Probability

Q1: Multiple Choice Type:

i. A coin is tossed three times, the number of possible outcomes are:

Step 1: Number of possible outcomes when a coin is tossed once = 2 (Head or Tail)
Step 2: For three tosses, total possible outcomes = \( 2 \times 2 \times 2 = 2^3 \)
Step 3: Calculating \( 2^3 = 8 \)
Answer: a. 8

ii. If P(A) denotes the probability of getting an event A, then P(not getting A) is:

Step 1: The sum of the probabilities of an event and its complement is always 1.
Mathematically, \( P(A) + P(\text{not } A) = 1 \)
Step 2: Rearranging, \( P(\text{not } A) = 1 – P(A) \)
Answer: a. 1 – P(A)

iii. A coin is tossed once. The probability of getting a tail is:

Step 1: Total possible outcomes when a coin is tossed once = 2 (Head, Tail)
Step 2: Number of favorable outcomes to get tail = 1 (Tail)
Step 3: Probability \( P(\text{tail}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{2} \)
Answer: c. \( \frac{1}{2} \)

iv. A coin is tossed two times. The probability of getting at least one tail is:

Step 1: Total possible outcomes when tossing a coin twice = \( 2^2 = 4 \)
These outcomes are: HH, HT, TH, TT
Step 2: Calculate probability of getting no tail (i.e., both heads) = Probability of HH = \( \frac{1}{4} \)
Step 3: Probability of getting at least one tail = \( 1 – \text{Probability of no tail} \)
So, \( P(\text{at least one tail}) = 1 – \frac{1}{4} = \frac{3}{4} \)
Answer: a. \( \frac{3}{4} \)

v. A card is drawn from a well shuffled pack of 52 playing cards. The probability of getting a face card is:

Step 1: Total cards in the pack = 52
Face cards are Jack, Queen, King of each suit
Number of face cards = \( 3 \times 4 = 12 \)
Step 2: Probability of drawing a face card = \( \frac{\text{Number of face cards}}{\text{Total cards}} = \frac{12}{52} \)
Step 3: Simplify \( \frac{12}{52} = \frac{3}{13} \)
Answer: c. \( \frac{3}{13} \)

Q2: A coin is tossed twice. Find probability of getting:

i. Exactly one head

Step 1: Total possible outcomes when a coin is tossed twice = \( 2^2 = 4 \)
The outcomes are: HH, HT, TH, TT
Step 2: Favorable outcomes for exactly one head = HT, TH
Number of favorable outcomes = 2
Step 3: Probability \( P(\text{exactly one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. Exactly one tail

Step 1: Possible outcomes when a coin is tossed twice = HH, HT, TH, TT
Step 2: Favorable outcomes for exactly one tail = HT, TH
Number of favorable outcomes = 2
Step 3: Probability \( P(\text{exactly one tail}) = \frac{2}{4} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

iii. Two tails

Step 1: Total outcomes = 4 (HH, HT, TH, TT)
Step 2: Favorable outcome for two tails = TT
Number of favorable outcomes = 1
Step 3: Probability \( P(\text{two tails}) = \frac{1}{4} \)
Answer: \( \frac{1}{4} \)

iv. Two heads

Step 1: Total possible outcomes = 4 (HH, HT, TH, TT)
Step 2: Favorable outcome for two heads = HH
Number of favorable outcomes = 1
Step 3: Probability \( P(\text{two heads}) = \frac{1}{4} \)
Answer: \( \frac{1}{4} \)

Q3: A letter is chosen from the word ‘PENCIL’. What is the probability that the letter chosen is a consonant?

Step 1: Write down all the letters in the word ‘PENCIL’: P, E, N, C, I, L
Step 2: Count total number of letters = 6
Step 3: Identify the consonants in the word:
Consonants are: P, N, C, L
Number of consonants = 4
Step 4: Probability of choosing a consonant = \[ P(\text{consonant}) = \frac{\text{Number of consonants}}{\text{Total letters}} = \frac{4}{6} \]Step 5: Simplify the fraction: \[ \frac{4}{6} = \frac{2}{3} \]Answer: \( \frac{2}{3} \)

Q4: A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:

i. a red ball

Step 1: Total number of balls in the bag = 3 (black, red, green)
Step 2: Number of favorable outcomes (drawing a red ball) = 1
Step 3: Probability \( P(\text{red ball}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3} \)
Answer: \( \frac{1}{3} \)

ii. not a red ball

Step 1: Total number of balls = 3
Step 2: Number of favorable outcomes (not red ball) = total balls – red balls = 3 – 1 = 2
Step 3: Probability \( P(\text{not red ball}) = \frac{2}{3} \)
Answer: \( \frac{2}{3} \)

iii. a white ball

Step 1: Total balls in the bag = 3 (black, red, green)
Step 2: Number of white balls = 0 (No white ball in the bag)
Step 3: Probability \( P(\text{white ball}) = \frac{0}{3} = 0 \)
Answer: 0

Q5: In a single throw of a die, find the probability of getting a number:

i. Greater than 2

Step 1: Total possible outcomes when a die is thrown = 6 (numbers 1, 2, 3, 4, 5, 6)
Step 2: Numbers greater than 2 are 3, 4, 5, 6
Number of favorable outcomes = 4
Step 3: Probability \( P(\text{number} \gt 2) = \frac{4}{6} \)
Step 4: Simplify the fraction: \[ \frac{4}{6} = \frac{2}{3} \]Answer: \( \frac{2}{3} \)

ii. Less than or equal to 2

Step 1: Numbers less than or equal to 2 are 1 and 2
Number of favorable outcomes = 2
Step 2: Probability \( P(\text{number} \leq 2) = \frac{2}{6} \)
Step 3: Simplify the fraction: \[ \frac{2}{6} = \frac{1}{3} \]Answer: \( \frac{1}{3} \)

iii. Not greater than 2

Step 1: “Not greater than 2” means the number is less than or equal to 2
Step 2: Numbers not greater than 2 are 1 and 2
Number of favorable outcomes = 2
Step 3: Probability \( P(\text{not greater than } 2) = \frac{2}{6} = \frac{1}{3} \)
Answer: \( \frac{1}{3} \)

Q6: A bag contains 3 white, 5 black and 2 red balls, all of the same size. A ball is drawn from the bag without looking into it. Find the probability that the ball drawn is:

i. a black ball

Step 1: Total number of balls = 3 (white) + 5 (black) + 2 (red) = 10
Step 2: Number of black balls = 5
Step 3: Probability \( P(\text{black ball}) = \frac{5}{10} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. a red ball

Step 1: Total number of balls = 10
Step 2: Number of red balls = 2
Step 3: Probability \( P(\text{red ball}) = \frac{2}{10} = \frac{1}{5} \)
Answer: \( \frac{1}{5} \)

iii. a white ball

Step 1: Total number of balls = 10
Step 2: Number of white balls = 3
Step 3: Probability \( P(\text{white ball}) = \frac{3}{10} \)
Answer: \( \frac{3}{10} \)

iv. not a red ball

Step 1: Total number of balls = 10
Step 2: Number of balls that are not red = Total balls – Red balls = 10 – 2 = 8
Step 3: Probability \( P(\text{not red ball}) = \frac{8}{10} = \frac{4}{5} \)
Answer: \( \frac{4}{5} \)

v. not a black ball

Step 1: Total number of balls = 10
Step 2: Number of balls that are not black = Total balls – Black balls = 10 – 5 = 5
Step 3: Probability \( P(\text{not black ball}) = \frac{5}{10} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

Q7: In a single throw of a die, find the probability that the number:

i. will be an even number

Step 1: Total possible outcomes when a die is thrown = 6 (numbers 1, 2, 3, 4, 5, 6)
Step 2: Even numbers on a die are 2, 4, 6
Number of favorable outcomes = 3
Step 3: Probability \( P(\text{even number}) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. will be an odd number

Step 1: Odd numbers on a die are 1, 3, 5
Number of favorable outcomes = 3
Step 2: Probability \( P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

iii. will not be an even number

Step 1: “Not an even number” means the number is odd
Step 2: Probability \( P(\text{not even number}) = P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

Q8: In a single throw of a die, find the probability of getting:

i. 8

Step 1: Total possible outcomes when a die is thrown = 6 (numbers 1, 2, 3, 4, 5, 6)
Step 2: Number 8 is not on a standard die
Step 3: Number of favorable outcomes = 0
Step 4: Probability \( P(8) = \frac{0}{6} = 0 \)
Answer: 0

ii. a number greater than 8

Step 1: Numbers greater than 8 do not exist on a standard die
Step 2: Number of favorable outcomes = 0
Step 3: Probability \( P(\text{number} \gt 8) = \frac{0}{6} = 0 \)
Answer: 0

iii. a number less than 8

Step 1: Numbers less than 8 on a die are 1, 2, 3, 4, 5, 6
Number of favorable outcomes = 6
Step 2: Probability \( P(\text{number} \lt 8) = \frac{6}{6} = 1 \)
Answer: 1

Q9: Which of the following cannot be the probability of an event?

i. \( \frac{2}{7} \)

Step 1: Probability of an event lies between 0 and 1 inclusive.
Check if \( \frac{2}{7} \) lies between 0 and 1: \[ \frac{2}{7} \approx 0.2857 \] Since \( 0 \leq 0.2857 \leq 1 \), it can be a probability.
Answer: Can be a probability

ii. 3.8

Step 1: Probability must be \( \geq 0 \) and \( \leq 1 \).
Since 3.8 \( > \) 1, it cannot be a probability.
Answer: Cannot be a probability

iii. 127%

Step 1: Convert percentage to decimal: \[ 127\% = \frac{127}{100} = 1.27 \] Since 1.27 \( > \) 1, it cannot be a probability.
Answer: Cannot be a probability

iv. -0.8

Step 1: Probability cannot be negative.
Since -0.8 \( < \) 0, it cannot be a probability.
Answer: Cannot be a probability

Q10: A bag contains six identical black balls. A boy withdraws one ball from the bag without looking into it. What is the probability that he takes out:

i. a white ball?

Step 1: Total number of balls in the bag = 6 (all black)
Step 2: Number of white balls = 0 (no white balls in the bag)
Step 3: Probability \( P(\text{white ball}) = \frac{0}{6} = 0 \)
Answer: 0

ii. a black ball?

Step 1: Total number of balls = 6
Step 2: Number of black balls = 6
Step 3: Probability \( P(\text{black ball}) = \frac{6}{6} = 1 \)
Answer: 1

Q11: Three identical coins are tossed together. What is the probability of obtaining:

i. all heads?

Step 1: Total possible outcomes when 3 coins are tossed = \( 2^3 = 8 \)
These outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Step 2: Number of favorable outcomes for all heads = 1 (HHH)
Step 3: Probability \( P(\text{all heads}) = \frac{1}{8} \)
Answer: \( \frac{1}{8} \)

ii. exactly two heads?

Step 1: Favorable outcomes with exactly two heads: HHT, HTH, THH
Number of favorable outcomes = 3
Step 2: Probability \( P(\text{exactly two heads}) = \frac{3}{8} \)
Answer: \( \frac{3}{8} \)

iii. exactly one head?

Step 1: Favorable outcomes with exactly one head: HTT, THT, TTH
Number of favorable outcomes = 3
Step 2: Probability \( P(\text{exactly one head}) = \frac{3}{8} \)
Answer: \( \frac{3}{8} \)

iv. no head?

Step 1: Favorable outcome with no head: TTT
Number of favorable outcomes = 1
Step 2: Probability \( P(\text{no head}) = \frac{1}{8} \)
Answer: \( \frac{1}{8} \)

Q12: A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?

Step 1: Total number of pages = 92
Step 2: Identify page numbers whose digits sum to 9.
Check page numbers from 1 to 92:
– For 1-digit pages (1 to 9), sum of digits = the number itself.
None equals 9 except page 9 itself.
– For 2-digit pages (10 to 92), sum digits as \(a + b = 9\), where \(a\) = tens digit, \(b\) = units digit.
Possible combinations:
\(a=1, b=8 \Rightarrow 18\)
\(a=2, b=7 \Rightarrow 27\)
\(a=3, b=6 \Rightarrow 36\)
\(a=4, b=5 \Rightarrow 45\)
\(a=5, b=4 \Rightarrow 54\)
\(a=6, b=3 \Rightarrow 63\)
\(a=7, b=2 \Rightarrow 72\)
\(a=8, b=1 \Rightarrow 81\)
\(a=9, b=0 \Rightarrow 90\)
All these pages are within 92 pages.
So the pages are:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Number of favorable pages = 10
Step 3: Probability \( P(\text{sum of digits} = 9) = \frac{\text{Number of favorable pages}}{\text{Total pages}} = \frac{10}{92} \)
Step 4: Simplify the fraction: \[ \frac{10}{92} = \frac{5}{46} \]Answer: \( \frac{5}{46} \)

Q13: Two coins are tossed together. What is the probability of getting:

i. at least one head?

Step 1: Total possible outcomes when two coins are tossed = \( 2^2 = 4 \)
The outcomes are: HH, HT, TH, TT
Step 2: Favorable outcomes for “at least one head” are: HH, HT, TH
Number of favorable outcomes = 3
Step 3: Probability \( P(\text{at least one head}) = \frac{3}{4} \)
Answer: \( \frac{3}{4} \)

ii. both heads or both tails?

Step 1: Favorable outcomes for both heads = HH (1 outcome)
Favorable outcomes for both tails = TT (1 outcome)
Step 2: Total favorable outcomes = 1 + 1 = 2
Step 3: Probability \( P(\text{both heads or both tails}) = \frac{2}{4} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

Q14: From 10 identical cards numbered 1, 2, 3, …, 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:

i. 2

Step 1: Total number of cards = 10
Step 2: Multiples of 2 between 1 and 10 are: 2, 4, 6, 8, 10
Number of multiples of 2 = 5
Step 3: Probability \( P(\text{multiple of 2}) = \frac{5}{10} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. 3

Step 1: Multiples of 3 between 1 and 10 are: 3, 6, 9
Number of multiples of 3 = 3
Step 2: Probability \( P(\text{multiple of 3}) = \frac{3}{10} \)
Answer: \( \frac{3}{10} \)

iii. 2 and 3

Step 1: Multiples of 6 between 1 and 10 are: 6
Number of multiples of 6 = 1
Step 2: Probability \( P(\text{multiple of 2 and 3}) = \frac{1}{10} \)
Answer: \( \frac{1}{10} \)

iv. 2 or 3

Step 1: Multiples of 2 are: 2, 4, 6, 8, 10 (5 numbers)
Multiples of 3 are: 3, 6, 9 (3 numbers)
Step 2: Numbers which are multiples of both 2 and 3 (i.e., multiples of 6) = 6 (1 number)
Step 3: Use the formula: \[ P(2 \text{ or } 3) = P(2) + P(3) – P(2 \text{ and } 3) \]\[ = \frac{5}{10} + \frac{3}{10} – \frac{1}{10} = \frac{7}{10} \]Answer: \( \frac{7}{10} \)

Q15: Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:

i. 0

Step 1: Minimum sum when two dice are thrown is 2 (1+1), so sum 0 is not possible.
Step 2: Number of favorable outcomes = 0
Step 3: Total possible outcomes = \(6 \times 6 = 36\)
Step 4: Probability \( P(\text{sum} = 0) = \frac{0}{36} = 0 \)
Answer: 0

ii. 12

Step 1: Sum 12 occurs only when both dice show 6.
Step 2: Number of favorable outcomes = 1 (6,6)
Step 3: Total possible outcomes = 36
Step 4: Probability \( P(\text{sum} = 12) = \frac{1}{36} \)
Answer: \( \frac{1}{36} \)

iii. less than 12

Step 1: Sum less than 12 means all sums from 2 to 11.
Step 2: Number of outcomes with sum 12 = 1 (already counted)
So number of outcomes with sum less than 12 = Total outcomes – outcomes with sum 12 = 36 – 1 = 35
Step 3: Probability \( P(\text{sum} < 12) = \frac{35}{36} \)
Answer: \( \frac{35}{36} \)

iv. less than or equal to 12

Step 1: Sum less than or equal to 12 means sum can be from 2 to 12.
Step 2: All possible sums of two dice fall between 2 and 12.
So number of favorable outcomes = Total outcomes = 36
Step 3: Probability \( P(\text{sum} \leq 12) = \frac{36}{36} = 1 \)
Answer: 1

Q16: A die is thrown once. Find the probability of getting:

i. a prime number

Step 1: Total possible outcomes when a die is thrown = 6 (numbers 1 to 6)
Step 2: Prime numbers between 1 and 6 are: 2, 3, 5
Number of favorable outcomes = 3
Step 3: Probability \( P(\text{prime number}) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. a number greater than 3

Step 1: Numbers greater than 3 on a die: 4, 5, 6
Number of favorable outcomes = 3
Step 2: Probability \( P(\text{number} > 3) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

iii. a number other than 3 and 5

Step 1: Total numbers on die = 6
Numbers excluded: 3, 5 (2 numbers)
Number of favorable outcomes = 6 – 2 = 4
Step 2: Probability \( P(\text{not 3 and not 5}) = \frac{4}{6} = \frac{2}{3} \)
Answer: \( \frac{2}{3} \)

iv. a number less than 6

Step 1: Numbers less than 6 on a die: 1, 2, 3, 4, 5
Number of favorable outcomes = 5
Step 2: Probability \( P(\text{number} < 6) = \frac{5}{6} \)
Answer: \( \frac{5}{6} \)

v. a number greater than 6

Step 1: Numbers greater than 6 on a die = none
Number of favorable outcomes = 0
Step 2: Probability \( P(\text{number} > 6) = \frac{0}{6} = 0 \)
Answer: 0

Q17: Two coins are tossed together. Find the probability of getting:

i. exactly one tail

Step 1: Total possible outcomes when two coins are tossed = \(2^2 = 4\)
Possible outcomes: HH, HT, TH, TT
Step 2: Favorable outcomes with exactly one tail: HT, TH
Number of favorable outcomes = 2
Step 3: Probability \(P(\text{exactly one tail}) = \frac{2}{4} = \frac{1}{2}\)
Answer: \( \frac{1}{2} \)

ii. at least one head

Step 1: Favorable outcomes for at least one head: HH, HT, TH
Number of favorable outcomes = 3
Step 2: Probability \(P(\text{at least one head}) = \frac{3}{4}\)
Answer: \( \frac{3}{4} \)

iii. no head

Step 1: Favorable outcome with no head: TT
Number of favorable outcomes = 1
Step 2: Probability \(P(\text{no head}) = \frac{1}{4}\)
Answer: \( \frac{1}{4} \)

iv. at most one head

Step 1: Outcomes with at most one head: HT, TH, TT
Number of favorable outcomes = 3
Step 2: Probability \(P(\text{at most one head}) = \frac{3}{4}\)
Answer: \( \frac{3}{4} \)

Q18: Two dice are thrown simultaneously, write all possible outcomes. Find:

Step 1: Each die has 6 faces numbered 1 to 6.
Total possible outcomes = \(6 \times 6 = 36\)
Step 2: List of all possible outcomes (ordered pairs):
(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)

i. Probability of getting same number on both dice

Step 1: Favorable outcomes where both dice show same number:
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
Number of favorable outcomes = 6
Step 2: Total possible outcomes = 36
Step 3: Probability \( P(\text{same number}) = \frac{6}{36} = \frac{1}{6} \)
Answer: \( \frac{1}{6} \)

ii. Probability of getting a sum 7 on the uppermost faces of both dice

Step 1: Favorable outcomes where sum = 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Number of favorable outcomes = 6
Step 2: Total possible outcomes = 36
Step 3: Probability \( P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6} \)
Answer: \( \frac{1}{6} \)