \frac{23}{60} \).
(i) Find the probability that the student opted for NCC or NSS. This is \( P(A \cup B) \).
Using the general addition rule for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the values:
\[ P(A \cup B) = \frac{30}{60} + \frac{32}{60} - \frac{23}{60} \]
\[ P(A \cup B) = \frac{30+32-23}{60} = \frac{62-23}{60} = \frac{39}{60} \]
Simplify the fraction:
\[ P(A \cup B) = \frac{13}{20} \]
(ii) Find the probability that the student has opted neither NCC nor NSS. This is \( P(A' \cap B') \).
Using De Morgan's Law, \( P(A' \cap B') = P((A \cup B)') \).
Using the complement rule, \( P((A \cup B)') = 1 - P(A \cup B) \).
Using the result from part (i):
\[ P(A' \cap B') = 1 - \frac{13}{20} = \frac{20}{20} - \frac{13}{20} = \frac{7}{20} \]
(iii) Find the probability that the student has opted NSS but not NCC. This is \( P(B \cap A') \).
The event "NSS but not NCC" means students who opted for NSS and *did not* opt for NCC.
This can be calculated as \( P(B) - P(A \cap B) \).
\[ P(B \cap A') = \frac{32}{60} - \frac{23}{60} \]
\[ P(B \cap A') = \frac{32-23}{60} = \frac{9}{60} \]
Simplify the fraction:
\[ P(B \cap A') = \frac{3}{20} \]
In simple words: First, find the total number of students and the number of students for each activity.
(i) For "NCC or NSS", add the number of NCC students and NSS students, then subtract the students who chose both (because they were counted twice). Divide this by the total students.
(ii) For "neither", subtract the "NCC or NSS" chance from 1.
(iii) For "NSS but not NCC", take the total NSS students and subtract those who also chose NCC. Divide this by the total students.

Exam Tip: For problems involving 'or', 'neither', and 'only', carefully draw a Venn diagram or use the correct probability formulas (addition rule, complement rule, De Morgan's Law). "NSS but not NCC" is equivalent to \( P(B \text{ only}) = P(B) - P(A \cap B) \).