Maharashtra Board Class 11 Maths Part 2 Chapter 7 Probability 7.2 Solutions

Question 1. A fair die is thrown two times. Find the chance that
(i) product of the numbers on the upper face is 12.
(ii) sum of the numbers on the upper face is 10.
(iii) sum of the numbers on the upper face is at least 10.
(iv) sum of the numbers on the upper face is 4.
(v) the first throw gives an odd number and the second throw gives a multiple of 3.
(vi) both the times die to show the same number (doublet).
Answer: Solution: If a fair die is thrown twice, the sample space is 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 product of the numbers on uppermost face is 12.
\( \therefore \) A = {(2, 6), (3, 4), (4, 3), (6, 2)}
\( \therefore \) n(A) = 4
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{4}{36} \) = \( \frac{1}{9} \)
(ii) Let B be the event that sum of the numbers on uppermost face is 10.
\( \therefore \) B = {(4, 6), (5, 5), (6, 4)}
\( \therefore \) n(B) = 3
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{3}{36} \) = \( \frac{1}{12} \)
(iii) Let C be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
\( \therefore \) C = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
\( \therefore \) n(C) = 6
\( \therefore \) P(C) = \( \frac{n(C)}{n(S)} \) = \( \frac{6}{36} \) = \( \frac{1}{6} \)
(iv) Let D be the event that sum of the numbers on uppermost face is 4.
\( \therefore \) D = {(1, 3), (2, 2), (3, 1)}
\( \therefore \) n(D) = 3
\( \therefore \) P(D) = \( \frac{n(D)}{n(S)} \) = \( \frac{3}{36} \) = \( \frac{1}{12} \)
(v) Let E be the event that 1st throw gives an odd number and 2nd throw gives multiple of 3.
\( \therefore \) E = {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6)}
\( \therefore \) n(E) = 6
\( \therefore \) P(E) = \( \frac{n(E)}{n(S)} \) = \( \frac{6}{36} \) = \( \frac{1}{6} \)
(vi) Let F be the event that both times die shows same number.
\( \therefore \) F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
\( \therefore \) n(F) = 6
\( \therefore \) P(F) = \( \frac{n(F)}{n(S)} \) = \( \frac{6}{36} \) = \( \frac{1}{6} \)
In simple words: This question calculates the probability of various outcomes when rolling a fair die twice by first defining the sample space and then identifying the favorable outcomes for each specific event.

🎯 Exam Tip: Clearly define the sample space (S) and the number of outcomes (n(S)) before calculating the number of favorable outcomes (n(Event)) for each part to avoid errors.

Question 2. Two cards are drawn from a pack of 52 cards. Find the probability that
(i) both are black.
(ii) both are diamonds.
(iii) both are ace cards.
(iv) both are face cards.
(v) one is a spade and the other is a non-spade.
(vi) both are from the same suit.
(vii) both are from the same denomination.
Answer: Solution: Two cards can be drawn from a pack of 52 cards in \( {^{52}}C_2 \) ways.
\( \therefore \) n(S) = \( {^{52}}C_2 \)
(i) Let A be the event that both the cards drawn are black. The pack of 52 cards contains 26 black cards.
\( \therefore \) 2 cards can be drawn from them in \( {^{26}}C_2 \) ways
\( \therefore \) n(A) = \( {^{26}}C_2 \)
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{{^{26}}C_2}{{^{52}}C_2} \)
(ii) Let B be the event that both the cards drawn are diamond. There are 13 diamond cards in a pack of 52 cards.
\( \therefore \) 2 diamond cards can be drawn from 13 diamond cards in \( {^{13}}C_2 \) ways
\( \therefore \) n(B) = \( {^{13}}C_2 \)
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{{^{13}}C_2}{{^{52}}C_2} \)
(iii) Let C be the event that both the cards drawn are aces. In a pack of 52 cards, there are 4 ace cards.
\( \therefore \) 2 ace cards can be drawn from 4 ace cards in \( {^{4}}C_2 \) ways
\( \therefore \) n(C) = \( {^{4}}C_2 \)
\( \therefore \) P(C) = \( \frac{n(C)}{n(S)} \) = \( \frac{{^{4}}C_2}{{^{52}}C_2} \)
(iv) Let D be the event that both the cards drawn are face cards. There are 12 face cards in a pack of 52 cards.
\( \therefore \) 2 face cards can be drawn from 12 face cards in \( {^{12}}C_2 \) ways.
\( \therefore \) n(D) = \( {^{12}}C_2 \)
\( \therefore \) P(D) = \( \frac{n(D)}{n(S)} \) = \( \frac{{^{12}}C_2}{{^{52}}C_2} \)
(v) Let E be the event that out of the two cards drawn one is a spade and other is non-spade. There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
\( \therefore \) One spade card can be drawn from 13 spade cards in \( {^{13}}C_1 \) ways and one non- spade card can be drawn from 39 non-spade cards in \( {^{39}}C_1 \) ways.
\( \therefore \) n(E) = \( {^{13}}C_1 \cdot {^{39}}C_1 \)
\( \therefore \) P(E) = \( \frac{n(E)}{n(S)} \) = \( \frac{{^{13}}C_1 \cdot {^{39}}C_1}{{^{52}}C_2} \)
(vi) Let F be the event that both the cards drawn are of the same suit. A pack of 52 cards consists of 4 suits each containing 13 cards. 2 cards can be drawn from a suit in \( {^{13}}C_2 \) ways. A suit can be selected in 4 ways.
\( \therefore \) n(F) = \( {^{13}}C_2 \) x 4
\( \therefore \) P(F) = \( \frac{n(F)}{n(S)} \) = \( \frac{4 \times {^{13}}C_2}{{^{52}}C_2} \)
(vii) Let G be the event that both the cards drawn are of same denominations. A pack of cards has 13 denominations and 4 different cards for each denomination
\( \therefore \) n(G) = 13 \( \times {^{4}}C_2 \)
\( \therefore \) P(G) = \( \frac{n(G)}{n(S)} \) = \( \frac{13 \times {^{4}}C_2}{{^{52}}C_2} \)
In simple words: This question uses combinations to find the probabilities of drawing specific types of cards (like black, diamond, ace, face cards, or cards of the same suit/denomination) from a standard 52-card deck.

🎯 Exam Tip: Remember the total number of items in each category (e.g., 26 black cards, 13 diamonds) and use the combination formula \( {^{n}}C_r \) for selecting items without regard to order.

Question 3. Four cards are drawn from a pack of 52 cards. Find the probability that
(i) 3 are Kings and 1 is Jack.
(ii) all the cards are from different suits.
(iii) at least one heart.
(iv) all cards are club and one of them is a jack.
Answer: Solution: 4 cards can be drawn out of 52 cards in \( {^{52}}C_4 \) ways.
\( \therefore \) n(S) = \( {^{52}}C_4 \)
(i) Let A be the event that out of the four cards drawn, 3 are kings and 1 is a jack. There are 4 kings and 4 jacks in a pack of 52 cards.
\( \therefore \) 3 kings can be drawn from 4 kings in \( {^{4}}C_3 \) ways. Similarly, 1 jack can be drawn out of 4 jacks in \( {^{4}}C_1 \) ways.
\( \therefore \) Total number of ways in which 3 kings and 1 jack can be drawn is \( {^{4}}C_3 \times {^{4}}C_1 \)
\( \therefore \) n(A) = \( {^{4}}C_3 \times {^{4}}C_1 \)
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{{^{4}}C_3 \times {^{4}}C_1}{{^{52}}C_4} \)
(ii) Let B be the event that all the cards drawn are of different suits. A pack of 52 cards consists of 4 suits each containing 13 cards.
\( \therefore \) A card can be drawn from each suit in \( {^{13}}C_1 \) ways.
\( \therefore \) 4 cards can be drawn from 4 different suits in \( {^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1 \) ways.
\( \therefore \) n(B) = \( {^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1 \)
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{{^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1 \times {^{13}}C_1}{{^{52}}C_4} \)
(iii) Let C be the event that out of the four cards drawn at least one is a heart.
\( \therefore \) C' is the event that all 4 cards drawn are non-heart cards. In a pack of 52 cards, there are 39 non-heart cards.
\( \therefore \) 4 non-heart cards can be drawn in \( {^{39}}C_4 \) ways.
\( \therefore \) n(C') = \( {^{39}}C_4 \)
\( \therefore \) P(C') = \( \frac{n(C')}{n(S)} \) = \( \frac{{^{39}}C_4}{{^{52}}C_4} \)
\( \therefore \) P(C) = 1 – P(C') = 1 \( - \frac{{^{39}}C_4}{{^{52}}C_4} \)
(iv) Let D be the event that all the 4 cards drawn are clubs and one of them is a jack. In a pack of 52 cards, there are 13 club cards having 1 jack card.
\( \therefore \) 1 jack can be drawn in \( {^{1}}C_1 \) way and the other 3 cards can be drawn from remaining 12 club cards in \( {^{12}}C_3 \) ways.
\( \therefore \) n(D) = \( {^{12}}C_3 \times {^{1}}C_1 \)
\( \therefore \) P(D) = \( \frac{n(D)}{n(S)} \) = \( \frac{{^{12}}C_3 \times {^{1}}C_1}{{^{52}}C_4} \)
In simple words: This problem uses combinations to determine probabilities of drawing specific card combinations, such as having a certain number of kings and jacks, drawing cards from different suits, having at least one heart, or drawing only clubs with a jack.

🎯 Exam Tip: For "at least one" probabilities, it's often easier to calculate the probability of the complementary event (none of that type) and subtract it from 1.

Question 4. A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is 1/3. The probability of yellow is 1/5.
(i) What is the probability of blackball?
(ii) How many balls are green, black, and yellow?
Answer: Solution:
(i) The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow (Y)
\( \therefore \) P(G) = \( \frac{1}{3} \) and P(Y) = \( \frac{1}{5} \) If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
\( \therefore \) P(G) + P(B) + P(Y) = 1
\( \frac{1}{3} \) + P(B) + \( \frac{1}{5} \) = 1
P(B) + \( \frac{8}{15} \) = 1
\( \therefore \) P(B) = 1 - \( \frac{8}{15} \) = \( \frac{7}{15} \)
\( \therefore \) Probability of black ball is \( \frac{7}{15} \)
(ii) Total number of balls = 15 and P(G) = \( \frac{1}{3} \), P(B) = \( \frac{7}{15} \), P(Y) = \( \frac{1}{5} \)
\( \therefore \) number of green balls = \( \frac{1}{3} \times 15 = 5 \) number of black balls = \( \frac{7}{15} \times 15 = 7 \) and number of yellow balls = \( \frac{1}{5} \times 15 = 3 \).
In simple words: This problem calculates the probability of drawing a black ball by subtracting the probabilities of green and yellow balls from 1, and then determines the actual count of each color of ball in the bag based on their probabilities and the total number of balls.

🎯 Exam Tip: Remember that the sum of probabilities of all possible outcomes in an event must equal 1. This rule is crucial for finding missing probabilities.

Question 5. A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that the
(i) number on the ticket is divisible by 6?
(ii) number on the ticket is a perfect square?
(iii) number on the ticket is prime?
(iv) number on the ticket is divisible by 3 and 5?
Answer: Solution: The box contains 75 tickets numbered 1 to 75.
\( \therefore \) 1 ticket can be drawn from the box in \( {^{75}}C_1 \) = 75 ways.
\( \therefore \) n(S) = 75
(i) Let A be the event that number on ticket is divisible by 6.
\( \therefore \) A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
\( \therefore \) n(A) = 12
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{12}{75} \) = \( \frac{4}{25} \)
(ii) Let B be the event that number on ticket is a perfect square.
\( \therefore \) B = {1, 4, 9, 16, 25, 36, 49, 64}
\( \therefore \) n(B) = 8
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{8}{75} \)
(iii) Let C be the event that the number on the ticket is a prime number.
\( \therefore \) C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
\( \therefore \) n(C) = 21
\( \therefore \) P(C) = \( \frac{n(C)}{n(S)} \) = \( \frac{21}{75} \) = \( \frac{7}{25} \)
(iv) Let D be the event that number on ticket is divisible by 3 and 5 i.e., divisible by L.C.M. of 3 and 5 i.e., 15
\( \therefore \) D = {15, 30, 45, 60, 75}
\( \therefore \) n(D) = 5
\( \therefore \) P(D) = \( \frac{n(D)}{n(S)} \) = \( \frac{5}{75} \) = \( \frac{1}{15} \)
In simple words: This question explores probabilities related to numbers on tickets, calculating the likelihood of a drawn ticket having a number divisible by 6, being a perfect square, being a prime number, or being divisible by both 3 and 5.

🎯 Exam Tip: When listing prime numbers, ensure you start from 2, as it is the smallest and only even prime number. For divisibility by two numbers, find their LCM.

Question 6. From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains
(i) 3 boys and 2 girls
(ii) at least 3 boys.
Answer: Solution: The group consists of 8 boys and 5 girls i.e., 8 + 5 = 13 persons. A committee of 5 is to be formed from this group.
\( \therefore \) 5 persons from 13 persons can be selected in \( {^{13}}C_5 \) ways
\( \therefore \) n(S) = \( {^{13}}C_5 \)
(i) Let A be the event that the committee contains 3 boys and 2 girls. 3 boys from 8 boys can be selected in \( {^{8}}C_3 \) ways and 2 girls from 5 girls can be selected in \( {^{5}}C_2 \) ways
\( \therefore \) n(A) = \( {^{8}}C_3 \cdot {^{5}}C_2 \)
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{{^{8}}C_3 \cdot {^{5}}C_2}{{^{13}}C_5} \)
(ii) Let B be the event that the committee contains at least 3 boys (i.e., 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
\( \therefore \) n(B) = \( {^{8}}C_3 \cdot {^{5}}C_2 + {^{8}}C_4 \cdot {^{5}}C_1 + {^{8}}C_5 \cdot {^{5}}C_0 \)
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{{^{8}}C_3 \cdot {^{5}}C_2 + {^{8}}C_4 \cdot {^{5}}C_1 + {^{8}}C_5 \cdot {^{5}}C_0}{{^{13}}C_5} \)
In simple words: This problem calculates the probability of forming a committee with specific boy-girl compositions from a larger group, covering both exact counts and "at least" conditions using combinations.

🎯 Exam Tip: For "at least" conditions, break down the event into mutually exclusive cases (e.g., exactly 3, exactly 4, exactly 5 boys) and sum their probabilities.

Question 7. A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in a socket. What is the probability that the room is lit?
Answer: Solution: Total number of bulbs = 10 Number of defective bulbs = 6
\( \therefore \) Number of non-defective bulbs = 4 3 bulbs can be selected out of 10 light bulbs in \( {^{10}}C_3 \) ways.
\( \therefore \) n(S) = \( {^{10}}C_3 \) Let A be the event that room is lit.
\( \therefore \) A' is the event that the room is not lit. For A' the bulbs should be selected from the 6 defective bulbs. This can be done in \( {^{6}}C_3 \) ways.
\( \therefore \) n(A') = \( {^{6}}C_3 \)
\( \therefore \) P(A') = \( \frac{n(A')}{n(S)} \) = \( \frac{{^{6}}C_3}{{^{10}}C_3} \)
\( \therefore \) P(Room is lit) = 1 – P(Room is not lit)
\( \therefore \) P(A) = 1 - P(A')
= 1 \( - \frac{{^{6}}C_3}{{^{10}}C_3} \)
= 1 \( - \frac{6 \times 5 \times 4}{10 \times 9 \times 8} \)
= 1 \( - \frac{1}{6} \)
= \( \frac{5}{6} \)
In simple words: To find the probability that a room is lit when 3 bulbs are chosen from a mix of working and defective ones, it's easier to calculate the probability that the room is NOT lit (meaning all chosen bulbs are defective) and subtract that from 1.

🎯 Exam Tip: When dealing with "at least one working" scenarios, calculate the probability of the complementary event (all defective) and subtract it from 1 for efficiency.

Question 8. The letters of the word LOGARITHM are arranged at random. Find the probability that
(i) Vowels are always together.
(ii) Vowels are never together.
(iii) Exactly 4 letters between G and H
(iv) begins with O and ends with T
(v) Start with a vowel and ends with a consonant.
Answer: Solution: There are 9 letters in the word LOGARITHM. These letters can be arranged among themselves in \( {^{9}}P_9 \) = 9! ways.
\( \therefore \) n(S) = 9!
(i) Let A be the event that vowels are always together. The word LOGARITHM consists of 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M). 3 vowels can be arranged among themselves in = \( {^{3}}P_3 \) = 3! ways. Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7) can be arranged in \( {^{7}}P_7 \) = 7! ways.
\( \therefore \) n(A) = 3! \( \times \) 7!
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \) = \( \frac{3! \times 7!}{9!} \)
(ii) Let B be the event that vowels are never together. Consider the following arrangement _C_C_C_C_C_C_ 6 consonants create 7 gaps.
\( \therefore \) 3 vowels can be arranged in 7 gaps in \( {^{7}}P_3 \) ways. Also 6 consonants can be arranged among themselves in \( {^{6}}P_6 \) = 6! ways.
\( \therefore \) n(B) = 6! \( \times {^{7}}P_3 \)
\( \therefore \) P(B) = \( \frac{n(B)}{n(S)} \) = \( \frac{6! \times {^{7}}P_3}{9!} \)
(iii) Let C be the event that exactly 4 letters are arranged between G and H. Consider the following arrangement 123456789
\( \therefore \) Out of 9 places, G and H can occupy any one of following 4 positions in 4 ways. 1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th Now, G and H can be arranged among themselves in \( {^{2}}P_2 \) = 2! =2 ways. Also, the remaining 7 letters can be arranged in remaining 7 places in \( {^{7}}P_7 \) = 7! ways.
\( \therefore \) n(C) = 4 \( \times \) 2 \( \times \) 7! = 8 \( \times \) 7! = 8!
\( \therefore \) P(C) = \( \frac{n(C)}{n(S)} \) = \( \frac{8!}{9!} \) = \( \frac{8!}{9 \times 8!} \) = \( \frac{1}{9} \)
(iv) Let D be the event that word begins with O and ends with T. Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in \( {^{7}}P_7 \) = 7! ways
\( \therefore \) n(D) = 7! \( \times \) 1 \( \times \) 1 = 7!
\( \therefore \) P(D) = \( \frac{n(D)}{n(S)} \) = \( \frac{7!}{9!} \)
(v) Let E be the event that word starts with vowel and ends with consonant. There are 3 vowels and 6 consonants in the word LOGARITHM.
\( \therefore \) The first place can be filled in 3 different ways and the last place can be filled in 6 ways. Now, remaining 7 letters can be arranged in 7 places in \( {^{7}}P_7 \) = 7! ways
\( \therefore \) n(E) = 3 \( \times \) 6 \( \times \) 7!
\( \therefore \) P(E) = \( \frac{n(E)}{n(S)} \) = \( \frac{3 \times 6 \times 7!}{9!} \)
In simple words: This question calculates the probabilities of different arrangements of letters in the word LOGARITHM based on conditions like vowels staying together, vowels never together, specific letters being separated by a certain count, fixed start and end letters, or starting with a vowel and ending with a consonant.

🎯 Exam Tip: Treat groups of letters (like vowels always together) as single units when calculating arrangements, and remember to arrange the letters within the group as well.

Question 9. The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Answer: Solution: The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T). 6 letters in which A repeats twice can be arranged among themselves in \( \frac{6!}{2!} \) ways.
\( \therefore \) n(S) = \( \frac{6!}{2!} \) ways. Let A be the event that vowels are always together. 3 vowels (A, A, I) can be arranged among themselves in \( \frac{3!}{2!} \) ways. Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in \( {^{4}}P_4 \) = 4! ways.
\( \therefore \) n(A) = 4! \( \times \frac{3!}{2!} \)
\( \therefore \) P(A) = \( \frac{n(A)}{n(S)} \)
= \( \frac{4! \times \frac{3!}{2!}}{\frac{6!}{2!}} \)
= \( \frac{4! \times 3!}{6!} \)
= \( \frac{1}{5} \)
In simple words: To find the probability that all vowels in the word SAVITA stay together when letters are arranged randomly, treat the vowels as a single block, arrange this block with the consonants, and also account for arrangements within the vowel block (considering repeated letters).

🎯 Exam Tip: When letters are repeated, remember to divide by the factorial of the count of repeated letters in both the total arrangements and favorable arrangements to ensure accurate probability calculations.