Class 11-NCERT Solutions-Chapter-14-Probability

Answer:

The sample space for rolling a die is \( S = \{1, 2, 3, 4, 5, 6\} \).

• Event E: "die shows 4" \( \Rightarrow E = \{4\} \)
• Event F: "die shows even number" \( \Rightarrow F = \{2, 4, 6\} \)

To check if they are mutually exclusive, we find their intersection:

\( E \cap F = \{4\} \cap \{2, 4, 6\} = \{4\} \)

Since \( E \cap F \ne \phi \), the events E and F are not mutually exclusive.

Answer:

Sample space \( S = \{1, 2, 3, 4, 5, 6\} \).

Event Descriptions:

• (i) \( A = \{1, 2, 3, 4, 5, 6\} \) (All numbers are less than 7)
• (ii) \( B = \phi \) (No number on a die is greater than 7)
• (iii) \( C = \{3, 6\} \)
• (iv) \( D = \{1, 2, 3\} \)
• (v) \( E = \{6\} \) (Only even number > 4 is 6)
• (vi) \( F = \{3, 4, 5, 6\} \) (Not less than 3 means \( \ge 3 \))

Calculations:

• \( A \cup B = \{1, 2, 3, 4, 5, 6\} \cup \phi = A \)
• \( A \cap B = A \cap \phi = \phi \)
• \( B \cup C = \phi \cup \{3, 6\} = \{3, 6\} = C \)
• \( E \cap F = \{6\} \cap \{3, 4, 5, 6\} = \{6\} \)
• \( D \cap E = \{1, 2, 3\} \cap \{6\} = \phi \)
• \( A - C = \{1, 2, 3, 4, 5, 6\} - \{3, 6\} = \{1, 2, 4, 5\} \)
• \( D - E = \{1, 2, 3\} - \{6\} = \{1, 2, 3\} = D \)
• To find \( E \cap F' \):
  First find \( F' = S - F = \{1, 2, 3, 4, 5, 6\} - \{3, 4, 5, 6\} = \{1, 2\} \)
  \( E \cap F' = \{6\} \cap \{1, 2\} = \phi \)
• \( F' = \{1, 2\} \)

Answer:

Sample Space: 36 pairs from (1,1) to (6,6).

Describing Sets:

• Event A (Sum > 8): Sums can be 9, 10, 11, 12.
  \( A = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\} \)
• Event B (2 on either die):
  \( B = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2)\} \)
• Event C (Sum \(\ge\) 7 and multiple of 3): Possible sums are 9 and 12.
  \( C = \{(3,6), (4,5), (5,4), (6,3), (6,6)\} \)

Checking for Mutually Exclusive Pairs:

• A and B: The maximum sum involving a 2 is \( 2+6=8 \) or \( 6+2=8 \). Event A requires sum > 8. Thus, \( A \cap B = \phi \).
  (A and B are mutually exclusive)
• B and C: Event C consists of sums 9 and 12. As established, B cannot have a sum greater than 8. Thus, \( B \cap C = \phi \).
  (B and C are mutually exclusive)
• A and C: Comparing the sets, every element of C is also in A (Sum 9 and 12 are > 8).
  \( A \cap C = C \ne \phi \).
  (A and C are not mutually exclusive)

Answer:

Sample Space \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \)

Events:

• \( A = \{HHH\} \)
• \( B = \{HHT, HTH, THH\} \)
• \( C = \{TTT\} \)
• \( D = \{HHH, HHT, HTH, HTT\} \) (First is H)

(i) Mutually Exclusive:

• \( A \cap B = \phi \)
• \( A \cap C = \phi \)
• \( B \cap C = \phi \)
• \( C \cap D = \phi \) (D has Heads, C is all Tails)
• Note: \( A \subset D \) and \( B \) overlaps with \( D \), so pairs with D (except C) are not exclusive.
• Pairs: (A, B), (A, C), (B, C), (C, D) are mutually exclusive.

(ii) Simple Events:

Events with only one sample point.

• A and C are simple events.

(iii) Compound Events:

Events with more than one sample point.

• B and D are compound events.

Answer: (Answers may vary; these are examples based on definitions)

Sample Space \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \)

1. Getting 3 heads (\( \{HHH\} \)) and Getting 3 tails (\( \{TTT\} \)).
2. Getting 0 heads, Getting exactly 1 head, and Getting at least 2 heads.
   (Union is S, Intersections are empty).
3. Getting at least 2 heads and Getting a head on the first coin.
   (Overlap at HHH, HHT, HTH).
4. Getting 3 heads and Getting 3 tails.
   (Intersection is \( \phi \), but Union is not S).
5. Getting exactly 1 head, Getting exactly 2 heads, and Getting 3 heads.
   (Exclusive, but misses "0 heads" so not exhaustive).

Answer:

Sets Definition:

• \( A = \{ (2,1)..(2,6), (4,1)..(4,6), (6,1)..(6,6) \} \)
• \( B = \{ (1,1)..(1,6), (3,1)..(3,6), (5,1)..(5,6) \} \)
• \( C = \{ (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1) \} \) (Sum \(\le 5\))

Descriptions:

1. \( A' \): Not getting even on first = Getting odd on first.
   \( A' = B \)
2. not B: \( B' = A \)
3. A or B (\( A \cup B \)): Even on first OR Odd on first. This covers all possibilities.
   \( A \cup B = S \)
4. A and B (\( A \cap B \)): Even on first AND Odd on first. Impossible.
   \( A \cap B = \phi \)
5. A but not C (\( A - C \)): Even on first die, excluding those with sum \(\le 5\).
   Excluded from A: \((2,1), (2,2), (2,3), (4,1)\).
   Result: \( \{ (2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \} \)
6. B or C (\( B \cup C \)): Odd on first OR Sum \(\le 5\).
   \( B \cup C = B \cup \{ (2,1), (2,2), (2,3), (4,1) \} \) (Adds even-first outcomes with sum \(\le 5\)).
7. B and C (\( B \cap C \)): Odd on first AND Sum \(\le 5\).
   \( \{ (1,1), (1,2), (1,3), (1,4), (3,1), (3,2) \} \)
8. \( A \cap B' \cap C' \):
   Since \( B' = A \), this is \( A \cap A \cap C' = A \cap C' = A - C \).
   (Same as result in v).

Answer:

1. True. \( A \cap B = \phi \) (A number cannot be both even and odd).
2. True. \( A \cap B = \phi \) and \( A \cup B = S \) (Every number is either even or odd).
3. True. \( B' \) (not odd) is Even (\( A \)).
4. False. \( A \cap C \) contains elements like \( (2,1), (2,2) \) etc. They overlap.
5. False. \( B' = A \). So \( A \cap B' = A \cap A = A \ne \phi \).
6. False.
   Sets are \( A'(=B) \), \( B'(=A) \), and \( C \).
   While they are exhaustive (\( A \cup B \cup C = S \)), they are not mutually exclusive. For example, \( B \cap C \ne \phi \) (Intersection of Odd first and Sum \(\le 5\)).

Answer:

For a valid probability assignment, two conditions must be met:

1. The probability of each outcome must be between 0 and 1 (inclusive): \( 0 \le P(\omega_i) \le 1 \).
2. The sum of probabilities of all outcomes must be exactly 1: \( \sum P(\omega_i) = 1 \).

Analysis:

• (a) Valid. All values are positive and \( 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0 \).
• (b) Valid. All values are \( 1/7 \) (positive) and sum is \( 7 \times (1/7) = 1 \).
• (c) Invalid. The sum is \( 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 \), which is greater than 1.
• (d) Invalid. Probabilities cannot be negative (\( -0.1 \) and \( -0.2 \)).
• (e) Invalid. The sum is \( (1+2+3+4+5+6+15)/14 = 36/14 > 1 \).

Conclusion: (c), (d), and (e) are not valid assignments.

Answer:

The sample space for tossing a coin twice is \( S = \{HH, HT, TH, TT\} \). Total outcomes = 4.

Let E be the event "at least one tail occurs".

\( E = \{HT, TH, TT\} \)

Number of favourable outcomes = 3.

\( P(E) = \frac{3}{4} \)

Answer:

Sample Space \( S = \{1, 2, 3, 4, 5, 6\} \). Total outcomes \( n(S) = 6 \).

1. Prime number: Favourable outcomes are \( \{2, 3, 5\} \). (3 outcomes).
   \( P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \).
2. Number \(\ge 3\): Favourable outcomes are \( \{3, 4, 5, 6\} \). (4 outcomes).
   \( P(\ge 3) = \frac{4}{6} = \frac{2}{3} \).
3. Number \(\le 1\): Favourable outcome is \( \{1\} \). (1 outcome).
   \( P(\le 1) = \frac{1}{6} \).
4. Number \(> 6\): Favourable outcomes \( \phi \). (0 outcomes).
   \( P(> 6) = \frac{0}{6} = 0 \).
5. Number \(< 6\): Favourable outcomes are \( \{1, 2, 3, 4, 5\} \). (5 outcomes).
   \( P(< 6) = \frac{5}{6} \).

Answer:

1. Total cards = 52. So, there are 52 points in the sample space.
2. There is only 1 Ace of Spades.
   \( P(\text{Ace of Spades}) = \frac{1}{52} \).
3. (i) An Ace: There are 4 aces (Spades, Clubs, Hearts, Diamonds).
   \( P(\text{Ace}) = \frac{4}{52} = \frac{1}{13} \).
   (ii) Black card: There are 26 black cards (13 Spades + 13 Clubs).
   \( P(\text{Black Card}) = \frac{26}{52} = \frac{1}{2} \).

Answer:

Outcomes for Coin: \( C = \{1, 6\} \). Outcomes for Die: \( D = \{1, 2, 3, 4, 5, 6\} \).

Total Sample Space consists of pairs \( (c, d) \):

\( S = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \} \)

Total outcomes = \( 2 \times 6 = 12 \).

1. Sum is 3:
   Possible pairs: \( (1, 2) \) (Since \( 1+2=3 \)). Note: \( (6, d) \) will always be \( > 3 \).
   Favourable outcomes = 1.
   \( P(\text{Sum } 3) = \frac{1}{12} \).
2. Sum is 12:
   Possible pairs: \( (6, 6) \) (Since \( 6+6=12 \)).
   Favourable outcomes = 1.
   \( P(\text{Sum } 12) = \frac{1}{12} \).

Answer:

Total members = \( 4 \text{ (men)} + 6 \text{ (women)} = 10 \).

Event E: Selecting a woman.

Number of women = 6.

\( P(E) = \frac{6}{10} = \frac{3}{5} \).

Answer:

Total tosses = 4. Sample Space size \( n(S) = 2^4 = 16 \).

Let's calculate the amount for each outcome (Win +1, Lose -1.50):

• 4 Heads (HHHH): \( 4 \times 1 = \textbf{Rs 4.00} \). (1 outcome)
• 3 Heads, 1 Tail: \( 3(1) - 1(1.50) = 3 - 1.50 = \textbf{Rs 1.50} \). (4 outcomes: HHHT, HHTH, HTHH, THHH)
• 2 Heads, 2 Tails: \( 2(1) - 2(1.50) = 2 - 3.00 = \textbf{Rs -1.00} \) (Loss of Re 1). (6 outcomes)
• 1 Head, 3 Tails: \( 1(1) - 3(1.50) = 1 - 4.50 = \textbf{Rs -3.50} \) (Loss of Rs 3.50). (4 outcomes)
• 4 Tails (TTTT): \( -4(1.50) = \textbf{Rs -6.00} \) (Loss of Rs 6). (1 outcome)

Probabilities:

• \( P(\text{Win Rs } 4) = \frac{1}{16} \)
• \( P(\text{Win Rs } 1.50) = \frac{4}{16} = \frac{1}{4} \)
• \( P(\text{Lose Re } 1) = \frac{6}{16} = \frac{3}{8} \)
• \( P(\text{Lose Rs } 3.50) = \frac{4}{16} = \frac{1}{4} \)
• \( P(\text{Lose Rs } 6) = \frac{1}{16} \)

Answer:

Sample Space \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \). Total = 8.

1. 3 heads: \( \{HHH\} \). \( P = 1/8 \).
2. 2 heads: \( \{HHT, HTH, THH\} \). \( P = 3/8 \).
3. At least 2 heads (2 or 3): \( \{HHT, HTH, THH, HHH\} \). \( P = 4/8 = 1/2 \).
4. At most 2 heads (0, 1, or 2): All except HHH. \( P = 7/8 \).
5. No head (All tails): \( \{TTT\} \). \( P = 1/8 \).
6. 3 tails: \( \{TTT\} \). \( P = 1/8 \).
7. Exactly 2 tails: \( \{HTT, THT, TTH\} \). \( P = 3/8 \).
8. No tail (All heads): \( \{HHH\} \). \( P = 1/8 \).
9. At most 2 tails (0, 1, or 2): All except TTT. \( P = 7/8 \).

Answer:

Given \( P(A) = \frac{2}{11} \).

We know that \( P(\text{not } A) = 1 - P(A) \).

\( P(\text{not } A) = 1 - \frac{2}{11} = \frac{11 - 2}{11} = \frac{9}{11} \).

Answer:

Word: ASSASSINATION. Total letters = 13.

Count of letters:

• A: 3
• S: 4
• I: 2
• N: 2
• T: 1
• O: 1

(i) A Vowel:

Vowels are A, I, O. Total vowels = \( 3(A) + 2(I) + 1(O) = 6 \).

\( P(\text{Vowel}) = \frac{6}{13} \).

(ii) A Consonant:

Consonants are S, N, T. Total consonants = \( 4(S) + 2(N) + 1(T) = 7 \).

\( P(\text{Consonant}) = \frac{7}{13} \).

Answer:

Total number of ways to choose 6 numbers out of 20 is given by combinations \( ^{20}C_6 \).

Total outcomes = \( \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38760 \)

There is only 1 winning combination.

Probability of winning = \( \frac{\text{Favourable Outcomes}}{\text{Total Outcomes}} = \frac{1}{38760} \)

Answer:

1. Not Consistently Defined.
   The intersection of two sets must be a subset of each set. Therefore, \( P(A \cap B) \) must be less than or equal to both \( P(A) \) and \( P(B) \).
   Here, \( P(A \cap B) = 0.6 \), which is greater than \( P(A) = 0.5 \). This is impossible.
2. Consistently Defined.
   We check the formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
   \( 0.8 = 0.5 + 0.4 - P(A \cap B) \)
   \( 0.8 = 0.9 - P(A \cap B) \)
   \( P(A \cap B) = 0.1 \)
   Since \( 0 \le 0.1 \le P(A) \) and \( P(B) \), the values are consistent.

Answer:

Formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

1. \( P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} = \frac{5+3-1}{15} = \frac{7}{15} \)
2. \( 0.6 = 0.35 + P(B) - 0.25 \)
   \( 0.6 = 0.10 + P(B) \Rightarrow P(B) = 0.5 \)
3. \( 0.7 = 0.5 + 0.35 - P(A \cap B) \)
   \( 0.7 = 0.85 - P(A \cap B) \Rightarrow P(A \cap B) = 0.15 \)

Answer:

If A and B are mutually exclusive, then \( P(A \cap B) = 0 \).

\( P(A \cup B) = P(A) + P(B) \)

\( P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} \)

Answer:

1. \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)
   \( = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8} \)
2. \( P(\text{not } E \text{ and not } F) = P(E' \cap F') \)
   By De Morgan's Law: \( P(E' \cap F') = P((E \cup F)') \)
   \( = 1 - P(E \cup F) = 1 - \frac{5}{8} = \frac{3}{8} \)

Answer:

Given \( P(E' \cup F') = 0.25 \).

By De Morgan's Law: \( E' \cup F' = (E \cap F)' \).

So, \( P((E \cap F)') = 0.25 \).

\( 1 - P(E \cap F) = 0.25 \Rightarrow P(E \cap F) = 0.75 \).

Since \( P(E \cap F) \ne 0 \), the events E and F are not mutually exclusive.

Answer:

1. \( P(\text{not } A) = 1 - P(A) = 1 - 0.42 = 0.58 \)
2. \( P(\text{not } B) = 1 - P(B) = 1 - 0.48 = 0.52 \)
3. \( P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
   \( = 0.42 + 0.48 - 0.16 = 0.90 - 0.16 = 0.74 \)

Answer:

Let M be the event "studying Mathematics" and B be "studying Biology".

• \( P(M) = 40\% = 0.40 \)
• \( P(B) = 30\% = 0.30 \)
• \( P(M \cap B) = 10\% = 0.10 \)

We need \( P(M \text{ or } B) = P(M \cup B) \).

\( P(M \cup B) = P(M) + P(B) - P(M \cap B) \)

\( = 0.40 + 0.30 - 0.10 = 0.60 \)

Probability is 0.6 (or 60%).

Answer:

Let A be "passing 1st exam" and B be "passing 2nd exam".

• \( P(A) = 0.8 \)
• \( P(B) = 0.7 \)
• \( P(A \cup B) = 0.95 \) (At least one)

We need \( P(A \cap B) \) (Passing both).

Using the formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

\( 0.95 = 0.8 + 0.7 - P(A \cap B) \)

\( 0.95 = 1.5 - P(A \cap B) \)

\( P(A \cap B) = 1.5 - 0.95 = 0.55 \)

Answer:

Let E be "passing English" and H be "passing Hindi".

• \( P(E \cap H) = 0.5 \) (Both)
• \( P(\text{neither}) = P(E' \cap H') = 0.1 \)
• \( P(E) = 0.75 \)

First, find \( P(E \cup H) \). We know \( P(E' \cap H') = 1 - P(E \cup H) \).

\( 0.1 = 1 - P(E \cup H) \Rightarrow P(E \cup H) = 0.9 \).

Now use the union formula:

\( P(E \cup H) = P(E) + P(H) - P(E \cap H) \)

\( 0.9 = 0.75 + P(H) - 0.5 \)

\( 0.9 = 0.25 + P(H) \)

\( P(H) = 0.9 - 0.25 = 0.65 \)

Answer:

Total students = 60. Let A = NCC, B = NSS.

• \( n(A) = 30 \Rightarrow P(A) = 30/60 = 0.5 \)
• \( n(B) = 32 \Rightarrow P(B) = 32/60 \)
• \( n(A \cap B) = 24 \Rightarrow P(A \cap B) = 24/60 = 0.4 \)

(i) NCC or NSS (\( A \cup B \)):

\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

\( = \frac{30}{60} + \frac{32}{60} - \frac{24}{60} = \frac{38}{60} = \frac{19}{30} \)

(ii) Neither NCC nor NSS (\( A' \cap B' \)):

\( P(A' \cap B') = 1 - P(A \cup B) \)

\( = 1 - \frac{19}{30} = \frac{11}{30} \)

(iii) NSS but not NCC (\( B - A \)):

\( P(B - A) = P(B) - P(A \cap B) \)

\( = \frac{32}{60} - \frac{24}{60} = \frac{8}{60} = \frac{2}{15} \)

Answer:

Total marbles = \( 10 + 20 + 30 = 60 \).

Number of marbles drawn = 5.

Total outcomes (Sample Space) \( n(S) = {^{60}C_5} \).

(i) All will be blue:

We need to select 5 marbles from the 20 blue marbles.

Favourable outcomes = \( ^{20}C_5 \).

\( P(\text{All blue}) = \frac{^{20}C_5}{^{60}C_5} \).

(ii) At least one will be green:

\( P(\text{At least one green}) = 1 - P(\text{No green}) \).

"No green" means all 5 marbles are selected from the non-green marbles (Red + Blue = 10 + 20 = 30).

Outcomes for no green = \( ^{30}C_5 \).

\( P(\text{No green}) = \frac{^{30}C_5}{^{60}C_5} \).

\( P(\text{At least one green}) = 1 - \frac{^{30}C_5}{^{60}C_5} \).

Answer:

Total cards = 52. Cards drawn = 4.

Total outcomes \( n(S) = {^{52}C_4} \).

Favourable Event:

Choosing 3 diamonds from 13 diamonds: \( ^{13}C_3 \).

Choosing 1 spade from 13 spades: \( ^{13}C_1 \).

Total favourable outcomes = \( ^{13}C_3 \times ^{13}C_1 \).

Probability = \( \frac{^{13}C_3 \times ^{13}C_1}{^{52}C_4} \).

Answer:

Total faces = 6. Sample Space configuration: \( \{1, 1, 2, 2, 2, 3\} \).

1. P(2): Number of faces with '2' is 3.
   \( P(2) = \frac{3}{6} = \frac{1}{2} \).
2. P(1 or 3): Number of faces with '1' is 2. Number of faces with '3' is 1. Total favourable = 3.
   \( P(1 \cup 3) = \frac{3}{6} = \frac{1}{2} \).
3. P(not 3): Number of faces NOT '3' (i.e., '1' or '2') is \( 6 - 1 = 5 \).
   \( P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6} \).

Answer:

Total tickets = 10,000.

Prize tickets = 10.

Non-prize tickets = \( 10,000 - 10 = 9,990 \).

(a) Buying one ticket:

P(not getting a prize) = \( \frac{\text{Non-prize tickets}}{\text{Total tickets}} = \frac{9990}{10000} = \frac{999}{1000} \).

(b) Buying two tickets:

We need to select 2 tickets from the 9,990 non-prize tickets.

P(not getting a prize) = \( \frac{^{9990}C_2}{^{10000}C_2} \).

(c) Buying 10 tickets:

We need to select 10 tickets from the 9,990 non-prize tickets.

P(not getting a prize) = \( \frac{^{9990}C_{10}}{^{10000}C_{10}} \).

Answer:

Total students = 100. Total ways to select 2 students (you and friend) = \( ^{100}C_2 \).

Section 1 (S1) has 40 students. Section 2 (S2) has 60 students.

(a) Both in the same section:

Case 1: Both in S1. Ways = \( ^{40}C_2 \).

Case 2: Both in S2. Ways = \( ^{60}C_2 \).

Total favourable = \( ^{40}C_2 + ^{60}C_2 \).

\( P(\text{Same}) = \frac{^{40}C_2 + ^{60}C_2}{^{100}C_2} \)

Calculation:

• \( ^{40}C_2 = \frac{40 \times 39}{2} = 780 \)
• \( ^{60}C_2 = \frac{60 \times 59}{2} = 1770 \)
• \( ^{100}C_2 = \frac{100 \times 99}{2} = 4950 \)

\( P(\text{Same}) = \frac{780 + 1770}{4950} = \frac{2550}{4950} = \frac{17}{33} \).

(b) Both in different sections:

\( P(\text{Different}) = 1 - P(\text{Same}) = 1 - \frac{17}{33} = \frac{16}{33} \).

Answer:

Total ways to arrange 3 letters in 3 envelopes = \( 3! = 6 \).

Let the correct arrangement be (1, 2, 3).

The total arrangements are: { (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) }.

Event: "At least one letter is in its proper envelope".

Let's check each case:

• (1, 2, 3): 3 matches (All correct) - Success
• (1, 3, 2): 1 match (Letter 1) - Success
• (2, 1, 3): 1 match (Letter 3) - Success
• (2, 3, 1): 0 matches (Derangement) - Failure
• (3, 1, 2): 0 matches (Derangement) - Failure
• (3, 2, 1): 1 match (Letter 2) - Success

Favourable outcomes = 4.

Probability = \( \frac{4}{6} = \frac{2}{3} \).

Answer:

1. \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
   \( = 0.54 + 0.69 - 0.35 = 1.23 - 0.35 = 0.88 \)
2. \( P(A' \cap B') = P((A \cup B)') \) (De Morgan's Law)
   \( = 1 - P(A \cup B) = 1 - 0.88 = 0.12 \)
3. \( P(A \cap B') \) (Only A)
   \( = P(A) - P(A \cap B) = 0.54 - 0.35 = 0.19 \)
4. \( P(B \cap A') \) (Only B)
   \( = P(B) - P(A \cap B) = 0.69 - 0.35 = 0.34 \)

Answer:

Total persons = 5.

Let E be the event "Male" and F be the event "Age > 35".

• Males (E): Harish, Rohan, Salim (3 persons).
• Over 35 (F): Sheetal (46), Salim (41) (2 persons).

We need \( P(E \text{ or } F) = P(E \cup F) \).

The set \( E \cup F \) contains persons who are Male OR > 35.

Members: Harish (M), Rohan (M), Salim (M, >35), Sheetal (>35).

Total unique members = 4.

Probability = \( \frac{4}{5} \).

Answer:

Digits available: 0, 1, 3, 5, 7. (Count = 5). We need 4-digit numbers \( > 5000 \).

(i) Repetition is allowed:

Total Outcomes:

• Thousands place: Must be 5 or 7 (2 options).
• Hundreds, Tens, Units: Any of the 5 digits (5 options each).
• Total = \( 2 \times 5 \times 5 \times 5 = 250 \).

Favourable Outcomes (Divisible by 5):

• Thousands place: 5 or 7 (2 options).
• Hundreds, Tens: Any of 5 (5 options each).
• Units place: Must be 0 or 5 (2 options).
• Favourable = \( 2 \times 5 \times 5 \times 2 = 100 \).

\( P = \frac{100}{250} = \frac{2}{5} = 0.4 \).

(ii) Repetition is not allowed:

Total Outcomes:

• Case 1: Starts with 5. Remaining digits available = 4. Positions left = 3. Ways = \( 4 \times 3 \times 2 = 24 \).
• Case 2: Starts with 7. Remaining digits available = 4. Positions left = 3. Ways = \( 4 \times 3 \times 2 = 24 \).
• Total = \( 24 + 24 = 48 \).

Favourable Outcomes (Divisible by 5 - ends in 0 or 5):

• Starts with 5: Must end in 0 (cannot end in 5 as no repetition).
  Format: \( 5 \_ \_ 0 \). Middle 2 spots from {1,3,7}. Ways = \( 3 \times 2 = 6 \).
• Starts with 7:
  Ends in 0 (\( 7 \_ \_ 0 \)): Middle from {1,3,5}. Ways = \( 3 \times 2 = 6 \).
  Ends in 5 (\( 7 \_ \_ 5 \)): Middle from {0,1,3}. Ways = \( 3 \times 2 = 6 \).
• Total Favourable = \( 6 + 6 + 6 = 18 \).

\( P = \frac{18}{48} = \frac{3}{8} \).

Answer:

Total digits = 10 (0-9). The code has 4 digits with no repeats.

Total possible sequences = Permutation of 10 digits taken 4 at a time (\( ^{10}P_4 \)).

\( \text{Total Outcomes} = 10 \times 9 \times 8 \times 7 = 5040 \).

There is only 1 correct sequence that opens the lock.

Probability = \( \frac{1}{5040} \).