GSEB Class 12 Statistics Solutions Chapter 1 Probability Exercise 1.2

GSEB Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.2

 

Question 1. A balanced coin is tossed three times. Find the probability of the following events:
(1) Getting all three heads
(2) Not getting a single head
(3) Getting at least one head
(4) Getting more than one head
(5) Getting at the most one head
(6) Getting less than two heads
(7) Getting head and tail alternately
(8) Getting more number of heads than tails.
Answer: When a fair coin is flipped three times, the total number of primary outcomes in its sample space is \( n = 2^3 = 8 \).
The sample space (U) includes all possible outcomes: \( U = \{HHH, HHT, HTH, THH, THT, HTT, TTH, TTT\} \)
(1) Let A be the event of getting all three heads.
Then \( A = \{HHH\} \).
The number of favorable outcomes for event A is \( m = 1 \).
Thus, the probability of event A is \( P(A) = \frac{m}{n} = \frac{1}{8} \).
(2) Let B be the event of not getting a single head. This means all three flips result in tails.
Then \( B = \{TTT\} \).
The number of favorable outcomes for event B is \( m = 1 \).
So, the probability of event B is \( P(B) = \frac{m}{n} = \frac{1}{8} \).
(3) Let B' be the event of getting at least one head. This means the outcome is not 'all tails'.
This can be found by taking the complement of event B (not getting a single head).
So, \( P(B') = 1 - P(B) = 1 - \frac{1}{8} = \frac{7}{8} \).
(4) Let C be the event of getting more than one head. This means getting either two heads or three heads.
Then \( C = \{HHH, HHT, HTH, THH\} \).
The number of favorable outcomes for event C is \( m = 4 \).
Hence, the probability of event C is \( P(C) = \frac{m}{n} = \frac{4}{8} = \frac{1}{2} \).
(5) Let D be the event of getting at most one head. This means getting either zero heads or one head.
Then \( D = \{HTT, THT, TTH, TTT\} \).
The number of favorable outcomes for event D is \( m = 4 \).
Therefore, the probability of event D is \( P(D) = \frac{m}{n} = \frac{4}{8} = \frac{1}{2} \).
(6) Let E be the event of getting less than two heads. This means getting zero heads or one head, which is the same as event D.
So, \( E = D \).
Thus, the probability of event E is \( P(E) = P(D) = \frac{1}{2} \).
(7) Let F be the event of getting head and tail alternately.
Then \( F = \{HTH, THT\} \).
The number of favorable outcomes for event F is \( m = 2 \).
Consequently, the probability of event F is \( P(F) = \frac{m}{n} = \frac{2}{8} = \frac{1}{4} \).
(8) Let G be the event of getting more heads than tails.
This includes outcomes with 3 heads (HHH) and 2 heads (HHT, HTH, THH).
Then \( G = \{HHH, HHT, HTH, THH\} \).
The number of favorable outcomes for event G is \( m = 4 \).
Therefore, the probability of event G is \( P(G) = \frac{m}{n} = \frac{4}{8} = \frac{1}{2} \).
In simple words: We list all possible outcomes for three coin flips. Then, for each specific scenario (like all heads or at least one head), we count how many times that scenario happens and divide it by the total number of outcomes to find its chance.

🎯 Exam Tip: Clearly defining the sample space (U) and the favorable outcomes (m) for each event is crucial for scoring. Always simplify the final probability fraction.

 

Question 2. Two balanced dice are thrown simultaneously. Find the probability of the following events:
(1) The sum of numbers on the dice is 6.
(2) The sum of numbers on the dice is not more than 10.
(3) The sum of numbers on the dice is a multiple of 3.
(4) The product of numbers on the dice is 12.
Answer: When two fair dice are rolled at the same time, the sample space for this random experiment is given by:
\( U = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} \)
The total number of primary outcomes in U is \( n = 36 \).
(1) Let A be the event that the sum of the numbers on the dice is 6.
Then \( A = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\} \).
The number of favorable outcomes for event A is \( m = 5 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{5}{36} \).
(2) Let B be the event that the sum of the numbers on the dice is more than 10. This means the sum is 11 or 12.
Then \( B = \{(5, 6), (6, 5), (6, 6)\} \).
The number of favorable outcomes for event B is \( m = 3 \).
So, the probability of event B is \( P(B) = \frac{m}{n} = \frac{3}{36} = \frac{1}{12} \).
Now, let B' be the event that the sum of the numbers on the dice is not more than 10. This is the complement of event B.
Therefore, \( P(B') = 1 - P(B) = 1 - \frac{1}{12} = \frac{11}{12} \).
(3) Let C be the event that the sum of the numbers on the dice is a multiple of 3. This means the sum can be 3, 6, 9, or 12.
Then \( C = \{(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\} \).
The number of favorable outcomes for event C is \( m = 12 \).
Hence, the probability of event C is \( P(C) = \frac{m}{n} = \frac{12}{36} = \frac{1}{3} \).
(4) Let D be the event that the product of the numbers on the dice is 12.
Then \( D = \{(2, 6), (3, 4), (4, 3), (6, 2)\} \).
The number of favorable outcomes for event D is \( m = 4 \).
Therefore, the probability of event D is \( P(D) = \frac{m}{n} = \frac{4}{36} = \frac{1}{9} \).
In simple words: When two dice are rolled, there are 36 possible pairs. For each event (like a sum of 6 or a product of 12), we count the pairs that match the event and divide by 36 to get the probability.

🎯 Exam Tip: Listing the full sample space helps avoid errors when counting favorable outcomes. Remember to use the complement rule ( \( P(A') = 1 - P(A) \) ) for "not more than" or "at least" type questions to simplify calculations.

 

Question 3. One family is randomly selected from the families having two children. Find the probability that
(1) One child is a girl and one child is a boy.
(2) At least one child is a girl among the two children of the selected family.
(Note: Assume that the chance of the child being a boy or girl is same.)
Answer: Let B represent a boy and G represent a girl.
The sample space for families with two children is: \( U = \{BB, BG, GB, GG\} \).
The total number of primary outcomes when selecting a family at random is \( n = 4C_1 = 4 \).
(1) Let A be the event that one child is a girl and one child is a boy.
Then \( A = \{BG, GB\} \).
The number of favorable outcomes for event A is \( m = 2 \).
Thus, the probability of event A is \( P(A) = \frac{m}{n} = \frac{2}{4} = \frac{1}{2} \).
(2) Let B be the event that at least one child is a girl among the two children.
This includes outcomes with one girl and one boy, or two girls.
Then \( B = \{GB, BG, GG\} \).
The number of favorable outcomes for event B is \( m = 3 \).
Therefore, the probability of event B is \( P(B) = \frac{m}{n} = \frac{3}{4} \).
In simple words: For two children, there are four possible combinations of genders (BB, BG, GB, GG). We count how many of these combinations fit the description (like one boy and one girl) to find the probability.

🎯 Exam Tip: Clearly listing all possible outcomes (sample space) is essential for probability problems involving combinations, especially when dealing with concepts like "at least one" or "exactly one".

 

Question 4. One number is selected at random from the first 100 natural numbers. Find the probability that this number is divisible by 7.
Answer: The set of the first 100 natural numbers is \( U = \{1, 2, 3, ..., 100\} \).
One number is chosen randomly from this set.
The total number of primary outcomes is \( n = 100C_1 = 100 \).
Let A be the event that the chosen number is divisible by 7.
The numbers divisible by 7 within the set U are: \( \{7, 14, 21, ..., 91, 98\} \).
To find the count of these numbers, we can divide the last number by 7: \( 98 \div 7 = 14 \).
So, the number of favorable outcomes for event A is \( m = 14 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{14}{100} = \frac{7}{50} \).
In simple words: We pick a number from 1 to 100. To find the chance it's a multiple of 7, we count how many numbers in that range are multiples of 7 (there are 14 of them) and divide that by the total 100 numbers.

🎯 Exam Tip: For divisibility problems, identify the smallest and largest multiples within the given range. Counting them accurately is key, often by dividing the largest multiple by the divisor.

 

Question 5. The sample space for a random experiment of selecting numbers is U = {1, 2, 3, .... 120} and all the outcomes in the sample space are equiprobable. Find the probability that the number selected is:
(1) a multiple of 3
(2) not a multiple of 3
(3) a multiple of 4
(4) not a multiple of 4
(5) a multiple of both 3 and 4
Answer: The given sample space is \( U = \{1, 2, 3, ..., 120\} \).
One number is chosen at random.
The total number of primary outcomes is \( n = 120C_1 = 120 \).
(1) Let A be the event that the chosen number is a multiple of 3.
The multiples of 3 in U are: \( \{3, 6, 9, 12, ..., 117, 120\} \).
The number of favorable outcomes for event A is \( m = \frac{120}{3} = 40 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{40}{120} = \frac{1}{3} \).
(2) Let A' be the event that the chosen number is not a multiple of 3. This is the complement of event A.
Thus, \( P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \).
(3) Let B be the event that the chosen number is a multiple of 4.
The multiples of 4 in U are: \( \{4, 8, 12, 16, ..., 116, 120\} \).
The number of favorable outcomes for event B is \( m = \frac{120}{4} = 30 \).
Hence, the probability of event B is \( P(B) = \frac{m}{n} = \frac{30}{120} = \frac{1}{4} \).
(4) Let B' be the event that the chosen number is not a multiple of 4. This is the complement of event B.
Therefore, \( P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} \).
(5) Let A \( \cap \) B be the event that the chosen number is a multiple of both 3 and 4.
This means the number must be a multiple of the Least Common Multiple (LCM) of 3 and 4.
The LCM of 3 and 4 is 12.
So, A \( \cap \) B consists of numbers that are multiples of 12: \( \{12, 24, 36, 48, ..., 108, 120\} \).
The number of favorable outcomes for event A \( \cap \) B is \( m = \frac{120}{12} = 10 \).
Hence, the probability of event A \( \cap \) B is \( P(A \cap B) = \frac{m}{n} = \frac{10}{120} = \frac{1}{12} \).
In simple words: From numbers 1 to 120, we find the chance of picking a number that follows specific rules, like being a multiple of 3, not a multiple of 4, or a multiple of both 3 and 4 (which means a multiple of 12). We count how many numbers fit each rule and divide by the total of 120 numbers.

🎯 Exam Tip: When finding probabilities for multiples, divide the total range by the multiple to find the count of favorable outcomes. For "not a multiple," use the complement rule. For "multiple of both," find the LCM first.

 

Question 6. Find the probability of getting R in the first place and M in the last place when all the letters of the word RANDOM are arranged in all possible ways.
Answer: The word RANDOM has 6 distinct letters: R, A, N, D, O, M.
The total number of ways to arrange these six letters is:
\( n = {}^6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \).
Let A be the event where R is in the first position and M is in the last position.
This leaves 4 remaining letters (A, N, D, O) to be arranged in the middle 4 positions.
The favorable outcomes for event A are visualized as:

R					M
\( {}^1P_1 \)	\( {}^4P_4 \)				\( {}^1P_1 \)

The number of favorable outcomes for event A is \( m = {}^1P_1 \times {}^4P_4 \times {}^1P_1 = 1! \times 4! \times 1! = 1 \times 24 \times 1 = 24 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{24}{720} = \frac{1}{30} \).
In simple words: For the word RANDOM, there are 720 ways to arrange its letters. If 'R' must be first and 'M' last, we only arrange the middle four letters. This gives 24 specific arrangements, so the chance is 24 out of 720.

🎯 Exam Tip: In permutation problems with fixed positions, first account for the fixed letters, then calculate permutations for the remaining letters in the remaining positions. Use factorials or permutation formulas correctly.

 

Question 7. Find the probability of getting vowels in the first, third and sixth place when all the letters of the word, ORANGE are arranged in nil possible ways.
Answer: The word ORANGE has 6 distinct letters: O, R, A, N, G, E.
The vowels are O, A, E (3 vowels). The consonants are R, N, G (3 consonants).
The total number of ways to arrange these six letters is:
\( n = {}^6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \).
Let A be the event of getting vowels in the first, third, and sixth positions.
This means the 3 vowels (O, A, E) must be arranged in these 3 specific vowel positions.
The remaining 3 consonants (R, N, G) must be arranged in the remaining 3 positions (second, fourth, and fifth).
The favorable outcomes for event A are visualized as:

Vowel	Cons.	Vowel	Cons.	Cons.	Vowel
O, A, E	R, N, G	O, A, E	R, N, G	R, N, G	O, A, E
\( {}^3P_3 \)					\( {}^3P_3 \)

The number of ways to arrange 3 vowels in 3 specified places is \( {}^3P_3 = 3! = 6 \).
The number of ways to arrange 3 consonants in the remaining 3 places is \( {}^3P_3 = 3! = 6 \).
So, the number of favorable outcomes for event A is \( m = {}^3P_3 \times {}^3P_3 = 3! \times 3! = 6 \times 6 = 36 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{36}{720} = \frac{1}{20} \).
In simple words: In the word ORANGE, there are 720 ways to arrange its letters. If vowels must go into the first, third, and sixth spots, we arrange the 3 vowels in those 3 spots and the 3 consonants in the remaining 3 spots. This results in 36 specific arrangements, so the probability is 36 out of 720.

🎯 Exam Tip: When letters of a word are arranged, distinguish between vowels and consonants. If specific positions are assigned to a group of letters, calculate permutations for that group separately, then for the remaining letters, and multiply the results.

 

Question 8. Five members of a family, husband, wife and three children are randomly arranged in a row for a family photograph. Find the probability that the husband and wife are seated next to each other.
Answer: There are five members in the family: Husband (H), Wife (W), and three Children (C1, C2, C3).
The total number of ways to arrange these five members in a row is:
\( n = {}^5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ways.
Let A be the event that the husband and wife are seated next to each other.
To calculate favorable outcomes, treat the husband and wife (HW) as a single unit.
Now, we have 4 units to arrange: (HW), C1, C2, C3.
These 4 units can be arranged in \( {}^4P_4 = 4! = 24 \) ways.
Within the (HW) unit, the husband and wife can swap positions (HW or WH). This can be done in \( {}^2P_2 = 2! = 2 \) ways.
The favorable outcomes for event A are visualized as:

Husband Wife		Child	Child	Child
Seated next to each other
\( {}^2P_2 \)		\( {}^4P_4 \)

So, the number of favorable outcomes for event A is \( m = {}^2P_2 \times {}^4P_4 = 2! \times 4! = 2 \times 24 = 48 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{48}{120} = \frac{2}{5} \).
In simple words: There are 120 ways to arrange five family members. If the husband and wife must sit together, we treat them as one item. This gives fewer items to arrange, but the husband and wife can still swap places within their pair. This results in 48 arrangements where they are together.

🎯 Exam Tip: For problems where items must be together, treat them as a single unit when arranging. Remember to multiply by the permutations of items within that unit.

 

Question 9. Seven speakers A, B, C, D, E, F, G are invited in a programme to deliver speech in random order. Find the probability that speaker B delivers speech immediately after speaker A.
Answer: There are seven speakers: A, B, C, D, E, F, G.
The total number of ways these seven speakers can deliver their speeches in a random order is:
\( n = {}^7P_7 = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \) ways.
Let A be the event that speaker B delivers speech immediately after speaker A (i.e., the pair AB speaks together in that order).
To find the number of favorable outcomes, consider the pair (AB) as a single unit.
Now, we have 6 units to arrange: (AB), C, D, E, F, G.
These 6 units can be arranged in \( {}^6P_6 = 6! = 720 \) ways.
Since B must deliver speech immediately *after* A, the internal order of (AB) is fixed (A then B), so we multiply by \( {}^1P_1 = 1! = 1 \).
The number of favorable outcomes for event A is \( m = {}^6P_6 \times {}^1P_1 = 6! \times 1 = 720 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{720}{5040} = \frac{1}{7} \).
In simple words: There are 5040 ways to order 7 speakers. If speaker B must follow speaker A directly, we treat 'AB' as one unit. This reduces the number of items to arrange, making 720 such arrangements. The chance is 720 out of 5040.

🎯 Exam Tip: When two items must be in a specific order and adjacent, treat them as one combined unit. Calculate permutations for the units, and multiply by 1 (or 1!) for the fixed internal order of the combined unit.

 

Question 10. Find the probability of having 5 Mondays in the month of February of a leap year.
Answer: In a leap year, February has 29 days.
We know that a week has 7 days.
So, 29 days = 4 weeks and 1 extra day ( \( 4 \times 7 = 28 \) days).
In 4 weeks, every day of the week (Monday, Tuesday, etc.) occurs exactly 4 times.
For February to have 5 Mondays, the extra day must be a Monday.
The sample space for this 1 extra day is:
\( U = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\} \).
The total number of primary outcomes (possible days for the extra day) is \( n = 7 \).
Let A be the event that the extra day is a Monday.
Then \( A = \{\text{Monday}\} \).
The number of favorable outcomes for event A is \( m = 1 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{1}{7} \).
In simple words: A leap year February has 29 days, which is 4 full weeks plus one extra day. For there to be 5 Mondays, that extra day must be a Monday. Since the extra day can be any of the 7 days of the week, there's a 1 in 7 chance it's a Monday.

🎯 Exam Tip: For calendar-based probability questions, identify the total number of days, convert to weeks and remaining days. The probability of a specific day occurring an extra time depends on the remaining days and the total days in a week.

 

Question 11. Find the probability of having 53 Fridays in a -year which is not a leap year.
Answer: A year that is not a leap year is a normal year, having 365 days.
We know that a week has 7 days.
So, 365 days = 52 weeks and 1 extra day ( \( 52 \times 7 = 364 \) days).
In 52 full weeks, every day of the week (including Friday) occurs exactly 52 times.
For the year to have 53 Fridays, the extra day must be a Friday.
The sample space for this 1 extra day is:
\( U = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\} \).
The total number of primary outcomes (possible days for the extra day) is \( n = 7 \).
Let A be the event that the extra day is a Friday.
Then \( A = \{\text{Friday}\} \).
The number of favorable outcomes for event A is \( m = 1 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{1}{7} \).
In simple words: A normal year has 365 days, which is 52 full weeks plus one extra day. For there to be 53 Fridays, that extra day must be a Friday. Since the extra day can be any of the 7 days, there's a 1 in 7 chance it's a Friday.

🎯 Exam Tip: Similar to leap year problems, for a normal year, the presence of an "extra" day determines the probability of a specific weekday occurring an additional time. Define the sample space for this single extra day.

 

Question 12. Find the probability of having 5 Tuesdays in the month of August of any year.
Answer: The month of August has 31 days.
We know that a week has 7 days.
So, 31 days = 4 weeks and 3 extra days ( \( 4 \times 7 = 28 \) days).
In 4 full weeks, every day of the week occurs exactly 4 times.
For August to have 5 Tuesdays, one of the 3 extra days must be a Tuesday.
The sample space for these 3 extra days, considering their sequence, is:
\( U = \{(\text{Sunday, Monday, Tuesday}), (\text{Monday, Tuesday, Wednesday}), (\text{Tuesday, Wednesday, Thursday}), (\text{Wednesday, Thursday, Friday}), (\text{Thursday, Friday, Saturday}), (\text{Friday, Saturday, Sunday}), (\text{Saturday, Sunday, Monday})\} \).
The total number of primary outcomes (possible combinations for the 3 extra days) is \( n = 7 \).
Let A be the event that one of these 3 extra days is a Tuesday.
The favorable outcomes for event A (combinations containing Tuesday) are:
\( A = \{(\text{Sunday, Monday, Tuesday}), (\text{Monday, Tuesday, Wednesday}), (\text{Tuesday, Wednesday, Thursday})\} \).
The number of favorable outcomes for event A is \( m = 3 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{3}{7} \).
In simple words: August has 31 days, which is 4 full weeks and 3 extra days. For August to have 5 Tuesdays, one of those 3 extra days must be a Tuesday. Listing the 7 possible groups of 3 consecutive days, we find that 3 of them contain a Tuesday.

🎯 Exam Tip: When dealing with multiple "extra days," list all possible sequences of those days in the sample space. Then, identify which sequences contain the target weekday to count favorable outcomes.

 

Question 13. 4 couples (husband-wife) attend a party. Two persons are randomly selected from these 8 persons. Find the probability that the selected persons are
(1) husband and wife,
(2) one man and one woman,
(3) one man and one woman who are not husband and wife.
Answer: There are 4 couples, meaning \( 4 \times 2 = 8 \) persons in total (4 men and 4 women).
Two persons are randomly selected from these 8 persons.
The total number of primary outcomes (ways to select 2 persons) is \( n = {}^8C_2 = \frac{8 \times 7}{2 \times 1} = 28 \).
(1) Let A be the event that the two selected persons are a husband and wife.
Since there are 4 couples, we can select one specific husband-wife pair in \( {}^4C_1 \) ways.
The number of favorable outcomes for event A is \( m = {}^4C_1 = 4 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{4}{28} = \frac{1}{7} \).
(2) Let B be the event that in the selected two persons, there is one man and one woman.
There are 4 men and 4 women.
To select one man and one woman, we choose 1 man from 4 men and 1 woman from 4 women.
The number of favorable outcomes for event B is \( m = {}^4C_1 \times {}^4C_1 = 4 \times 4 = 16 \).
Hence, the probability of event B is \( P(B) = \frac{m}{n} = \frac{16}{28} = \frac{4}{7} \).
(3) Let C be the event that the selected persons are one man and one woman who are not husband and wife.
We can denote the 4 couples as M1F1, M2F2, M3F3, M4F4.
The total number of ways to select one man and one woman (event B) is 16.
From these 16 combinations, we need to subtract the cases where the selected man and woman are a couple. There are 4 such cases: (M1F1), (M2F2), (M3F3), (M4F4).
So, the number of favorable outcomes for event C is \( m = (\text{total ways to pick one man and one woman}) - (\text{ways to pick a husband-wife pair}) = 16 - 4 = 12 \).
Hence, the probability of event C is \( P(C) = \frac{m}{n} = \frac{12}{28} = \frac{3}{7} \).
Alternatively, using set theory: C = B - A (Event C is Event B excluding Event A).
Since A is a subset of B (all husband-wife pairs are also one man and one woman),
\( P(C) = P(B) - P(A) = \frac{4}{7} - \frac{1}{7} = \frac{3}{7} \).
In simple words: When picking two people from 4 couples (8 people total), there are 28 ways to do this. We then calculate the chance of picking a husband and wife (4 ways), one man and one woman (16 ways), or one man and one woman who are not a couple (12 ways, by removing the 4 couple pairs from the 16 one man-one woman pairs).

🎯 Exam Tip: For selection problems, use combinations (\( {}^nC_r \)). When dealing with "not a pair" scenarios, calculate the total possible combinations and subtract the "pair" combinations.

 

Question 14. 8 workers are employed in a factory and of them are excellent in efficiency where as the rest of them are moderate in efficiency. 2 workers are randomly selected from these 8 workers. Find the probability that
(1) both the workers have excellent efficiency
(2) both the workers have moderate efficiency
(3) one worker is excellent and one worker is moderate in efficiency.
Answer: There are 8 workers in total.
3 workers have excellent efficiency.
The remaining \( 8 - 3 = 5 \) workers have moderate efficiency.
2 workers are randomly selected from these 8 workers.
The total number of primary outcomes (ways to select 2 workers) is \( n = {}^8C_2 = \frac{8 \times 7}{2 \times 1} = 28 \).
(1) Let A be the event that both selected workers have excellent efficiency.
To select 2 excellent workers, we choose 2 from the 3 excellent workers and 0 from the 5 moderate workers.
The number of favorable outcomes for event A is \( m = {}^3C_2 \times {}^5C_0 = 3 \times 1 = 3 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{3}{28} \).
(2) Let B be the event that both selected workers have moderate efficiency.
To select 2 moderate workers, we choose 0 from the 3 excellent workers and 2 from the 5 moderate workers.
The number of favorable outcomes for event B is \( m = {}^3C_0 \times {}^5C_2 = 1 \times 10 = 10 \).
Hence, the probability of event B is \( P(B) = \frac{m}{n} = \frac{10}{28} = \frac{5}{14} \).
(3) Let C be the event that one selected worker is excellent and one worker is moderate in efficiency.
To select one excellent and one moderate worker, we choose 1 from the 3 excellent workers and 1 from the 5 moderate workers.
The number of favorable outcomes for event C is \( m = {}^3C_1 \times {}^5C_1 = 3 \times 5 = 15 \).
Hence, the probability of event C is \( P(C) = \frac{m}{n} = \frac{15}{28} \).
In simple words: Out of 8 workers (3 excellent, 5 moderate), we pick 2. There are 28 ways to do this. We then calculate the chance of picking two excellent workers (3 ways), two moderate workers (10 ways), or one of each (15 ways).

🎯 Exam Tip: When selecting items from distinct categories (e.g., excellent and moderate workers), use the multiplication rule for combinations ( \( {}^aC_x \times {}^bC_y \) ) to find favorable outcomes for mixed selections.

 

Question 15. Two cards are drawn from a well shuffled pace of 52 cards. Find the probability that
(1) both the cards are of different colour
(2) both the cards are face cards
(3) one of the two cards is a king.
Answer: A standard deck has 52 cards.
Two cards are drawn from the deck.
The total number of primary outcomes (ways to draw 2 cards) is \( n = {}^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 1326 \).
(1) Let A be the event that both drawn cards are of different colors.
In a pack of 52 cards, there are 26 black cards and 26 red cards.
To draw two cards of different colors, we must select 1 black card from 26 and 1 red card from 26.
The number of favorable outcomes for event A is \( m = {}^{26}C_1 \times {}^{26}C_1 = 26 \times 26 = 676 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{676}{1326} = \frac{26}{51} \).
(2) Let B be the event that both drawn cards are face cards.
In a pack of 52 cards, there are 12 face cards (3 per suit: Jack, Queen, King).
To draw two face cards, we select 2 from the 12 face cards.
The number of favorable outcomes for event B is \( m = {}^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66 \).
Hence, the probability of event B is \( P(B) = \frac{m}{n} = \frac{66}{1326} = \frac{11}{221} \).
(3) Let C be the event that one of the two drawn cards is a king.
In a pack of 52 cards, there are 4 kings and \( 52 - 4 = 48 \) non-king cards.
To draw one king and one other card, we select 1 king from 4 kings and 1 non-king from 48 non-king cards.
The number of favorable outcomes for event C is \( m = {}^4C_1 \times {}^{48}C_1 = 4 \times 48 = 192 \).
Hence, the probability of event C is \( P(C) = \frac{m}{n} = \frac{192}{1326} = \frac{32}{221} \).
In simple words: When drawing two cards from 52, there are 1326 ways to do it. We find the probability of getting cards of different colors (one red, one black), two face cards, or exactly one king, by counting the specific ways for each and dividing by the total ways.

🎯 Exam Tip: Remember card distribution: 26 red/black, 12 face cards, 4 kings, etc. Clearly define the groups from which cards are selected for combinations.

 

Question 16. 3 bulbs are defective in a box of 10 bulbs. 2 bulbs are randomly selected from this box. These bulbs are fixed in two bulb-holders installed in a room. Find the probability that the room will be lighted after starting the electric supply.
Answer: There are 10 bulbs in total.
3 bulbs are defective.
The number of non-defective bulbs is \( 10 - 3 = 7 \).
2 bulbs are randomly selected from the 10 bulbs.
The total number of primary outcomes (ways to select 2 bulbs) is \( n = {}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45 \).
Let A be the event that the room will be lighted.
The room will be lighted if at least one of the two selected bulbs is non-defective. This can happen in two ways:
1. Both selected bulbs are non-defective. (Choose 2 from 7 non-defective, 0 from 3 defective).
Number of ways = \( {}^7C_2 \times {}^3C_0 = \frac{7 \times 6}{2 \times 1} \times 1 = 21 \times 1 = 21 \).
2. One selected bulb is non-defective and one is defective. (Choose 1 from 7 non-defective, 1 from 3 defective).
Number of ways = \( {}^7C_1 \times {}^3C_1 = 7 \times 3 = 21 \).
The total number of favorable outcomes for event A is \( m = 21 + 21 = 42 \).
Hence, the probability of event A is \( P(A) = \frac{m}{n} = \frac{42}{45} = \frac{14}{15} \).
Alternatively, using the complement rule:
Let A' be the event that the room will NOT be lighted. This means both selected bulbs are defective.
Number of ways to select 2 defective bulbs = \( {}^3C_2 = \frac{3 \times 2}{2 \times 1} = 3 \).
\( P(A') = \frac{3}{45} = \frac{1}{15} \).
Then \( P(A) = 1 - P(A') = 1 - \frac{1}{15} = \frac{14}{15} \).
In simple words: We pick 2 bulbs from a box of 10 (3 bad, 7 good). The room lights up if at least one bulb works. This means either both bulbs are good, or one is good and one is bad. We calculate these ways and add them, then divide by the total ways to pick 2 bulbs.

🎯 Exam Tip: For "at least one" type problems, it's often simpler to calculate the probability of the complementary event ("none") and subtract it from 1. Ensure you correctly identify the number of defective and non-defective items.

 

Question 17. For two events A and B in the sample space of a random experiment, P (A) = 0.6, P(B) = 0.5 and P(A \( \cap \) B) = 0.15. Find
(1) P(A')
(2) P (B – A)
(3) P (A \( \cap \) B')
(4) P(A' \( \cap \) B)
(5) P(A \( \cup \) B)
Answer: We are given the following probabilities: \( P(A) = 0.6 \), \( P(B) = 0.5 \), and \( P(A \cap B) = 0.15 \).
(1) To find the probability of the complement of A, P(A'):
\( P(A') = 1 - P(A) = 1 - 0.6 = 0.4 \).
(2) To find the probability of event (B - A), which means B occurs but A does not:
\( P(B - A) = P(B) - P(A \cap B) = 0.5 - 0.15 = 0.35 \).
(3) To find the probability of event (A \( \cap \) B'), which means A occurs but B does not:
\( P(A \cap B') = P(A) - P(A \cap B) = 0.6 - 0.15 = 0.45 \).
(4) To find the probability of event (A' \( \cap \) B), which means B occurs but A does not (same as B - A):
\( P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.15 = 0.35 \).
(5) To find the probability of the union of A and B, P(A \( \cup \) B), using the addition rule:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.15 = 1.1 - 0.15 = 0.95 \).
Additionally, we can find:
\( P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B) = 1 - 0.95 = 0.05 \).
\( P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B) = 1 - 0.15 = 0.85 \).
In simple words: Given the chances of A, B, and both A and B happening, we use basic probability rules to find other chances. For example, the chance A doesn't happen is 1 minus the chance A does happen. The chance only B happens is the chance of B minus the chance of both. The chance of A or B happening is the sum of their individual chances minus the chance of both.

🎯 Exam Tip: Master the fundamental probability formulas for complements ( \( P(A') = 1 - P(A) \) ), differences ( \( P(A-B) = P(A) - P(A \cap B) \) ), and unions ( \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) ). De Morgan's Laws are useful for \( P(A' \cap B') \) and \( P(A' \cup B') \).

 

Question 18. For two events A and B in the sample space of a random experiment, P (A) = 2P(B') = 3P (A \( \cap \) B) = 0.6. Find the probability of difference events A – B and B – A.
Answer: We are given the relationship: \( P(A) = 2P(B') = 3P(A \cap B) = 0.6 \).
From this, we can determine the individual probabilities:
1. \( P(A) = 0.6 \).
2. \( 2P(B') = 0.6 \implies P(B') = \frac{0.6}{2} = 0.3 \).
Then, \( P(B) = 1 - P(B') = 1 - 0.3 = 0.7 \).
3. \( 3P(A \cap B) = 0.6 \implies P(A \cap B) = \frac{0.6}{3} = 0.2 \).
Now, we need to find the probability of the difference events A - B and B - A.
For event A - B (A occurs, B does not):
\( P(A - B) = P(A \cap B') \)
\( = P(A) - P(A \cap B) \)
\( = 0.6 - 0.2 = 0.4 \).
For event B - A (B occurs, A does not):
\( P(B - A) = P(A' \cap B) \)
\( = P(B) - P(A \cap B) \)
\( = 0.7 - 0.2 = 0.5 \).
In simple words: We are given an equation that links the probabilities of A, not B, and both A and B. From this, we figure out the individual chances for P(A), P(B), and P(A and B). Then, we use these values to find the chance that only A happens (A minus B) and the chance that only B happens (B minus A).

🎯 Exam Tip: When given an equation relating multiple probabilities, first solve for each individual probability term. Then, apply the standard formulas for difference events: \( P(A-B) = P(A) - P(A \cap B) \) and \( P(B-A) = P(B) - P(A \cap B) \).

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