OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Exercise 22 (C)

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(c)

Question 1. A die is thrown once. Find
(i) P (an ace),
(ii) P (an even number),
(iii) P (a number < 3),
(iv) P (a number ≥ 4),
(v) P (a number < 7),
(vi) P (a number > 8).
Answer:
When a die is thrown once, the possible outcomes are 1, 2, 3, 4, 5, 6. So, the total number of outcomes, \( n(S) = 6 \).
(i) An ace means getting the number 1.
The number of favourable outcomes for an ace is \( n(\text{ace}) = 1 \).
So, \( P(\text{an ace}) = \frac{1}{6} \).
(ii) An even number means getting 2, 4, or 6.
The number of favourable outcomes for an even number is \( n(\text{even number}) = 3 \).
So, \( P(\text{an even number}) = \frac{3}{6} = \frac{1}{2} \).
(iii) A number less than 3 means getting 1 or 2.
Let E be the event of getting a number < 3. So \( E = \{1, 2\} \), and \( n(E) = 2 \).
Therefore, \( P(\text{a number} < 3) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3} \).
(iv) A number greater than or equal to 4 means getting 4, 5, or 6.
Let F be the event of getting a number ≥ 4. So \( F = \{4, 5, 6\} \), and \( n(F) = 3 \).
Therefore, \( P(\text{a number} \ge 4) = \frac{n(F)}{n(S)} = \frac{3}{6} = \frac{1}{2} \).
(v) A number less than 7 means getting 1, 2, 3, 4, 5, or 6.
Let G be the event of getting a number < 7. So \( G = \{1, 2, 3, 4, 5, 6\} \), and \( n(G) = 6 \).
Therefore, \( P(\text{a number} < 7) = \frac{n(G)}{n(S)} = \frac{6}{6} = 1 \). This means it is a certain event.
(vi) A number greater than 8 means no possible outcome when rolling a standard die.
Let H be the event of getting a number > 8. So \( H = \{\} \), and \( n(H) = 0 \).
Therefore, \( P(\text{a number} > 8) = \frac{n(H)}{n(S)} = \frac{0}{6} = 0 \). This is an impossible event.
In simple words: When you roll one die, there are 6 possible results. For each part, we count how many of these results match the condition and divide that by 6 to find the chance. A probability of 1 means it will definitely happen, and 0 means it will definitely not happen.

🎯 Exam Tip: Always clearly list the sample space and the favourable outcomes for each event. Simplify all probability fractions to their lowest terms.

Question 2. A card is drawn from a well shuffled pack of 52 cards. Find the probability of
(i) an ace
(ii) a spade card
(iii) a black card
(iv) a face card
(v) Jack, queen or king
(vi) 3 of heart or diamond.
Answer:
When a card is drawn from a well-shuffled pack of 52 cards, the total number of outcomes is \( n(S) = 52 \). A standard deck has 4 suits, each with 13 cards.
(i) An ace: There are 4 ace cards in a deck (one for each suit).
Let A be the event of getting an ace card. So \( n(A) = 4 \).
Therefore, \( P(\text{an ace}) = \frac{n(A)}{n(S)} = \frac{4}{52} = \frac{1}{13} \).
(ii) A spade card: There are 13 spade cards in a deck.
Let S be the event of getting a spade card. So \( n(S_{\text{favourable}}) = 13 \).
Therefore, \( P(\text{a spade card}) = \frac{13}{52} = \frac{1}{4} \).
(iii) A black card: There are 13 spade cards and 13 club cards, both of which are black. So, there are \( 13 + 13 = 26 \) black cards.
Let B be the event of getting a black card. So \( n(B) = 26 \).
Therefore, \( P(\text{a black card}) = \frac{26}{52} = \frac{1}{2} \).
(iv) A face card: Face cards are Jack, Queen, and King. There are 4 of each. So, \( 4 + 4 + 4 = 12 \) face cards.
Let F be the event of getting a face card. So \( n(F) = 12 \).
Therefore, \( P(\text{a face card}) = \frac{12}{52} = \frac{3}{13} \).
(v) A Jack, Queen or King: These are the same as face cards. There are \( 4 + 4 + 4 = 12 \) such cards.
Let JQK be the event of getting a Jack, Queen or King. So \( n(JQK) = 12 \).
Therefore, \( P(\text{Jack, Queen or King}) = \frac{12}{52} = \frac{3}{13} \).
(vi) A 3 of heart or diamond: There is one '3 of Heart' and one '3 of Diamond' card. So, there are \( 1 + 1 = 2 \) such cards.
Let TDH be the event of getting a 3 of heart or diamond. So \( n(TDH) = 2 \).
Therefore, \( P(\text{3 of heart or diamond}) = \frac{2}{52} = \frac{1}{26} \).
In simple words: When you pick one card from a full deck, the chance of getting a specific type of card depends on how many of that type are in the deck. We divide that number by the total of 52 cards. Cards like aces, spades, and face cards each have a different count that affects their probability.

🎯 Exam Tip: Remember the composition of a standard 52-card deck (4 suits, 13 cards per suit, 3 face cards per suit, colors) to quickly identify favourable outcomes for card probability questions.

Question 3. One card is drawn from a pack of 52 cards being equally likely to be drawn. Find the probability of
(i) the card drawn to be red
(ii) the card drawn to be a king
(iii) the card drawn to be red and a king
(iv) the card drawn to be either red or a king.
Answer:
The total number of outcomes when drawing one card from a pack is \( n(S) = 52 \).
(i) Card drawn to be red: There are 26 red cards in a deck (13 hearts and 13 diamonds).
Let A be the event that the card is red. So \( n(A) = 26 \).
Therefore, \( P(\text{red card}) = \frac{n(A)}{n(S)} = \frac{26}{52} = \frac{1}{2} \).
(ii) Card drawn to be a king: There are 4 king cards in a deck (King of Hearts, Diamonds, Clubs, Spades).
Let B be the event that the card is a king. So \( n(B) = 4 \).
Therefore, \( P(\text{a king}) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13} \).
(iii) Card drawn to be red and a king: This means getting the King of Hearts or the King of Diamonds. There are 2 such cards.
Let C be the event that the card is red and a king. So \( n(C) = 2 \).
Therefore, \( P(\text{red and king}) = \frac{n(C)}{n(S)} = \frac{2}{52} = \frac{1}{26} \).
(iv) Card drawn to be either red or a king: We use the formula \( P(Red \cup King) = P(Red) + P(King) - P(Red \cap King) \).
There are 26 red cards and 4 kings. Out of the 4 kings, 2 are red (King of Hearts, King of Diamonds).
So, \( n(\text{red or king}) = 26 (\text{red cards}) + 2 (\text{black kings}) = 28 \).
Let E be the event that the card is either red or a king. So \( n(E) = 28 \).
Therefore, \( P(\text{red or king}) = \frac{n(E)}{n(S)} = \frac{28}{52} = \frac{7}{13} \).
In simple words: When drawing a card, we look at how many cards fit our condition out of 52. For "red or king," we count all red cards and then add the kings that are not red (the black kings) to avoid counting the red kings twice. This formula helps us count combined events correctly.

🎯 Exam Tip: For "or" probability questions, always consider if the events overlap. If they do, subtract the probability of the overlap to avoid double-counting.

Question 4. A book contains 100 pages. A page is chosen at random. What is the chance that the sum of digits on the page is equal to 9?
Answer:
The total number of pages in the book is 100. So, the total number of outcomes, \( n(S) = 100 \).
We need to find pages where the sum of the digits equals 9. These pages are:
\(\{9 (0+9), 18 (1+8), 27 (2+7), 36 (3+6), 45 (4+5), 54 (5+4), 63 (6+3), 72 (7+2), 81 (8+1), 90 (9+0)\}\).
Note that page 9 can be thought of as 09, so \( 0+9=9 \).
Let A be the event that the sum of digits on the page is 9. So, \( n(A) = 10 \).
Therefore, the required probability is \( P(A) = \frac{n(A)}{n(S)} = \frac{10}{100} = \frac{1}{10} \). The digit sum rule is a fun way to explore numbers.
In simple words: Out of 100 pages, we count how many have digits that add up to 9. Since there are 10 such pages, the chance of picking one is 10 out of 100, which simplifies to 1 out of 10.

🎯 Exam Tip: When counting numbers with specific digit properties, remember to systematically list them or identify a pattern to ensure accuracy.

Question 5. From 25 tickets, marked with the first 25 numerals, one is drawn at random. Find the probability that
(i) it is a multiple of 5 or 7
(ii) it is a multiple of 3 or 7.
Answer:
The total number of tickets is 25, marked from 1 to 25. So, \( n(S) = 25 \).
(i) Multiple of 5 or 7:
Multiples of 5 between 1 and 25 are: \(\{5, 10, 15, 20, 25\}\).
Multiples of 7 between 1 and 25 are: \(\{7, 14, 21\}\).
There are no common multiples of 5 and 7 in this range.
Let E be the event that the ticket is a multiple of 5 or 7.
So, \( E = \{5, 7, 10, 14, 15, 20, 21, 25\} \), and \( n(E) = 8 \).
Therefore, \( P(E) = \frac{n(E)}{n(S)} = \frac{8}{25} \).
(ii) Multiple of 3 or 7:
Multiples of 3 between 1 and 25 are: \(\{3, 6, 9, 12, 15, 18, 21, 24\}\).
Multiples of 7 between 1 and 25 are: \(\{7, 14, 21\}\).
The common multiple of 3 and 7 is 21.
Let F be the event that the ticket is a multiple of 3 or 7.
So, \( F = \{3, 6, 7, 9, 12, 14, 15, 18, 21, 24\} \).
There are 10 such numbers, so \( n(F) = 10 \). (Alternatively, \( n(3) = 8, n(7) = 3, n(3 \cap 7) = n(21) = 1 \). So \( n(F) = 8 + 3 - 1 = 10 \)).
Therefore, \( P(F) = \frac{n(F)}{n(S)} = \frac{10}{25} = \frac{2}{5} \).
In simple words: When picking a ticket from 1 to 25, we find how many numbers can be divided by 5 or 7 (part i), or by 3 or 7 (part ii). We list all unique numbers that fit and divide that count by 25 to get the probability. We must be careful not to count numbers twice if they are multiples of both.

🎯 Exam Tip: When calculating the probability of "A or B," remember to list all individual multiples first, then check for common multiples to avoid double-counting, especially when using the inclusion-exclusion principle \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

Question 6. What is the probability that a number selected from the numbers 1, 2, 3,..., 25 is a prime number? You may assume that each of the 25 numbers is equally likely to be selected.
Answer:
The total number of outcomes is 25, as we are selecting from numbers 1 to 25. So, \( n(S) = 25 \).
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
The prime numbers between 1 and 25 are: \(\{2, 3, 5, 7, 11, 13, 17, 19, 23\}\).
Let E be the event that the selected number is a prime number. So, \( n(E) = 9 \).
Therefore, the required probability of getting a prime number is \( P(E) = \frac{n(E)}{n(S)} = \frac{9}{25} \). Remember that 1 is not a prime number.
In simple words: We need to find how many numbers between 1 and 25 are prime. There are 9 prime numbers. So, the chance of picking a prime number out of 25 is 9 out of 25.

🎯 Exam Tip: Be careful to correctly identify prime numbers; remember that 1 is not prime, and 2 is the only even prime number. Listing them out ensures accuracy.

Question 7. In a simultaneous throw of two coins find the probability of
(i) two heads,
(ii) exactly one tail,
(iii) at least one tail.
Answer:
When two coins are thrown simultaneously, the total possible outcomes are: \(\{HH, HT, TH, TT\}\).
So, the total number of outcomes, \( n(S) = 4 \).
(i) Two heads:
Let E be the event of getting two heads. So, \( E = \{HH\} \), and \( n(E) = 1 \).
Therefore, \( P(\text{two heads}) = \frac{n(E)}{n(S)} = \frac{1}{4} \).
(ii) Exactly one tail:
Let F be the event of getting exactly one tail. So, \( F = \{HT, TH\} \), and \( n(F) = 2 \).
Therefore, \( P(\text{exactly one tail}) = \frac{n(F)}{n(S)} = \frac{2}{4} = \frac{1}{2} \).
(iii) At least one tail: This means one tail or two tails.
Let G be the event of getting at least one tail. So, \( G = \{HT, TH, TT\} \), and \( n(G) = 3 \).
Therefore, \( P(\text{at least one tail}) = \frac{n(G)}{n(S)} = \frac{3}{4} \). This is often an easier way to think about "not zero" events.
In simple words: When flipping two coins, there are four possible results. We count how many of these results match "two heads," "one tail," or "at least one tail" and divide by four to find the chance.

🎯 Exam Tip: When dealing with "at least one" probabilities, it's often simpler to calculate the probability of the complementary event (e.g., "no tails" for "at least one tail") and subtract it from 1.

Question 8. In a throw of two dice, find the probability of getting:
(i) a total of 10 or 11
(ii) an odd number on one die and a multiple of 3 on the other
(iii) a sum as 6
(iv) a multiple of 3 as the sum
(v) the sum as a prime number
(vi) throwing:
    (a) a number > 4 on each die
    (b) an odd number on one die and 5 on the other
    (c) a multiple of 2 on one and a multiple of 3 on the other
    (d) an even number as the sum
    (e) an odd number as the sum.
(vii) the probability of:
    (a) two aces
    (b) at least one ace
    (c) doublets
    (d) a total less than 10
    (e) a total of 11
    (f) a total of 12
    (g) a total of at least 10
    (h) a doublet of even number
(viii) the product a perfect square (square of a natural number).
Answer:
When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). Each outcome is an ordered pair (die 1 result, die 2 result).
(i) Getting a total of 10 or 11:
Sums that are 10: \(\{ (4,6), (5,5), (6,4) \}\)
Sums that are 11: \(\{ (5,6), (6,5) \}\)
The total favorable outcomes for sum 10 or 11 are \(\{ (4,6), (5,5), (6,4), (5,6), (6,5) \}\).
So, \( n(E) = 5 \).
The probability is \( P(E) = \frac{5}{36} \).
(ii) Getting an odd number on one die and a multiple of 3 on the other:
Odd numbers are \(\{1,3,5\}\). Multiples of 3 are \(\{3,6\}\).
Pairs where die 1 is odd and die 2 is a multiple of 3: \(\{ (1,3), (1,6), (3,3), (3,6), (5,3), (5,6) \}\). (6 outcomes)
Pairs where die 1 is a multiple of 3 and die 2 is odd: \(\{ (3,1), (3,5), (6,1), (6,3), (6,5) \}\). (5 outcomes, after removing (3,3) which is already counted).
The unique favorable outcomes are \(\{ (1,3), (1,6), (3,1), (3,3), (3,5), (3,6), (5,1), (5,3), (5,6), (6,1), (6,3), (6,5) \}\).
So, \( n(F) = 11 \).
The probability is \( P(F) = \frac{11}{36} \). This calculation requires careful listing to avoid duplicates.
(iii) Getting a sum as 6:
The pairs that sum to 6 are \(\{ (1,5), (2,4), (3,3), (4,2), (5,1) \}\).
So, \( n(G) = 5 \).
The probability is \( P(G) = \frac{5}{36} \).
(iv) Getting a multiple of 3 as the sum:
Possible sums that are multiples of 3 are 3, 6, 9, 12.
Sum 3: \(\{ (1,2), (2,1) \}\)
Sum 6: \(\{ (1,5), (2,4), (3,3), (4,2), (5,1) \}\)
Sum 9: \(\{ (3,6), (4,5), (5,4), (6,3) \}\)
Sum 12: \(\{ (6,6) \}\)
The total favorable outcomes are \( 2 + 5 + 4 + 1 = 12 \).
So, \( n(A) = 12 \).
The probability is \( P(A) = \frac{12}{36} = \frac{1}{3} \).
(v) Getting the sum as a prime number:
Possible prime sums are 2, 3, 5, 7, 11.
Sum 2: \(\{ (1,1) \}\)
Sum 3: \(\{ (1,2), (2,1) \}\)
Sum 5: \(\{ (1,4), (2,3), (3,2), (4,1) \}\)
Sum 7: \(\{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \}\)
Sum 11: \(\{ (5,6), (6,5) \}\)
The total favorable outcomes are \( 1 + 2 + 4 + 6 + 2 = 15 \).
So, \( n(E) = 15 \).
The probability is \( P(E) = \frac{15}{36} = \frac{5}{12} \). Prime numbers are fundamental in number theory.
(vi) In a single throw of two dice, find the probability of throwing:
    (a) A number > 4 on each die:
    This means each die must show 5 or 6.
    The favorable outcomes are \(\{ (5,5), (5,6), (6,5), (6,6) \}\).
    So, \( n(A) = 4 \).
    The probability is \( P(A) = \frac{4}{36} = \frac{1}{9} \).
    (b) An odd number on one die and 5 on the other:
    Odd numbers are \(\{1,3,5\}\).
    If the first die is odd and the second is 5: \(\{ (1,5), (3,5), (5,5) \}\).
    If the first die is 5 and the second is odd: \(\{ (5,1), (5,3) \}\) (excluding (5,5) which is already counted).
    The unique favorable outcomes are \(\{ (1,5), (3,5), (5,1), (5,3), (5,5) \}\).
    So, \( n(B) = 5 \).
    The probability is \( P(B) = \frac{5}{36} \).
    (c) A multiple of 2 on one and a multiple of 3 on the other:
    Multiples of 2 are \(\{2,4,6\}\). Multiples of 3 are \(\{3,6\}\).
    If die 1 is a multiple of 2 and die 2 is a multiple of 3: \(\{ (2,3), (2,6), (4,3), (4,6), (6,3), (6,6) \}\). (6 outcomes)
    If die 1 is a multiple of 3 and die 2 is a multiple of 2: \(\{ (3,2), (3,4), (3,6), (6,2), (6,4) \}\) (excluding (6,6) which is already counted). (5 outcomes)
    The unique favorable outcomes are \(\{ (2,3), (2,6), (3,2), (3,4), (3,6), (4,3), (4,6), (6,2), (6,3), (6,4), (6,6) \}\).
    So, \( n(C) = 11 \).
    The probability is \( P(C) = \frac{11}{36} \).
    (d) An even number as the sum:
    The sum of two numbers is even if both are even (E+E) or both are odd (O+O).
    Number of even results per die is 3 \(\{2,4,6\}\). Number of odd results is 3 \(\{1,3,5\}\).
    Number of (E,E) outcomes = \( 3 \times 3 = 9 \).
    Number of (O,O) outcomes = \( 3 \times 3 = 9 \).
    Total favorable outcomes = \( 9 + 9 = 18 \).
    The probability is \( P(D) = \frac{18}{36} = \frac{1}{2} \).
    (e) An odd number as the sum:
    The sum of two numbers is odd if one is even and the other is odd (E+O or O+E).
    Number of (E,O) outcomes = \( 3 \times 3 = 9 \).
    Number of (O,E) outcomes = \( 3 \times 3 = 9 \).
    Total favorable outcomes = \( 9 + 9 = 18 \).
    The probability is \( P(E) = \frac{18}{36} = \frac{1}{2} \).
(vii) In a single throw of two dice, what is the probability of:
    (a) Two aces:
    An ace on a die is the number 1. So, \(\{ (1,1) \}\).
    So, \( n(A) = 1 \).
    The probability is \( P(A) = \frac{1}{36} \).
    (b) At least one ace:
    This means die 1 is 1 (and die 2 is any number) OR die 2 is 1 (and die 1 is any number), but not double counting (1,1).
    Favorable outcomes: \(\{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1) \}\).
    So, \( n(B) = 11 \).
    The probability is \( P(B) = \frac{11}{36} \).
    (c) Doublets:
    Doublets are when both dice show the same number.
    Favorable outcomes: \(\{ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \}\).
    So, \( n(C) = 6 \).
    The probability is \( P(C) = \frac{6}{36} = \frac{1}{6} \).
    (d) A total less than 10:
    This means the sum can be 2, 3, 4, 5, 6, 7, 8, or 9.
    The complementary event is "total of at least 10" (sums 10, 11, 12).
    Number of outcomes for "total at least 10" is 6 (from part (vii)(g)).
    So, \( n(\text{total} < 10) = 36 - 6 = 30 \).
    The probability is \( P(D) = \frac{30}{36} = \frac{5}{6} \).
    (e) A total of 11:
    Favorable outcomes: \(\{ (5,6), (6,5) \}\).
    So, \( n(E) = 2 \).
    The probability is \( P(E) = \frac{2}{36} = \frac{1}{18} \).
    (f) A total of 12:
    Favorable outcomes: \(\{ (6,6) \}\).
    So, \( n(F) = 1 \).
    The probability is \( P(F) = \frac{1}{36} \).
    (g) A total of at least 10:
    This means the sum is 10, 11, or 12.
    Sum 10: \(\{ (4,6), (5,5), (6,4) \}\) (3 outcomes)
    Sum 11: \(\{ (5,6), (6,5) \}\) (2 outcomes)
    Sum 12: \(\{ (6,6) \}\) (1 outcome)
    Total favorable outcomes = \( 3 + 2 + 1 = 6 \).
    So, \( n(G) = 6 \).
    The probability is \( P(G) = \frac{6}{36} = \frac{1}{6} \).
    (h) A doublet of even number:
    This means both dice show the same even number.
    Favorable outcomes: \(\{ (2,2), (4,4), (6,6) \}\).
    So, \( n(H) = 3 \).
    The probability is \( P(H) = \frac{3}{36} = \frac{1}{12} \).
(viii) Getting the product a perfect square:
Perfect squares on a die are 1, 4. Products can be 1, 4, 9, 16, 25, 36.
Product 1: \(\{ (1,1) \}\)
Product 4: \(\{ (1,4), (2,2), (4,1) \}\)
Product 9: \(\{ (3,3) \}\)
Product 16: \(\{ (4,4) \}\)
Product 25: \(\{ (5,5) \}\)
Product 36: \(\{ (6,6) \}\)
The total favorable outcomes are \( 1 + 3 + 1 + 1 + 1 + 1 = 8 \).
So, \( n(P) = 8 \).
The probability is \( P(P) = \frac{8}{36} = \frac{2}{9} \).
In simple words: When rolling two dice, there are 36 possible outcomes. For each part, we carefully list all the pairs that match the condition and then divide this count by 36. Complex conditions like "odd on one, multiple of 3 on other" need extra care to list all unique pairs without missing any or counting any twice.

🎯 Exam Tip: For problems involving two dice, it's very helpful to visualize the 6x6 grid of outcomes. This helps in systematically listing favourable events for sums, products, or specific combinations, ensuring no outcome is missed or double-counted.

Question 9. In a single throw of three dice, find the probability of getting a total of 17 or 18.
Answer:
When three dice are thrown, the total number of possible outcomes is \( 6 \times 6 \times 6 = 6^3 = 216 \).
We need to find the outcomes where the sum of the numbers on the three dice is 17 or 18.
For a sum of 17: The only combinations of numbers from 1 to 6 that add up to 17 are (5, 6, 6) in any order. These are \(\{ (5,6,6), (6,5,6), (6,6,5) \}\).
For a sum of 18: The only combination is (6, 6, 6).
Let E be the event of getting a total of 17 or 18.
The favorable outcomes are \(\{ (5,6,6), (6,5,6), (6,6,5), (6,6,6) \}\).
So, \( n(E) = 4 \).
The probability is \( P(E) = \frac{n(E)}{n(S)} = \frac{4}{216} = \frac{1}{54} \). Listing specific combinations is key for three dice.
In simple words: When rolling three dice, there are 216 possible outcomes. To get a total of 17 or 18, there are only a few ways this can happen: (5,6,6) in different orders, and (6,6,6). Counting these 4 ways and dividing by 216 gives the probability.

🎯 Exam Tip: For probabilities involving three or more dice, systematically list the combinations that sum to a specific value. Remember that permutations of the same numbers (e.g., 5,6,6) count as different outcomes unless the dice are identical and unordered.

Question 10. In a single throw of three dice, find the probability of getting:
(i) a total of 5
(ii) a total of at most 5
(iii) a total of at least 5<
(iv) the same number on all the dice
(v) not getting the same number on all the dice.
Answer:
When three dice are thrown, the total number of possible outcomes is \( 6 \times 6 \times 6 = 6^3 = 216 \).
(i) A total of 5:
The combinations that sum to 5 are:
\(\{ (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) \}\).
So, \( n(A) = 6 \).
The probability is \( P(A) = \frac{6}{216} = \frac{1}{36} \).
(ii) A total of at most 5:
This means the sum is 3, 4, or 5.
Sum 3: \(\{ (1,1,1) \}\) (1 outcome)
Sum 4: \(\{ (1,1,2), (1,2,1), (2,1,1) \}\) (3 outcomes)
Sum 5: \(\{ (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) \}\) (6 outcomes)
So, \( n(B) = 1 + 3 + 6 = 10 \).
The probability is \( P(B) = \frac{10}{216} = \frac{5}{108} \). It is important to list all unique permutations.
(iii) A total of at least 5:
This is the complementary event to "getting a total of at most 4".
Total of at most 4 (from calculations above for part ii):
Sum 3: \(\{ (1,1,1) \}\) (1 outcome)
Sum 4: \(\{ (1,1,2), (1,2,1), (2,1,1) \}\) (3 outcomes)
So, \( n(\text{at most 4}) = 1 + 3 = 4 \).
The probability of getting at most 4 is \( P(\text{at most 4}) = \frac{4}{216} \).
Therefore, \( P(\text{at least 5}) = 1 - P(\text{at most 4}) = 1 - \frac{4}{216} = \frac{216 - 4}{216} = \frac{212}{216} = \frac{53}{54} \).
(iv) The same number on all the dice:
This means getting triplets like (1,1,1), (2,2,2), etc.
Favorable outcomes: \(\{ (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6) \}\).
So, \( n(D) = 6 \).
The probability is \( P(D) = \frac{6}{216} = \frac{1}{36} \).
(v) Not getting the same number on all the dice:
This is the complementary event to "getting the same number on all the dice" (from part iv).
\( P(\text{not same number}) = 1 - P(\text{same number}) = 1 - \frac{1}{36} = \frac{36 - 1}{36} = \frac{35}{36} \).
In simple words: With three dice, there are 216 total outcomes. For each part, we count the specific combinations that fit the rule (like summing to 5, or all dice showing the same number). For "at least" or "not," sometimes it's easier to find the opposite case first and subtract its probability from 1.

🎯 Exam Tip: When calculating probabilities for "at least" or "not" events, consider using the complement rule: \( P(A') = 1 - P(A) \). This can often simplify calculations, especially with many possible outcomes.

Question 11. There are four events E1, E2, E3, and E4, one of which must and only one can happen. The odds are 2 : 5 in favour of E₁, 3 : 4 in favour of E2 and 1 : 3 in favour of E3. Find the odds against E4.
Answer:
When "odds in favour" are given as \( a:b \), the probability is \( P = \frac{a}{a+b} \).
For \( E_1 \): Odds in favour are 2 : 5.
So, \( P(E_1) = \frac{2}{2+5} = \frac{2}{7} \).
For \( E_2 \): Odds in favour are 3 : 4.
So, \( P(E_2) = \frac{3}{3+4} = \frac{3}{7} \).
For \( E_3 \): Odds in favour are 1 : 3.
So, \( P(E_3) = \frac{1}{1+3} = \frac{1}{4} \).
Since only one of the four events \( E_1, E_2, E_3, E_4 \) can happen and one must happen, their probabilities must sum to 1.
\( P(E_1) + P(E_2) + P(E_3) + P(E_4) = 1 \)
\( \frac{2}{7} + \frac{3}{7} + \frac{1}{4} + P(E_4) = 1 \)
\( P(E_4) = 1 - \frac{2}{7} - \frac{3}{7} - \frac{1}{4} \)
\( P(E_4) = 1 - \left( \frac{2}{7} + \frac{3}{7} \right) - \frac{1}{4} \)
\( P(E_4) = 1 - \frac{5}{7} - \frac{1}{4} \)
To subtract, find a common denominator, which is 28:
\( P(E_4) = \frac{28}{28} - \frac{5 \times 4}{28} - \frac{1 \times 7}{28} \)
\( P(E_4) = \frac{28 - 20 - 7}{28} = \frac{1}{28} \).
Now, we need to find the odds against \( E_4 \).
Odds against an event are given by \( P(\overline{E_4}) : P(E_4) \).
First, calculate the probability of \( E_4 \) not happening: \( P(\overline{E_4}) = 1 - P(E_4) = 1 - \frac{1}{28} = \frac{27}{28} \).
So, the odds against \( E_4 \) are \( \frac{27}{28} : \frac{1}{28} \), which simplifies to \( 27 : 1 \).
In simple words: When we know the chances of three events happening, and we know only one of four events can happen in total, we can find the chance of the fourth event. "Odds against" simply means comparing the chance of something not happening to the chance of it happening.

🎯 Exam Tip: Remember that if events are mutually exclusive and exhaustive, their probabilities sum to 1. Convert odds to probabilities carefully ( \( a:b \) in favour means \( P = a/(a+b) \) ) and vice versa (odds against \( P:P' \) means \( P' : P \) ).

Question 12. In a simultaneous toss of 4 coins, what is the probability of getting exactly 3 heads?
Answer:
When 4 coins are tossed simultaneously, the total number of possible outcomes is \( 2^4 = 16 \).
The sample space (S) is:
S = \(\{ \text{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} \}\).
So, \( n(S) = 16 \).
We are looking for the event of getting exactly 3 heads.
Let E be the event of getting exactly 3 heads.
The favorable outcomes are \(\{ \text{HHHT, HHTH, HTHH, THHH} \}\).
So, \( n(E) = 4 \).
The required probability is \( P(E) = \frac{n(E)}{n(S)} = \frac{4}{16} = \frac{1}{4} \). This is a simple application of combinations.
In simple words: When you flip four coins, there are 16 different results. We want to find the chance of getting exactly three heads. By listing all the ways to get three heads, we find there are 4 such ways. So the probability is 4 out of 16, which simplifies to 1 out of 4.

🎯 Exam Tip: For coin toss problems, remember that the number of outcomes is \( 2^n \) where 'n' is the number of coins. Listing outcomes systematically (e.g., using a tree diagram or combinations) helps ensure accuracy for specific head/tail counts.