Maharashtra Board Class 10 Maths Chapter 5 Probability Set 5.4 Solutions

Question 1. If two coins are tossed, find the probability of the following events.
(i) Getting at least one head.
(ii) Getting no head.
Answer: Solution:
Sample space,
S = {HH, HT, TH, TT}
\( \therefore \) n(S) = 4
(i) Let A be the event of getting at least one head.
\( \therefore \) A = {HT, TH, HH}
\( \therefore \) n(A) = 3
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \)
\( \therefore \) P(A) = \( \frac{3}{4} \)
(ii) Let B be the event of getting no head.
\( \therefore \) B = {TT}
\( \therefore \) n(B) = 1
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \)
\( \therefore \) P(B) = \( \frac{1}{4} \)
\( \therefore \) P(A) = \( \frac{3}{4} \); P(B) = \( \frac{1}{4} \)
In simple words: When two coins are tossed, we list all possible outcomes to find the total sample space. Then, we identify outcomes corresponding to "at least one head" and "no head" to calculate their respective probabilities.

🎯 Exam Tip: Clearly define the sample space (S) and the number of outcomes n(S) before identifying specific events. Use correct notation for probability P(Event) = n(Event)/n(S).

Question 2. If two dice are rolled simultaneously, find the probability of the following events.
(i) The sum of the digits on the upper faces is at least 10.
(ii) The sum of the digits on the upper faces is 33.
(iii) The digit on the first die is greater than the digit on second die.
Answer: Solution:
Sample space,
s = {(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)}
\( \therefore \) n(S) = 36
(i) Let A be the event that the sum of the digits on the upper faces is at least 10.
\( \therefore \) A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
\( \therefore \) n(A) = 6
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} = \frac{6}{36} \)
\( \therefore \) P(A) = \( \frac{1}{6} \)
(ii) Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
\( \therefore \) Event B is an impossible event.
\( \therefore \) B = { }
\( \therefore \) n(B) = 0
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} = \frac{0}{36} \)
\( \therefore \) P(B) = 0
(iii) Let C be the event that the digit on the first die is greater than the digit on the second die.
C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5)}
\( \therefore \) n(C) = 15
\( \therefore \) P(C) = \( \frac{n(C)}{n(S)} = \frac{15}{36} \)
\( \therefore \) P(C) = \( \frac{5}{12} \)
\( \therefore \) P(A) = \( \frac{1}{6} \); P(B) = 0; P(C) = \( \frac{5}{12} \)
In simple words: For two dice rolls, we define the 36 possible outcomes. Then, we find the outcomes for sums of at least 10, an impossible sum of 33, and the first die being greater than the second, to calculate their respective probabilities.

🎯 Exam Tip: For problems involving two dice, accurately listing the 36-element sample space is crucial. Pay attention to "at least," "at most," "impossible event," and "greater than" conditions carefully.

Question 3. There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn:
(i) shows an even number.
(ii) shows a number which is a multiple of 5.
Answer: Solution:
Sample space,
S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15}
\( \therefore \) n(S) = 15
(i) Let A be the event that the ticket drawn shows an even number.
\( \therefore \) A = {2, 4, 6, 8, 10, 12, 14}
\( \therefore \) n(A) = 7
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \)
\( \therefore \) P(A) = \( \frac{7}{15} \)
(ii) Let B be the event that the ticket drawn shows a number which is a multiple of 5.
\( \therefore \) B = {5, 10, 15}
\( \therefore \) n(B) = 3
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} = \frac{3}{15} \)
\( \therefore \) P(B) = \( \frac{1}{5} \)
\( \therefore \) P(A) = \( \frac{7}{15} \); P(B) = \( \frac{1}{5} \)
In simple words: From a set of 15 numbered tickets, we first define the total possibilities. Then, we list the numbers that are even and those that are multiples of 5 to determine their individual probabilities.

🎯 Exam Tip: When drawing from a limited set, ensure all possible outcomes (sample space) are correctly identified. Distinguish between even numbers and multiples of a specific number for accurate event listing.

Question 4. A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
(i) an odd number?
(ii) a multiple of 5?
Answer: Solution:
Sample space
(S) = {23, 25, 27, 29,
32, 35, 37, 39,
52, 53, 57, 59,
72, 73, 75, 79,
92, 93, 95, 97}
\( \therefore \) n(S) = 20
(i) Let A be the event that the number formed is an odd number.
\( \therefore \) A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75,79,93,95,97}
\( \therefore \) n(A) = 16
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} = \frac{16}{20} \)
\( \therefore \) P(A) = \( \frac{4}{5} \)
(ii) Let B be the event that the number formed is a multiple of 5.
\( \therefore \) B = {25,35,75,95}
\( \therefore \) n(B) = 4
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} = \frac{4}{20} \)
\( \therefore \) P(B) = \( \frac{1}{5} \)
\( \therefore \) P(A) = \( \frac{4}{5} \); P(B) = \( \frac{1}{5} \)
In simple words: We first list all two-digit numbers that can be formed using the given digits without repetition. Then, we identify which of these numbers are odd and which are multiples of 5 to calculate their respective probabilities.

🎯 Exam Tip: When forming numbers with given digits, ensure all unique permutations are listed to define the sample space accurately. Carefully apply the conditions for odd numbers (ending in 3, 5, 7, 9) and multiples of 5 (ending in 5).

Question 5. A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is
(i) an ace.
(ii) a spade.
Answer: Solution:
There are 52 playing cards.
\( \therefore \) n(S) = 52
(i) Let A be the event that the card drawn is an ace.
\( \therefore \) n(A) = 4
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} = \frac{4}{52} \)
\( \therefore \) P(A) = \( \frac{1}{13} \)
(ii) Let B be the event that the card drawn is a spade.
\( \therefore \) n(B) = 13
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} = \frac{13}{52} \)
\( \therefore \) P(B) = \( \frac{1}{4} \)
\( \therefore \) P(A) = \( \frac{1}{13} \); P(B) = \( \frac{1}{4} \)
In simple words: From a standard deck of 52 cards, we identify the total number of aces and spades to calculate the probability of drawing each type of card.

🎯 Exam Tip: Remember the standard composition of a 52-card deck: 4 suits of 13 cards each, and 4 cards of each rank (e.g., 4 Aces). This knowledge is fundamental for card-related probability questions.