CBSE Class 10 Maths HOTs Probability Set 01

Important Terms Related to Probability

Random Experiment: An experiment whose outcome is from the events that are completely known but whose exact outcome is unknown is called a random experiment.

Sample space: All possible outcomes of a random experiment forms the sample space of the random experiment. For example, if a coin is tossed twice the possible outcomes are HH, HT, TH or TT. These four possible outcomes forms the sample space of this random experiments.

Event: A desired outcome of the random experiment is known as an event of the random experiment.

Probability of an event: Let E is an event of a random experiment. The probability of the event E is written as P(E) and
\( P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of possible outcomes of the random experiment}} \)

Sure Event: Those events whose probability is one.

Impossible Event: Those events whose probability is zero.

Important Results related to probability: If E is any event of a random experiment then,
(i) \( 0 \le P(E) \le 1 \)
(ii) \( P(\text{not E}) = 1 - P(E) \)
(iii) \( P(E) + P(\text{not E}) = 1 \)
(iv) \( P(E) \) is never negative

Question. A box has 10 equal size cards. Of the 10 cards, 4 are blue, 3 are green, 2 are yellow and 1 is red. If a card is randomly drawn from the box, which is the colour that the card is most likely to have? 
(a) Red
(b) Blue
(c) Green
(d) Yellow
Answer: (b) Blue

Question. A coin is tossed twice. The probability of getting both heads is 
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{4} \)
(d) 1
Answer: (c) \( \frac{1}{4} \)
Sol. Sample space = {HH, HT, TH, TT}
Number of total possible outcomes = 4
Number of favourable outcome (both heads) = 1
\( \therefore \) Probability of getting both heads = \( \frac{1}{4} \)

Question. A fair dice is rolled. Probability of getting a number x such that \( 1 \le x \le 6 \), is
(a) 0
(b) > 1
(c) between 0 and 1
(d) 1
Answer: (d) 1
Sol. 1, \( \because \) It is a sure event.

Question. The sum of the probabilities of all elementary events of an experiment is p, then
(a) \( 0 < p < 1 \)
(b) \( 0 \le p < 1 \)
(c) \( p = 1 \)
(d) \( p = 0 \)
Answer: (c) \( p = 1 \)
Sol. \( p = 1 \)

Question. Match the columns:
(1) probability of sure event | (A) \( \frac{1}{2} \)
(2) Probability of an impossible event | (B) 0
(3) A and B are complementary events | (C) 1
| (D) \( P(B) = 1 - P(A) \)
| (E) \( P(A) = P(B) \)
(a) (1) → (A), (2) → (B), (3) → (C)
(b) (1) → (B), (2) → (A), (3) → (C)
(c) (1) → (C), (2) → (B), (3) → (E)
(d) (1) → (C), (2) → (B), (3) → (D)
Answer: (d) (1) → (C), (2) → (B), (3) → (D)
Sol. Probability facts.

Question. A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected randomwise, the probability that it is a black or red ball is
(a) \( \frac{5}{12} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{5}{7} \)
Answer: (c) \( \frac{2}{3} \)
Sol. Number of black or red balls = 5 + 3 = 8
\( \therefore \) Required probability = \( \frac{8}{12} = \frac{2}{3} \)

Question. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. Then, the number of rotten apples in the heap is
(a) 180
(b) 170
(c) 175
(d) 162
Answer: (d) 162
Sol. Total apples in the heap = 900
\( \therefore \) Total number of elementary events = 900
One rotten apple is randomly selected from this heap
\( P(\text{a rotten apple}) = \frac{\text{Favourable events}}{\text{Total elementary events}} \)
\( 0.18 = \frac{\text{Favourable events}}{900} \)

\( \implies \) Favourable events = \( 0.18 \times 900 = 162 \)
Hence, there are 162 rotten apples in the heap.

Question. From a well shuffled pack of cards, a card is drawn at random. The probability of getting a black queen is
(a) \( \frac{1}{13} \)
(b) \( \frac{2}{13} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{3}{13} \)
Answer: (c) \( \frac{1}{26} \)
Sol. Total number of ways to draw a card = 52
Number of ways to draw a black queen = 2
\( \therefore \) Probability of getting a black queen = \( \frac{2}{52} = \frac{1}{26} \).

Question. If three different coins are tossed together, then find the probability of getting two heads is
(a) \( \frac{1}{8} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{5}{8} \)
(d) \( \frac{1}{4} \)
Answer: (b) \( \frac{3}{8} \)
Sol. Three coins are tossed together.
Possible outcomes
{HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
Number of 2 heads together = 3
Probability of getting two heads = \( \frac{3}{8} \)

Question. A die is thrown once. Find the probability of getting a number less than 3 is
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{3} \)
Answer: (d) \( \frac{1}{3} \)
Sol. Total number of ways = 6
Number of ways to get a number less than 3 = 2
\( \therefore \) Required probability = \( \frac{2}{6} = \frac{1}{3} \)

Question. Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random. The probability that the number on the card taken out is an even number is
(a) \( \frac{1}{5} \)
(b) \( \frac{1}{18} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{6} \)
Answer: (c) \( \frac{1}{2} \)
Sol. Total number of cards = 18
Even numbers from 3 to 20 are 4, 6, 8, 10, 12, 14, 16, 18, 20 = 9 numbers
Probability that the number on the card taken out is an even number
\( = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{9}{18} = \frac{1}{2} \)

Question. Two coins are tossed simultaneously. The probability of getting exactly one head is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{3}{4} \)
(d) 1
Answer: (a) \( \frac{1}{2} \)
Sol. When two coins are tossed simultaneously
total number of outcomes = {HH, HT, TH, TT}
total number of outcomes = 4
Favourable outcomes = {HT, TH} = 2
Probability of getting exactly one head = \( \frac{2}{4} = \frac{1}{2} \)

Question. An unbiased die is thrown, the probability of getting an even number is
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{5} \)
Answer: (c) \( \frac{1}{2} \)
Sol. Total outcomes = {1, 2, 3, 4, 5, 6}
Favourable outcomes = {2, 4, 6}
\( \therefore \) Probability of getting an even number = \( \frac{3}{6} = \frac{1}{2} \)

Question. If the probability of winning a game is 0.07. The probability of losing it is
(a) 0.3
(b) 0.91
(c) 0.93
(d) 0.13
Answer: (c) 0.93
Sol. Probability of winning a game = 0.07
\( \therefore \) Probability of losing the game = 1 - Probability of winning the game = 1 - 0.07 = 0.93

Very Short Answer Type Questions 

Question. Two different dice are tossed together. Find the probability:
(i) of getting a doublet
(ii) of getting a sum 10, of the numbers on the two dice. 
Answer: Sol. Possible outcomes are
{(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)}
Total elementary events = 36
(i) Outcomes of doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, i.e. 6
P(getting a doublet) = \( \frac{6}{36} = \frac{1}{6} \)
(ii) Outcomes of getting a sum 10 = {(4, 6), (5, 5), (6, 4)}, i.e. 3
P(getting a sum 10) = \( \frac{3}{36} = \frac{1}{12} \)

Question. An integer is chosen at random between 1 and 100. Find the probability that it is:
(i) divisible by 8.
(ii) not divisible by 8. 
Answer: Sol. (i) The integers divisible by 8 between 1 and 100 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, i.e. 12
Total outcomes = 98
P(divisible by 8) = \( \frac{12}{98} = \frac{6}{49} \)
(ii) P(not divisible by 8) = \( \frac{98 - 12}{98} = \frac{86}{98} = \frac{43}{49} \)

Question. Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken card is
(i) a prime number less than 10.
(ii) a number which is a perfect square.
Answer: Sol. Total no. of cards = 46
Total no. of ways to select a card = 46
(i) Prime no. less than 10 in these cards are 5, 7
\( \therefore \) No. of ways to select a prime no. less than 10 = 2.
\( \therefore \) Probability that the number on the card is prime = \( \frac{2}{46} = \frac{1}{23} \).
(ii) No. which is a perfect square, i.e. 9, 16, 25, 36, 49.
No. of ways to select a card with perfect square = 5.
\( \therefore \) Probability = \( \frac{5}{46} \).

Question. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is \( \frac{1}{5} \). The probability of selecting a black marble at random from the same jar is \( \frac{1}{4} \). If the jar contains 11 green marbles, find the total number of marbles in the jar. 
Answer: Sol. Let the total number of marbles in the jar = x
Probability of selecting a blue marble + Probability of selecting a black marble + Probability of selecting a green marble = 1
\( \frac{1}{5} + \frac{1}{4} + \frac{11}{x} = 1 \)

\( \implies \) \( \frac{11}{x} = 1 - \frac{1}{5} - \frac{1}{4} \)

\( \implies \) \( \frac{11}{x} = \frac{20 - 4 - 5}{20} \)

\( \implies \) \( \frac{11}{x} = \frac{11}{20} \)

\( \implies \) \( x = 20 \)

Question. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Answer: Sol. Total number of cards = 52
Number of aces and kings = 4 + 4 = 8
Number of cards which are neither ace nor king = 44
Probability that the card drawn is neither an ace nor a king = \( \frac{44}{52} = \frac{11}{13} \)

Question. A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6. 
Answer: Sol. For a die, number of possible outcomes = 6
Prime numbers are 2, 3 and 5.
(i) Probability of getting a prime number = \( \frac{3}{6} = \frac{1}{2} \)
(ii) Probability of getting a number lies between 2 and 6 (i.e. 3, 4, 5).
= \( \frac{3}{6} = \frac{1}{2} \)

Question. One vowel is selected from letter of English alphabets. Find the probability that it is vowel of the word Permutation.
Answer: Sol. There are five vowels in English alphabets.
\( \therefore \) Total number of outcomes = 5
Now, vowels in word Permutation are E, U, A, I, O.
\( \therefore \) Favourable outcome = 5
\( \therefore \) Required probability = \( \frac{5}{5} = 1 \)

Short Answer Type Questions

Question. Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd number. 
Answer: Sol. Total events associated with the random experiment of throwing two dice are:
(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)
Total number of elementary events = 6 × 6 = 36
(i) Let A be the event of getting a sum less than 7 in the numbers obtained on the two dice.
\( \therefore \) Elementary events favourable to event A are:
(1, 1), (2, 1), (3,1), (4, 1), (5, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (4, 2)
Favourable number of elementary events = 15
Hence, required probability = \( \frac{15}{36} = \frac{5}{12} \)
(ii) Let A be the event of getting a product less than 16.
\( \therefore \) Elementary events favourable to event A are
(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), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2)

\( \implies \) Favourable number of elementary events = 25
Hence, required probability = \( \frac{25}{36} \)
(iii) Let A be the event of doublet of odd numbers.
Elementary events favourable to the event A are (1, 1), (3, 3), (5, 5)
\( \therefore \) Favourable number of elementary events = 3
\( \therefore \) Required probability = \( \frac{3}{36} = \frac{1}{12} \)

Question. A box contains 19 balls bearing numbers 1, 2, 3, ....., 19. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) a prime number?
(ii) divisible by 3 or 5?
(iii) neither divisible by 5 nor by 10?
(iv) an even number?
Answer: Sol. Total number of balls = 19
(i) Prime numbers from 1 to 19 are 2, 3, 5, 7, 11, 13, 17, 19 = Total 8 prime numbers
\( \therefore \) Probability of drawing a prime number = \( \frac{8}{19} \)
(ii) Numbers divisible by 3 or 5 are 3, 6, 9, 12, 15, 18, 10, 5 = Total 8 numbers
\( \therefore \) Probability of drawing a number divisible by 3 or 5 = \( \frac{8}{19} \)
(iii) Numbers divisible by 5 or 10 are 5, 10, 15 = Total 3
\( \therefore \) Numbers which are neither divisible by 5 nor 10 are 19 - 3 = 16
\( \therefore \) Probability of drawing a number which is neither divisible by 5 nor by 10 = \( \frac{16}{19} \)
(iv) Even numbers from 1 - 19 are 2, 4, 6, 8, 10, 12, 14, 16, 18 [Total 9 even numbers]
\( \therefore \) Probability of drawing an even number = \( \frac{9}{19} \)

Question. From a pack of 52 playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is
(i) a black queen (ii) a red card
(iii) a black jack
(iv) a picture card (jacks, queens and kings are picture cards)
Answer: Sol. From the total playing 52 cards, red coloured jacks, queen, kings and aces are removed (i.e., 2 jacks, 2 queens, 2 kings, 2 aces)
\( \therefore \) Remaining cards = 52 - 8 = 44
(i) Favourable cases for a black queen are 2 (i.e., queen of club or spade)
\( \therefore \) Probability of drawing a black queen = \( \frac{\text{Favourable cases for black queen}}{\text{Total possible cases}} = \frac{2}{44} = \frac{1}{22} \)
(ii) Favourable cases for red cards are 26 - 8 = 18 (as 8 cards have been removed) (i.e. 9 diamonds + 9 hearts)
\( \therefore \) Probability of drawing a red card = \( \frac{\text{Favourable cases for a red card}}{\text{Total possible cases}} = \frac{18}{44} = \frac{9}{22} \)
(iii) Favourable cases for a black jack are 2 (i.e. jacks of club or spade)
\( \therefore \) Probability of drawing a black jack = \( \frac{\text{Favourable cases for a black jack}}{\text{Total possible cases}} = \frac{2}{44} = \frac{1}{22} \)
(iv) Favourable cases for a picture card are 6 (i.e. 2 black jacks, queens and kings each)
\( \therefore \) Probability of drawing a picture card = \( \frac{\text{Favourable cases for a picture card}}{\text{Total possible cases}} = \frac{6}{44} = \frac{3}{22} \).

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Answer: Sol. Let the number of blue balls be x.
Total number of balls in the bag = 5 + x
\( \therefore \) Probability of drawing a red ball = \( \frac{5}{5 + x} \) and probability of drawing a blue ball = \( \frac{x}{5 + x} \)
Given probability of drawing a blue ball = 3 × probability of drawing a red ball

\( \implies \) \( 3 \times \frac{5}{5 + x} = \frac{x}{5 + x} \)

\( \implies \) 15 = x

\( \implies \) Number of blue balls = 15

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing:
(i) an ace
(ii) ‘2’ of spades
(iii) ‘10’ of a black suit.
Answer: Sol. Total number of cards = 52
(i) Number of ace = 4
\( \therefore \) Probability of drawing an ace = \( \frac{4}{52} = \frac{1}{13} \)
(ii) There is only one ‘2’ of spades
\( \therefore \) Probability of drawing a ‘2’ of spade = \( \frac{1}{52} \)
(iii) ‘10’ of a black suit there are two cards
\( \therefore \) Probability of drawing 10 of black suit = \( \frac{2}{52} = \frac{1}{26} \)

Question. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14
(iii) a number which is a perfect square
(iv) a prime number less than 20.
Answer: Sol. Total number of cards = 100
(i) Number of cards bearing even number = 50
\( \therefore \) Probability of drawing an even number = \( \frac{50}{100} = \frac{1}{2} \)
(ii) Total cards with number less than 14 = 12
\( \therefore \) Probability of drawing a number less than 14 = \( \frac{12}{100} = \frac{3}{25} \)
(iii) Total perfect squares {4, 9, 16, 25, 36, 49, 64, 81, 100} = 9
\( \therefore \) Probability of drawing a perfect square = \( \frac{9}{100} \)
(iv) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
\( \therefore \) Probability of drawing a prime number less than 20 is = \( \frac{8}{100} = \frac{2}{25} \)

Question. A two digit number is selected at random. Find the probability that it is multiple of 4 but not divisible by 5.
Answer: Sol. Total number of two digit numbers = 90
Numbers multiple of 4 but not divisible by 5 are 12, 16, 24, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, and 96
Favourable outcomes = 18
\( \therefore \) Required probability = \( \frac{18}{90} = \frac{1}{5} \).

Question. A number is selected at random from the numbers 3, 5, 5, 7, 7, 7, 9, 9, 9, 9. Find the probability that the selected number is their average. 
Answer: Sol. Average of 3, 5, 5, 7, 7, 7, 9, 9, 9, 9 is
\( = \frac{3 + 5 + 5 + 7 + 7 + 7 + 9 + 9 + 9 + 9}{10} \)
\( = \frac{3 + 10 + 21 + 36}{10} = \frac{70}{10} = 7 \)
P(the selected number is the average i.e. 7) = \( \frac{3}{10} \)

Question. If a number x is chosen from the number 1, 2, 3 and a number y is selected from the numbers 1, 4, 9. Find the probability that xy = 10. 
Answer: Sol. x can take values 1, 2 and 3.
y can take values 1, 4 and 9.
xy can take one of these values
[1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 2 × 9, 3 × 1, 3 × 4, 3 × 9]
= [1, 4, 9, 2, 8, 18, 3, 12, 27]
In none of these cases xy = 10
\( \therefore \) xy = 10 is an impossible event

\( \implies \) P(xy = 10) = 0

Question. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What is probability that \( x^2 \le 4 \)? 
Answer: Sol. If \( x^2 \le 4 \), then x can take values –2, –1, 0, 1, 2
\( \therefore \) Probability of x such that \( x^2 \le 4 = \frac{5}{7} \)

Long Answer Type Questions 

Question. Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25? 
Answer: Sol. Peter throws two different dice
Rina throws only one die
Both Peter and Rina want the same outcome 25 but their approach is not the same.
Peter will multiply the outcomes on the two dice whereas Rina will square the number obtained on throwing only one die.
let us calculate the probability in each case when a pair of die is thrown there are 36 elementary events as given below:
(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 product of two numbers on the two dice will be 25 if both the dice show number 5.
So, there is only one elementary event favourable to to get a ‘5’ on each die.
\( \therefore \) P(Peter getting 25) = \( \frac{1}{36} \)
Rina throws only one die on which she can get one of the six numbers 1, 2, 3, 4, 5, 6.
If she gets number 5 on the upper face of the die thrown, then the square of the number is 25.
\( \therefore \) \( P_2 \)(Rina getting a number whose square is 25) = \( \frac{1}{6} \)
Clearly \( P_2 > P_1 \).
So, Rina has better chance of getting the number 25.

Question. In a single throw of two dice, find the probability of:
(i) getting a total of 10
(ii) getting a total of 9 or 11
(iii) getting a sum greater than 9
(iv) getting a doublet of even numbers
(v) not getting the same number on the two dice.
Answer: Sol. The sample space for single throw of two dice is
{ (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) }
Now,
(i) P(Total of 10) = \( \frac{n\{(6, 4), (5, 5), (4, 6)\}}{36} = \frac{3}{36} = \frac{1}{12} \)
(ii) P(Total of 9 or 11) = \( \frac{n\{(6, 3), (5, 4), (4, 5), (3, 6), (6, 5), (5, 6)\}}{36} = \frac{6}{36} = \frac{1}{6} \)
(iii) P(Total greater than 9) = P(Total of 10 or 11 or 12) = \( \frac{n\{(6, 4), (5, 5), (4, 6), (6, 5), (5, 6), (6, 6)\}}{36} = \frac{6}{36} = \frac{1}{6} \)
(iv) P(Doublet of even numbers) = \( \frac{n\{(2, 2), (4, 4), (6, 6)\}}{36} = \frac{3}{36} = \frac{1}{12} \)
(v) P(not getting the same number on two dice) = 1 – P(getting the same number on two dice)
= \( 1 - \frac{n\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}}{36} = 1 - \frac{6}{36} = 1 - \frac{1}{6} = \frac{5}{6} \)

Question. A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is
(i) a prime number
(ii) a composite number 
(iii) a number divisible by 3
Answer: Sol. (i) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Number of prime numbers (favourable cases) = 8
Total possible outcomes = 20
P(prime number) = \( \frac{8}{20} = \frac{2}{5} \)
(ii) Composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20
Number of composite numbers (Favourable outcomes) = 11
Total possible outcomes = 20
P(composite numbers) = \( \frac{11}{20} \)
(iii) Numbers divisible by 3 are 3, 6, 9, 12, 15 and 18
Number of favourable outcomes = 6
Total possible outcomes = 20
P(number divisible by 3) = \( \frac{6}{20} = \frac{3}{10} \)

Question. The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well shuffled. A card is selected from the remaining cards. Find the probability of getting a card
(i) of spade (ii) of black king
(iii) of club (iv) of jacks 
Answer: Sol. King, Queen and Jack of clubs are removed from a pack of 52 cards.
Remaining cards = 52 – 3 = 49
(i) Number of spades = 13
P(spade) = \( \frac{13}{49} \)
(ii) Number of black king = 1
P(black king) = \( \frac{1}{49} \)
(iii) Number of clubs = 10
P(clubs) = \( \frac{10}{49} \)
(iv) Number of Jacks = 3
P(Jacks) = \( \frac{3}{49} \)

Question. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 2, 2, 3, 3, 4 respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting (i) sum 7 (ii) sum is a perfect square. 
Answer: Sol. When these two dice are thrown then total possible outcomes are
\( \therefore \) Total possible outcomes = 36
(i) Sum of number = 7
Favourable outcomes are (3, 4), (4, 3), (4, 3), (5, 2), (5, 2), (6, 1)
\( \therefore \) Favourable ways = 6
\( \therefore \) Probability that sum of number is 7 = \( \frac{6}{36} = \frac{1}{6} \)
(ii) Sum is a perfect square i.e., sum is 4 or 9 (\( \because \) Maximum sum is 10)
Favourable outcomes are (1, 3), (1, 3), (3, 1), (2, 2), (2, 2), (5, 4), (6, 3), (6, 3) = 8 outcomes.
\( \therefore \) Probability = \( \frac{8}{36} = \frac{2}{9} \)

Question. A box contains cards, number from 1 to 90. A card is drawn at random from the box. Find the probability that the selected card bears a
(i) two digit number.
(ii) perfect square number. 
Answer: Sol. Total cards = 90
Two digit numbers are from 10 to 90
Number of two digit cards = 81
(i) Probability of getting 2-digit number = \( \frac{81}{90} = \frac{9}{10} \)
(ii) Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.
\( \therefore \) number of perfect square = 9
Probability of getting perfect square = \( \frac{9}{90} = \frac{1}{10} \)

Question. What is the probability that a leap year will have 53 Mondays and 53 Tuesdays?
Answer: Sol. Number of days in a leap year = 366
366 days = 52 weeks + 2 days
Now, probability of having 53 Mondays and 53 Tuesdays is equal to probability of having a Monday and Tuesday in these 2 days.
The remains two days can be
(Sunday and Monday)
(Monday and Tuesday), (Thursday and Friday)
(Tuesday and Wednesday), (Friday and Saturday)
(Wednesday and Thursday), (Saturday and Sunday)
Total outcomes = 7
Favourable outcome = 1
\( \therefore \) Required probability = \( \frac{1}{7} \)

Question. What is the probability that month of December had exactly 5 Sundays, 5 Mondays and 5 Tuesdays?
Answer: Sol. Number of days in the month of December = 31
31 days = 4 weeks + 3 days
Now, probability of 5 Sundays, 5 Mondays and 5 Tuesdays is equal to probability of getting a Sunday, Monday and Tuesday in remaining 3 days.
The remaining 3 days can be
(Sunday, Monday and Tuesday),
(Monday, Tuesday and Wednesday),
(Tuesday, Wednesday and Thursday),
(Wednesday, Thursday and Friday),
(Thursday, Friday and Saturday),
(Friday, Saturday and Sunday),
(Saturday, Sunday and Monday),
No. of total outcomes = 7
No. of favourable outcome = 1
\( \therefore \) Required probability = \( \frac{1}{7} \)

Question. A three digit number is chosen. Find the probability at all three digits are not same.
Answer: Sol. Total number of three digits numbers = 900
\( \therefore \) Total number of outcomes = 900
Three digits number in which all the digits are same 111, 222, 333, 444, 555, 666, 777, 888, 999.
Number of outcomes when all three digits are same = 9
P(three digit number when all three digits are same) = \( \frac{9}{900} = \frac{1}{100} \)
\( \therefore \) P(three digit number when all three digits are not same)
= 1 – P(three digit number when all three digits are same)
= \( 1 - \frac{1}{100} = \frac{99}{100} \)