Maharashtra Board Class 10 Maths Chapter 5 Probability Set 5 Solutions

Question 1. Choose the correct alternative answer for each of the following questions.
(i) Which number cannot represent a probability?
(A) \( \frac{2}{3} \)
(B) 1.5
(C) 15%
(D) 0.7
Answer: (B) 1.5
In simple words: Probability values must always be between 0 and 1 (inclusive), or 0% and 100%. A value like 1.5 is greater than 1, so it cannot be a probability.

🎯 Exam Tip: Remember the fundamental range for probability: it's a value between 0 and 1. Any number outside this range (like a negative number or a number greater than 1) cannot be a valid probability.

Question 1. Choose the correct alternative answer for each of the following questions.
(ii) A die is rolled. What is the probability that the number appearing on upper face is less than 3?
(A) \( \frac{1}{6} \)
(B) \( \frac{1}{3} \)
(C) \( \frac{1}{2} \)
(D) 0
Answer: (B) \( \frac{1}{3} \)
In simple words: When a die is rolled, there are 6 possible outcomes. Numbers less than 3 are 1 and 2, so there are 2 favorable outcomes. The probability is the number of favorable outcomes divided by the total outcomes, which is \( \frac{2}{6} = \frac{1}{3} \).

🎯 Exam Tip: For problems involving dice, always list the complete sample space first, then identify the favorable outcomes. This systematic approach reduces errors.

Question 1. Choose the correct alternative answer for each of the following questions.
(iii) What is the probability of the event that a number chosen from 1 to 100 is a prime number?
(A) \( \frac{1}{5} \)
(B) 25
(C) \( \frac{1}{4} \)
(D) \( \frac{13}{50} \)
Answer: (C) \( \frac{1}{4} \)
Solution:
n(S) = 100
Let A be the event that the number chosen is a prime number.
Therefore, A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
Therefore, n(A) = 25
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{25}{100} = \frac{1}{4} \)
In simple words: There are 25 prime numbers between 1 and 100. Since there are 100 numbers in total, the probability of selecting a prime number is 25 out of 100, which simplifies to 1/4.

🎯 Exam Tip: When dealing with prime numbers, it's helpful to have the list of primes up to a certain range memorized or know how to quickly identify them to count favorable outcomes accurately.

Question 1. Choose the correct alternative answer for each of the following questions.
(iv) There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5?
(A) \( \frac{1}{5} \)
(B) \( \frac{3}{5} \)
(C) \( \frac{4}{5} \)
(D) \( \frac{1}{3} \)
Answer: (A) \( \frac{1}{5} \)
In simple words: Out of 40 cards, numbers that are multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40 (which are 8 numbers). The probability is 8 favorable outcomes out of 40 total outcomes, simplifying to 1/5.

🎯 Exam Tip: To find multiples of a number within a range, divide the upper limit by the number and take the floor (integer part) to get the count of multiples.

Question 1. Choose the correct alternative answer for each of the following questions.
(v) If n(A) = 2, P(A) = \( \frac{1}{5} \), then n(S) = ?
(A) 10
(B) \( \frac{5}{2} \)
(C) \( \frac{2}{5} \)
(D) \( \frac{1}{3} \)
Answer: (A) 10
Solution:
We know that \( P(A) = \frac{n(A)}{n(S)} \)

\( \implies \frac{1}{5} = \frac{2}{n(S)} \)

\( \implies n(S) = 2 \times 5 \)

\( \implies n(S) = 10 \)
In simple words: Probability is calculated as the ratio of favorable outcomes (n(A)) to the total possible outcomes (n(S)). Given n(A) and P(A), we can rearrange the formula to find n(S).

🎯 Exam Tip: Understand the relationship between n(A), n(S), and P(A). Any two of these values can be used to find the third using the probability formula.

Question 2. Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are \( \frac{4}{5} \), 0.83 and 58% respectively. Who had the greatest probability of success ?
Answer: Solution:
The probability that the ball is dropped in the basket by John = \( \frac{4}{5} \) = 0.80
The probability that the ball is dropped in the basket by Vasim = 0.83
The probability that the ball is dropped in the basket by Akash = 58% = \( \frac{58}{100} \) = 0.58
Comparing the probabilities:
0.83 > 0.80 > 0.58
Therefore, Vasim has the greatest probability of success.
In simple words: To compare probabilities given in different formats (fraction, decimal, percentage), convert all of them to a common format, usually decimal, then simply compare the decimal values to find the greatest one.

🎯 Exam Tip: Always convert all given probability values into the same format (decimal is usually easiest) before comparing them to avoid errors.

Question 3. In a hockey team there are 6 defenders, 4 offenders and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that:
(i) The goalie will be selected.
(ii) A defender will be selected.
Answer: Solution:
Total number of players in the hockey team
= 6 + 4 + 1 = 11
Therefore, n(S) = 11

(i) Let A be the event that the captain selected will be a goalie.
There is only one goalie in the hockey team.
Therefore, n(A) = 1
\( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{1}{11} \)

(ii) Let B be the event that the captain selected will be a defender.
There are 6 defenders in the hockey team.
Therefore, n(B) = 6
\( P(B) = \frac{n(B)}{n(S)} \)
Therefore, \( P(B) = \frac{6}{11} \)
Therefore, \( P(A) = \frac{1}{11} \); \( P(B) = \frac{6}{11} \)
In simple words: The total number of players is 11. The probability of selecting a goalie is 1 out of 11. The probability of selecting a defender is 6 out of 11.

🎯 Exam Tip: Clearly define the sample space (total possible outcomes) and the event space (favorable outcomes) for each sub-question. This organized approach ensures accurate calculations.

Question 4. Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?
Answer: Solution:
Each card bears an English alphabet.
Therefore, n(S) = 26
Let A be the event that the card drawn is a vowel card.
There are 5 vowels in English alphabets.
Therefore, A = {a, e, i, o, u}
Therefore, n(A) = 5
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{5}{26} \)
Therefore, The probability that the card drawn is a vowel card is \( \frac{5}{26} \).
In simple words: There are 26 alphabets in total, and 5 of them are vowels. So, the probability of drawing a vowel is 5 out of 26.

🎯 Exam Tip: When dealing with alphabets, remember the count of vowels and consonants. Clearly list the favorable outcomes for accurate counting.

Question 5. A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
(i) a red balloon.
(ii) a blue balloon,
(iii) a green balloon.
Answer: Solution:
Let the 2 red balloons be R1, R2,
3 blue balloons be B1, B2, B3, and
4 green balloons be G1, G2, G3, G4.
Therefore, Sample space
S = {R1, R2, B1, B2, B3, G1, G2, G3, G4}
Therefore, n(S) = 9

(i) Let A be the event that Pranali gets a red balloon.
Therefore, A = {R1, R2}
Therefore, n(A) = 2
\( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{2}{9} \)
Therefore, The probability that Pranali gets a red balloon is \( \frac{2}{9} \).

(ii) Let B be the event that Pranali gets a blue balloon.
Therefore, B = {B1, B2, B3}
Therefore, n(B) = 3
Therefore, \( P(B) = \frac{n(B)}{n(S)} = \frac{3}{9} \)
Therefore, \( P(B) = \frac{1}{3} \)
Therefore, The probability that Pranali gets a blue balloon is \( \frac{1}{3} \).

(iii) Let C be the event that Pranali gets a green balloon.
Therefore, C = {G1, G2, G3, G4}
Therefore, n(C) = 4
Therefore, \( P(C) = \frac{n(C)}{n(S)} \)
Therefore, \( P(C) = \frac{4}{9} \)
Therefore, The probability that Pranali gets a green balloon is \( \frac{4}{9} \).
In simple words: The total number of balloons is 9. The probability of getting a red balloon is 2/9, a blue balloon is 3/9 (or 1/3), and a green balloon is 4/9.

🎯 Exam Tip: When dealing with multiple categories of items, ensure the total sample space (n(S)) is correctly calculated by summing all items. Then, count favorable outcomes for each specific event.

Question 6. A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Answer: Solution:
Let 5 red pens be R1, R2, R3, R4, R5.
8 blue pens be B1, B2, B3, B4, B5, B6, B7, B8. and
3 green pens be G1, G2, G3.
Therefore, Sample space
S = {R1, R2, R3, R4, R5, B1, B2, B3, B4, B5, B6, B7, B8, G1, G2, G3}
Therefore, n(S) = 16
Let A be the event that Rutuja picks a blue pen.
Therefore, A = {B1, B2, B3, B4, B5, B6, B7, B8}
Therefore, n(A) = 8
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{8}{16} \)
Therefore, \( P(A) = \frac{1}{2} \)
Therefore, The probability that Rutuja picks a blue pen is \( \frac{1}{2} \).
In simple words: There are 5 + 8 + 3 = 16 pens in total. Since 8 of these pens are blue, the probability of picking a blue pen is 8 out of 16, which simplifies to 1/2.

🎯 Exam Tip: Clearly list the components of the sample space and the favorable event. Simplification of fractions is a common step and should always be done for final answers.

Question 7. Six faces of a die are as shown below.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक पासा (डाई) दिखाया गया है जिसके छह फलकों पर अक्षर अंकित हैं: A, B, C, D, E, और एक बार फिर A। यह दर्शाता है कि अक्षर 'A' दो बार पासे पर आता है, जबकि अन्य अक्षर एक-एक बार आते हैं।

If the die is rolled once, find the probability of
(i) 'A' appears on upper face.
(ii) 'D' appears on upper face.
Answer: Solution:
Sample space
S = {A, B, C, D, E, A}
Therefore, n(S) = 6

(i) Let R be the event that 'A' appears on the upper face.
Therefore, R = {A, A}
Therefore, n(R) = 2
\( P(R) = \frac{n(R)}{n(S)} = \frac{2}{6} \)
Therefore, \( P(R) = \frac{1}{3} \)

(ii) Let Q be the event that 'D' appears on the upper face.
Total number of faces having 'D' on it = 1
Q = {D}
Therefore, n(Q) = 1
Therefore, \( P(Q) = \frac{n(Q)}{n(S)} \)
Therefore, \( P(Q) = \frac{1}{6} \)
Therefore, \( P(R) = \frac{1}{3} \); \( P(Q) = \frac{1}{6} \)
In simple words: The die has 6 faces. The letter 'A' appears on 2 faces, so the probability of 'A' is 2/6 or 1/3. The letter 'D' appears on 1 face, so the probability of 'D' is 1/6.

🎯 Exam Tip: Carefully list all outcomes in the sample space, especially when faces of a die are non-standard. Double-counting or missing outcomes can lead to incorrect probabilities.

Question 8. A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(i) an odd number.
(ii) a complete square number.
Answer: Solution:
Sample space,
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
Therefore, n(S) = 30

(i) Let A be the event that the ticket drawn bears an odd number.
Therefore, A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}
Therefore, n(A) = 15
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{15}{30} \)
Therefore, \( P(A) = \frac{1}{2} \)

(ii) Let B be the event that the ticket drawn bears a complete square number.
Therefore, B = {1, 4, 9, 16, 25}
Therefore, n(B) = 5
Therefore, \( P(B) = \frac{n(B)}{n(S)} = \frac{5}{30} \)
Therefore, \( P(B) = \frac{1}{6} \)
Therefore, \( P(A) = \frac{1}{2} \); \( P(B) = \frac{1}{6} \)
In simple words: Out of 30 tickets, 15 are odd numbers, so the probability is 15/30 or 1/2. There are 5 perfect square numbers (1, 4, 9, 16, 25), so the probability is 5/30 or 1/6.

🎯 Exam Tip: For problems involving sequences of numbers, systematically list out the numbers in the sample space and then identify those that satisfy the event conditions (e.g., odd, prime, square) to ensure accuracy.

Question 9. Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयताकार बगीचा दिखाया गया है जिसकी लंबाई 77 मीटर और चौड़ाई 50 मीटर है। बगीचे के अंदर एक वृत्ताकार झील है जिसका व्यास 14 मीटर है। यह चित्र बगीचे और उसके अंदर झील के स्थानिक संबंध को दर्शाता है, जिसका उपयोग प्रायिकता की गणना के लिए किया जाएगा।

Answer: Solution:
Area of the rectangular garden
= length - breadth
= 77 - 50
Therefore, Area of the rectangular garden = 3850 sq.m
Radius of the lake = \( \frac{14}{2} \) = 7 m
Therefore, Area of circular lake = \( \pi r^2 = \frac{22}{7} \times 7 \times 7 \)
Therefore, Area of circular lake = 154 sq.m
Probability that the towel fell in the lake
= \( \frac{\text{Area of the lake}}{\text{Area of the garden}} = \frac{154}{3850} = \frac{1}{25} \)
Therefore, The probability of the event that the towel fell in the lake is \( \frac{1}{25} \).
In simple words: The total area of the garden is 3850 sq.m (77x50). The area of the circular lake is 154 sq.m (pi*r^2 with r=7). The probability of the towel falling in the lake is the ratio of the lake's area to the garden's area.

🎯 Exam Tip: For geometric probability problems, calculate the area of the favorable region and divide it by the total area of the sample space. Ensure units are consistent.

Question 10. In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at
(i) 8.
(ii) an odd number.
(iii) a number greater than 2.
(iv) a number less than 9.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक चक्रिका (स्पिनर) दिखाया गया है जिस पर 1 से 8 तक की संख्याएँ अंकित हैं। एक तीर (एरो) केंद्र से निकलता है जो घूमने के बाद इन संख्याओं में से किसी एक पर रुकता है। यह चित्र 8 समान रूप से संभावित परिणामों के साथ एक संभाव्यता खेल का प्रतिनिधित्व करता है।

Answer: Solution:
Sample space (S) = {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, n(S) = 8

(i) Let A be the event that the spinning arrow comes to rest at 8.
Therefore, A = {8}
Therefore, n(A) = 1
Therefore, \( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{1}{8} \)

(ii) Let B be the event that the spinning arrow comes to rest at an odd number.
Therefore, B = {1, 3, 5, 7}
Therefore, n(B) = 4
Therefore, \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{8} \)
Therefore, \( P(B) = \frac{1}{2} \)

(iii) Let C be the event that the spinning arrow comes to rest at a number greater than 2.
C = {3, 4, 5, 6, 7, 8}
n(C) = 6
Therefore, \( P(C) = \frac{n(C)}{n(S)} = \frac{6}{8} \)
Therefore, \( P(C) = \frac{3}{4} \)

(iv) Let D be the event that the spinning arrow comes to rest at a number less than 9.
Therefore, D = {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, n(D) = 8
Therefore, \( P(D) = \frac{n(D)}{n(S)} = \frac{8}{8} \)
Therefore, \( P(D) = 1 \)
Therefore, \( P(A) = \frac{1}{8} \); \( P(B) = \frac{1}{2} \); \( P(C) = \frac{3}{4} \); \( P(D) = 1 \)
In simple words: The total outcomes are 8. The probability for '8' is 1/8. For odd numbers (1,3,5,7), it's 4/8 or 1/2. For numbers greater than 2 (3,4,5,6,7,8), it's 6/8 or 3/4. For numbers less than 9 (all of them), it's 8/8 or 1.

🎯 Exam Tip: Always clearly define the sample space (S) and the specific event (A, B, C, D) before calculating probabilities for each scenario. This helps in correctly identifying n(S) and n(Event).

Question 11. There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
(i) a natural number.
(ii) a number less than 1.
(iii) a whole number.
(iv) a number greater than 5.
Answer: Solution:
Sample space (S) = {0, 1, 2, 3, 4, 5}
Therefore, n(S) = 6

(i) Let A be the event that the card drawn shows a natural number.
Therefore, A = {1, 2, 3, 4, 5}
Therefore, n(A) = 5
Therefore, \( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{5}{6} \)

(ii) Let B be the event that the card drawn shows a number less than 1.
Therefore, B = {0}
Therefore, n(B) = 1
Therefore, \( P(B) = \frac{n(B)}{n(S)} \)
Therefore, \( P(B) = \frac{1}{6} \)

(iii) Let C be the event that the card drawn shows a whole number.
Therefore, C = {0, 1, 2, 3, 4, 5}
Therefore, n(C) = 6
Therefore, \( P(C) = \frac{n(C)}{n(S)} = \frac{6}{6} \)
Therefore, \( P(C) = 1 \)

(iv) Let D be the event that the card drawn shows a number greater than 5.
Here, the greatest number is 5.
Therefore, Event D is an impossible event.
Therefore, D = { }
Therefore, n(D) = 0
Therefore, \( P(D) = \frac{n(D)}{n(S)} = \frac{0}{6} \)
Therefore, \( P(D) = 0 \)
Therefore, \( P(A) = \frac{5}{6} \); \( P(B) = \frac{1}{6} \); \( P(C) = 1 \); \( P(D) = 0 \)
In simple words: The numbers are 0, 1, 2, 3, 4, 5 (total 6). Natural numbers (1-5) give probability 5/6. Number less than 1 (only 0) gives 1/6. Whole numbers (0-5) give 6/6 or 1. Number greater than 5 (none) gives 0/6 or 0.

🎯 Exam Tip: Distinguish between natural numbers (starting from 1), whole numbers (starting from 0), and integers. An impossible event always has a probability of 0, and a certain event has a probability of 1.

Question 12. A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is:
(i) red.
(ii) not red.
(iii) either red or white.
Answer: Solution:
Let the three red balls be R1, R2, R3, three white balls be W1, W2, W3 and three green balls be G1, G2, G3.
Therefore, Sample space,
S = {R1, R2, R3, W1, W2, W3, G1, G2, G3}
Therefore, n(S) = 9

(i) Let A be the event that the ball drawn is red.
Therefore, A = {R1, R2, R3}
Therefore, n(A) = 3
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{9} \)
Therefore, \( P(A) = \frac{1}{3} \)

(ii) Let B be the event that the ball drawn is not red.
B = {W1, W2, W3, G1, G2, G3}
Therefore, n(B) = 6
Therefore, \( P(B) = \frac{n(B)}{n(S)} = \frac{6}{9} \)
Therefore, \( P(B) = \frac{2}{3} \)

(iii) Let C be the event that the ball drawn is red or white.
Therefore, C = {R1, R2, R3, W1, W2, W3}
Therefore, n(C) = 6
Therefore, \( P(C) = \frac{n(C)}{n(S)} = \frac{6}{9} \)
Therefore, \( P(C) = \frac{2}{3} \)
Therefore, \( P(A) = \frac{1}{3} \); \( P(B) = \frac{2}{3} \); \( P(C) = \frac{2}{3} \)
In simple words: With 9 balls total, 3 red, 3 white, 3 green: probability of red is 3/9 or 1/3. Probability of not red (white or green) is 6/9 or 2/3. Probability of red or white is 6/9 or 2/3.

🎯 Exam Tip: For "not an event" probabilities, calculate 1 minus the probability of the event. For "either/or" (union of events), sum the favorable outcomes and divide by the total, ensuring no overlap is double-counted.

Question 13. Each card bears one letter from the word 'mathematics'. The cards are placed on a table upside down. Find the probability that a card drawn bears the letter 'm'.
Answer: Solution:
Sample space
= {m, a, t, h, e, m, a, t, i, c, s}
Therefore, n(S) = 11
Let A be the event that the card drawn bears the letter 'm'
Therefore, A = {m, m}
Therefore, n(A) = 2
Therefore, \( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{2}{11} \)
Therefore, The probability that a card drawn bears letter 'm' is \( \frac{2}{11} \).
In simple words: The word 'mathematics' has 11 letters in total. The letter 'm' appears twice. So, the probability of drawing an 'm' is 2 out of 11.

🎯 Exam Tip: When counting letters from a word, remember to count repeated letters for the total sample space (n(S)) and for the specific event (n(A)).

Question 14. Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected doesn't like Kabaddi.
Answer: Solution:
Total number of students in the school = 200
Therefore, n(S) = 200
Number of students who like Kabaddi = 135
Therefore, Number of students who do not like Kabaddi
= 200 - 135 = 65
Let A be the event that the student selected does not like Kabaddi.
Therefore, n(A) = 65
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{65}{200} \)
Therefore, \( P(A) = \frac{13}{40} \)
Therefore, The probability that the student selected doesn't like kabaddi is \( \frac{13}{40} \).
In simple words: There are 200 students in total. 65 students (200 - 135) do not like Kabaddi. So, the probability of selecting a student who doesn't like Kabaddi is 65 out of 200, which simplifies to 13/40.

🎯 Exam Tip: When given total and favorable outcomes for one event, calculate the "not favorable" outcomes by subtracting from the total. Always simplify the resulting probability fraction to its lowest terms.

Question 15. A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a:
(i) prime number.
(ii) multiple of 4.
(iii) multiple of 11.
Answer: Solution:
Sample space
(S) = {10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}
Therefore, n(S) = 20

(i) Let A be the event that the number so formed is a prime number.
Therefore, A = {11, 13, 23, 31, 41, 43}
Therefore, n(A) = 6
Therefore, \( P(A) = \frac{n(A)}{n(S)} = \frac{6}{20} \)
Therefore, \( P(A) = \frac{3}{10} \)

(ii) Let B be the event that the number so formed is a multiple of 4.
Therefore, B = {12, 20, 24, 32, 40, 44}
Therefore, n(B) = 6
Therefore, \( P(B) = \frac{n(B)}{n(S)} = \frac{6}{20} \)
Therefore, \( P(B) = \frac{3}{10} \)

(iii) Let C be the event that the number so formed is a multiple of 11.
Therefore, C = {11, 22, 33, 44}
Therefore, n(C) = 4
Therefore, \( P(C) = \frac{n(C)}{n(S)} \)
Therefore, \( P(C) = \frac{4}{20} \)
Therefore, \( P(C) = \frac{1}{5} \)
Therefore, \( P(A) = \frac{3}{10} \); \( P(B) = \frac{3}{10} \); \( P(C) = \frac{1}{5} \)
In simple words: 20 two-digit numbers can be formed. 6 are prime (3/10 probability). 6 are multiples of 4 (3/10 probability). 4 are multiples of 11 (4/20 or 1/5 probability).

🎯 Exam Tip: When forming numbers with repetition allowed, be systematic in listing the sample space. For events like prime numbers or multiples, carefully check each number in the sample space against the criteria.

Question 16. The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Answer: Solution:
Sample space,
S = {(0, 0), (0, 1), (0, 2),
(1, 0), (1, 1), (1, 2),
(2, 0), (2, 1), (2, 2),
(3, 0), (3, 1), (3, 2),
(4, 0), (4, 1), (4, 2),
(5, 0), (5, 1), (5, 2),
...
Therefore, n(S) = 36
Let A be the event that the product of digits on the upper face is zero.
Therefore, A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)}
Therefore, n(A) = 11
Therefore, \( P(A) = \frac{n(A)}{n(S)} \)
Therefore, \( P(A) = \frac{11}{36} \)
Therefore, The probability that the product of the digits on the upper face is zero is \( \frac{11}{36} \).
In simple words: When a die with faces 0-5 is rolled twice, there are 36 possible outcomes. The product of the two rolls is zero if at least one of the rolls is zero. There are 11 such outcomes. So the probability is 11/36.

🎯 Exam Tip: When dealing with multiple rolls of a die, construct the full sample space as ordered pairs. For product-related events, systematically check which pairs satisfy the condition. The product is zero if and only if one of the numbers is zero.

P(A) = \( \frac{n(A)}{n(S)} \)
P(A) = \( \frac{11}{36} \)
The probability that the product of the digits on the upper face is zero is \( \frac{11}{36} \).
In simple words: When a die with faces numbered 0-5 is rolled twice, the total possible outcomes are 36. The outcomes where the product of the two numbers is zero occur when at least one of the rolls is zero, leading to 11 such possibilities. Therefore, the probability is 11/36.

🎯 Exam Tip: Remember to list all possible outcomes for the sample space and the favorable outcomes for the event carefully. A clear understanding of the sample space is crucial for accurate probability calculations.