9th Mathematics

When the dice is rolled, the sample space is:

S = {L, M, N, O, P, U}

The total number of outcomes is:

n(S) = 6

Let A be the event of getting a consonant. The consonants in the sample space are L, M, N, and P:

A = {L, M, N, P}

The number of favorable outcomes is:

n(A) = 4

Using the probability formula:

P(A) = n(A)n(S) = 46 = 23

Thus, the probability of getting a consonant is 23.

When a pair of fair dice is rolled, the sample space S consists of 36 outcomes represented as follows:

Total outcomes: n(S) = 36

• (i) Sum of dots is at least 4:
  Let A be the event that the sum of dots is at least 4. Instead of counting all outcomes with sum ≥ 4, we count the complement (sum less than 4):
  Outcomes with sum < 4 are: A' = {(1, 1), (1, 2), (2, 1)} with n(A') = 3.
  Therefore, n(A) = 36 − 3 = 33 outcomes.
  P(A) = n(A)n(S) = 3336 = 1112.
• (ii) Product of both dots is between 5 to 10 (inclusive):
  Let B be the event that the product of both dots is between 5 and 10.
  Listing all such pairs:
  B = {(1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (4, 2), (5, 1), (5, 2), (6, 1)}
  Their respective products are: 5, 6, 6, 8, 10, 6, 9, 8, 5, 10, 6 (all within [5, 10]).
  The number of outcomes is: n(B) = 11.
  P(B) = n(B)n(S) = 1136.
• (iii) The difference between both the dots is equal to 4:
  Let C be the event that the difference between the two dots is 4.
  Listing all such pairs: C = {(1, 5), (2, 6), (5, 1), (6, 2)} with n(C) = 4.
  P(C) = n(C)n(S) = 436 = 19.
• (iv) Number at least 5 on the first dice and at least 4 on the second dice:
  Let E be this event. First die must show 5 or 6; second die must show 4, 5, or 6.
  Listing the pairs: E = {(5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} with n(E) = 6.
  P(E) = n(E)n(S) = 636 = 16.

The letters of the word MATHEMATICS are:

S = {M, A, T, H, E, M, A, T, I, C, S}

Total letters: n(S) = 11

• (i) Vowel:
  Vowels are {A, E, A, I}. Let V be the event of choosing a vowel. n(V) = 4.
  P(V) = 411.
• (ii) Consonant:
  Consonants are {M, T, H, M, T, C, S}. Let C be the event of choosing a consonant. n(C) = 7.
  P(C) = 711.
• (iii) an E:
  The letter E occurs once: {E}. Let E_event be the event of choosing E. n(E_event) = 1.
  P(E_event) = 111.
• (iv) an A:
  The letter A occurs twice: {A, A}. Let A_event be the event of choosing A. n(A_event) = 2.
  P(A_event) = 211.
• (v) not M:
  The letter M occurs twice: {M, M}. The probability of getting M is:
  P(M) = 211.
  The probability of not getting M is:
  P(not M) = 1 − P(M) = 1 − 211 = 911.
• (vi) not T:
  The letter T occurs twice: {T, T}. The probability of getting T is:
  P(T) = 211.
  The probability of not getting T is:
  P(not T) = 1 − P(T) = 1 − 211 = 911.

When a single dice is rolled, the sample space is:

S = {1, 2, 3, 4, 5, 6},   n(S) = 6

Part 1: Probability of getting 3 or 4:

Let A be the event of getting 3 or 4: A = {3, 4}, with n(A) = 2.

P(A) = n(A)n(S) = 26 = 13.

Part 2: Probability of getting not 3 or 4:

This is the complement of event A:

P(A') = 1 − P(A) = 1 − 13 = 23.

The sample space contains cards labelled from 1 to 30:

S = {1, 2, 3, ..., 30},   n(S) = 30

• (i) The number 25:
  Let A be the event of selecting card 25: A = {25}, n(A) = 1.
  P(A) = 130.
• (ii) Number between 17 to 22 (inclusive in the textbook key):
  Let B be this event: B = {17, 18, 19, 20, 21, 22}, n(B) = 6.
  P(B) = 630 = 15.
• (iii) Number at least 20:
  Let C be this event: C = {20, 21, 22, ..., 30}, n(C) = 11.
  P(C) = 1130.
• (iv) Number not 27 and 29:
  Let D be the event of choosing card 27 or 29: D = {27, 29}, n(D) = 2.
  The probability of choosing 27 or 29 is: P(D) = 230 = 115.
  The probability of NOT choosing 27 or 29 is the complement:
  P(D') = 1 − P(D) = 1 − 115 = 1415.
• (v) Number not between 12 to 15:
  Let E be the event of selecting a number between 12 and 15: E = {12, 13, 14, 15}, n(E) = 4.
  The probability of getting a card between 12 and 15 is: P(E) = 430 = 215.
  The probability of NOT getting a card between 12 and 15 is:
  P(E') = 1 − P(E) = 1 − 215 = 1315.

Let A be the event that Ayesha will pass the examination.

P(A) = 0.85

The event that Ayesha will not pass the examination is the complement of A, denoted as A':

P(A') = 1 − P(A)

Substituting the given value:

P(A') = 1 − 0.85 = 0.15

Thus, the probability that Ayesha will not pass the examination is 0.15.

When a fair coin is tossed and a fair dice is rolled once, the sample space S consists of 12 outcomes:

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

Total outcomes: n(S) = 12

• (i) Tail on coin and at least 4 on dice:
  Let A be this event: A = {(T, 4), (T, 5), (T, 6)}, with n(A) = 3.
  P(A) = 312 = 14.
• (ii) Head on coin and the number 2 or 3 on dice:
  Let B be this event: B = {(H, 2), (H, 3)}, with n(B) = 2.
  P(B) = 212 = 16.
• (iii) Head or tail on coin and the number 6 on dice:
  Note: The book states "head and tail" (physically impossible simultaneously), but resolves it as "head or tail".
  Let C be this event: C = {(H, 6), (T, 6)}, with n(C) = 2.
  P(C) = 212 = 16.
• (iv) Not tail on coin and the number 5 on dice:
  
  • Correct Logical Answer: "Not tail" means Head. The event is {(H, 5)}, with 1 outcome. Probability = 112.
  • Textbook Solution: The textbook defines the base event as "tail and 5 on dice" i.e. {(T, 5)} (prob = 112), and then takes the complement of this entire joint event, calculating 1 − 112 = 1112.
• (v) Not head on coin and the number 5 or 2 on dice:
  
  • Correct Logical Answer: "Not head" means Tail. The event is {(T, 2), (T, 5)}, with 2 outcomes. Probability = 212 = 16.
  • Textbook Solution: The textbook defines the base event as "head on coin and (5 or 2) on dice" i.e. {(H, 2), (H, 5)} (prob = 212 = 16), and then calculates the complement: 1 − 16 = 56.

Total cards in a standard deck: n(S) = 52

• (i) Selecting a queen:
  There are 4 queens in a deck. Let A be the event of selecting a queen: n(A) = 4.
  P(A) = n(A)n(S) = 452 = 113.
• (ii) Selecting neither a queen nor a jack:
  Let B be the event of selecting a queen or a jack. Since there are 4 queens and 4 jacks:
  n(B) = 4 + 4 = 8 cards.
  The probability of selecting a queen or a jack is:
  P(B) = 852 = 213.
  The probability of selecting neither is the complement event B':
  P(B') = 1 − P(B) = 1 − 213 = 1113.

Total cards in a standard deck: n(S) = 52

• (i) Getting a Jack:
  There are 4 jacks in a deck. Let A be the event of selecting a jack: n(A) = 4.
  P(A) = 452 = 113.
• (ii) Getting no diamond:
  Let B be the event of selecting a diamond card. There are 13 diamonds in a deck:
  n(B) = 13.
  The probability of selecting a diamond is:
  P(B) = 1352 = 14.
  The probability of selecting no diamond is the complement event B':
  P(B') = 1 − P(B) = 1 − 14 = 34.