Maharashtra Board Class 11 Maths Part 2 Chapter 7 Probability 7.3 Solutions

Question 1. Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?
Answer: Solution:
When two dice are thrown, the sample space is
S = \(\{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \} \)
\( \therefore n(S) = 36 \)
Let A be the event that sum of numbers on two dice is 5.
\( \therefore A = \{ (1, 4), (2, 3), (3, 2), (4, 1) \} \)
\( \therefore n(A) = 4 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} \)
Let B be the event that number on second die is greater than or equal to number on first die.
\( B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6) \} \)
\( \therefore n(B) = 21 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{21}{36} \)
Now, \( A \cap B = \{ (1, 4), (2, 3) \} \)
\( \therefore n(A \cap B) = 2 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{4}{36} + \frac{21}{36} - \frac{2}{36} \)
\( = \frac{23}{36} \)
In simple words: We calculated the probabilities of getting a sum of 5 and the second die being greater than or equal to the first, then used the formula for the union of two events to find the probability of either occurring.

🎯 Exam Tip: Clearly define your sample space (S) and events (A, B) with their respective counts (n(S), n(A), n(B)) before applying probability formulas. Remember the formula for \( P(A \cup B) \).

Question 2. A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either red or face card?
Answer: Solution:
One card can be drawn from the pack of 52 cards in \( ^{52}C_1 = 52 \) ways
\( \therefore n(S) = 52 \)
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that a red card is drawn Red card can be drawn in \( ^{26}C_1 = 26 \) ways
\( \therefore n(A) = 26 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{26}{52} \)
Let B be the event that a black card is drawn
\( \therefore \) Black card can be drawn in \( ^{26}C_1 = 26 \) ways.
\( \therefore n(B) = 26 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{26}{52} \)
Since A and B are mutually exclusive and exhaustive events
\( \therefore P(A \cap B) = 0 \)
\( \therefore \) required probability = \( P(A \cup B) \)
\( \therefore P(A \cup B) = P(A) + P(B) \)
\( = \frac{26}{52} + \frac{26}{52} \)
\( = \frac{52}{52} \)
\( = 1 \)
(ii) Let A be the event that a red card is drawn
\( \therefore \) red card can be drawn in \( ^{26}C_1 = 26 \) ways
\( \therefore n(A) = 26 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{26}{52} \)
Let B be the event that a face card is drawn There are 12 face cards in the pack of 52 cards
\( \therefore 1 \) face card can be drawn in \( ^{12}C_1 = 12 \) ways
\( \therefore n(B) = 12 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{12}{52} \)
There are 6 red face cards.
\( \therefore n(A \cap B) = 6 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{6}{52} \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{26}{52} + \frac{12}{52} - \frac{6}{52} \)
\( = \frac{32}{52} \)
\( = \frac{8}{13} \)
In simple words: For part (i), since a card must be either red or black, the probability is 1. For part (ii), we calculate the probability of drawing a red card, a face card, and a card that is both (red face card), then use the union formula.

🎯 Exam Tip: Differentiate between mutually exclusive events (like red and black cards) and overlapping events (like red and face cards). Correctly identifying the intersection is key for overlapping events.

Question 3. Two cards are drawn from a pack of 52 cards. What is the probability that,
(i) both the cards are of the same colour?
(ii) both the cards are either black or queens?
Answer: Solution:
Two cards can be drawn from 52 cards in \( ^{52}C_2 \) ways.
\( \therefore n(S) = ^{52}C_2 \)
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that both cards are red.
\( \therefore 2 \) red cards can be drawn in \( ^{26}C_2 \) ways.
\( \therefore n(A) = ^{26}C_2 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{^{26}C_2}{^{52}C_2} = \frac{26 \times 25}{52 \times 51} = \frac{25}{102} \)
Let B be the event that both cards are black.
\( \therefore 2 \) black cards can be drawn in \( ^{26}C_2 \) ways
\( \therefore n(B) = ^{26}C_2 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{^{26}C_2}{^{52}C_2} = \frac{26 \times 25}{52 \times 51} = \frac{25}{102} \)
Since A and B are mutually exclusive and exhaustive events
\( \therefore P(A \cap B) = 0 \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( \therefore P(A \cup B) = P(A) + P(B) \)
\( = \frac{25}{102} + \frac{25}{102} \)
\( = \frac{50}{102} \)
\( = \frac{25}{51} \)
(ii) Let A be the event that both cards are black.
\( \therefore 2 \) black cards can be drawn in \( ^{26}C_2 \) ways.
\( \therefore n(A) = ^{26}C_2 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{^{26}C_2}{^{52}C_2} = \frac{26 \times 25}{52 \times 51} = \frac{25}{102} \)
Let B be the event that both cards are queens.
There are 4 queens in a pack of 52 cards
\( \therefore 2 \) queen cards can be drawn in \( ^{4}C_2 \) ways.
\( \therefore n(B) = ^{4}C_2 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{^{4}C_2}{^{52}C_2} = \frac{4 \times 3}{52 \times 51} = \frac{12}{2652} = \frac{1}{221} \)
There are two black queen cards.
\( \therefore n(A \cap B) = ^{2}C_2 = 1 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{^{52}C_2} = \frac{1 \times 2 \times 1}{52 \times 51} = \frac{1}{1326} \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{25}{102} + \frac{1}{221} - \frac{1}{1326} \)
\( = \frac{325 + 6 - 1}{1326} \)
\( = \frac{330}{1326} \)
\( = \frac{55}{221} \)
In simple words: For same color, we found the probability of both red and both black cards separately and added them. For both black or queens, we calculated the probabilities of both black, both queens, and both black queens, then applied the union formula considering the overlap.

🎯 Exam Tip: Use combinations \( (^{n}C_r) \) for drawing multiple items without replacement. Always check for mutual exclusivity; if events overlap, subtract the probability of their intersection.

Question 4. A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that
(i) number on the ticket is a perfect square or divisible by 4?
(ii) number on the ticket is a prime number or greater than 30?
Answer: Solution:
Out of the 50 tickets, a ticket can be drawn in \( ^{50}C_1 = 50 \) ways.
\( \therefore n(S) = 50 \)
(i) Let A be the event that the number on the ticket is a perfect square.
\( \therefore A = \{1, 4, 9, 16, 25, 36, 49\} \)
\( \therefore n(A) = 7 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{7}{50} \)
Let B be the event that the number on the ticket is divisible by 4.
\( \therefore B = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48\} \)
\( \therefore n(B) = 12 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{12}{50} \)
Now, \( A \cap B = \{4, 16, 36\} \)
\( \therefore n(A \cap B) = 3 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{50} \)
Required probability = \( P(A \cup B) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{7}{50} + \frac{12}{50} - \frac{3}{50} \)
\( = \frac{7+12-3}{50} \)
\( = \frac{16}{50} \)
\( = \frac{8}{25} \)
(ii) Let A be the event that the number on the ticket is a prime number.
\( \therefore A = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\} \)
\( \therefore n(A) = 15 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{15}{50} \)
Let B be the event that the number is greater than 30.
\( \therefore B = \{31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50\} \)
\( \therefore n(B) = 20 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{20}{50} \)
Now, \( A \cap B = \{31, 37, 41, 43, 47\} \)
\( \therefore n(A \cap B) = 5 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{5}{50} \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{15}{50} + \frac{20}{50} - \frac{5}{50} \)
\( = \frac{15+20-5}{50} \)
\( = \frac{30}{50} \)
\( = \frac{3}{5} \)
In simple words: For each part, we listed the numbers satisfying each condition and their intersection. Then, we applied the formula for the probability of the union of two events.

🎯 Exam Tip: Carefully list out the elements of each set (A, B, and \( A \cap B \)) to avoid errors in counting \( n(A) \), \( n(B) \), and \( n(A \cap B) \). This is crucial for accurate probability calculation.

Question 5. A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random
(i) passed at least one examination.
(ii) passed in exactly one examination.
(iii) failed in both examinations.
Answer: Solution:
Out of hundred students 1 student can be selected in \( ^{100}C_1 = 100 \) ways.
\( \therefore n(S) = 100 \)
Let A be the event that the student passed in the first examination.
Let B be the event that student passed in second examination.
\( \therefore n(A) = 60 \), \( n(B) = 50 \) and \( n(A \cap B) = 30 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{60}{100} = \frac{6}{10} \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{50}{100} = \frac{5}{10} \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{30}{100} = \frac{3}{10} \)
(i) P(student passed in at least one examination) = \( P(A \cup B) \)
\( = P(A) + P(B) - P(A \cap B) \)
\( = \frac{6}{10} + \frac{5}{10} - \frac{3}{10} \)
\( = \frac{6+5-3}{10} \)
\( = \frac{8}{10} = \frac{4}{5} \)
(ii) P(student passed in exactly one examination) = \( P(A) + P(B) - 2 \cdot P(A \cap B) \)
\( = \frac{6}{10} + \frac{5}{10} - 2 \left( \frac{3}{10} \right) \)
\( = \frac{11}{10} - \frac{6}{10} \)
\( = \frac{5}{10} = \frac{1}{2} \)
(iii) P(student failed in both examinations) = \( P(A' \cap B') \)
\( = P(A \cup B)' \) .....[De Morgan's law]
\( = 1 - P(A \cup B) \)
\( = 1 - \frac{4}{5} \)
\( = \frac{1}{5} \)
In simple words: We used the given probabilities of passing individual exams and both to calculate the probability of passing at least one, exactly one, and failing both, applying standard probability rules and De Morgan's law.

🎯 Exam Tip: Understand De Morgan's laws for complements: \( P(A' \cap B') = P(A \cup B)' \) and \( P(A' \cup B') = P(A \cap B)' \). Also, remember that "exactly one" means \( P(A) + P(B) - 2P(A \cap B) \).

Question 6. If \( P(A) = \frac{1}{4} \), \( P(B) = \frac{2}{5} \) and \( P(A \cup B) = \frac{1}{2} \). Find the values of the following probabilities.
(i) P(A ∩ B)
(ii) P(A ∩ B')
(iii) P(A' ∩ B)
(iv) P(A' ∪ B')
(v) P(A' ∩ B')
Answer: Solution:
Here, \( P(A) = \frac{1}{4} \), \( P(B) = \frac{2}{5} \) and \( P(A \cup B) = \frac{1}{2} \)
(i) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \therefore P(A \cap B) = P(A) + P(B) - P(A \cup B) \)
\( = \frac{1}{4} + \frac{2}{5} - \frac{1}{2} \)
\( = \frac{5+8-10}{20} \)
\( = \frac{3}{20} \)
(ii) \( P(A \cap B') = P(A) - P(A \cap B) \)
\( = \frac{1}{4} - \frac{3}{20} \)
\( = \frac{5-3}{20} \)
\( = \frac{2}{20} \)
\( = \frac{1}{10} \)
(iii) \( P(A' \cap B) = P(B) - P(A \cap B) \)
\( = \frac{2}{5} - \frac{3}{20} \)
\( = \frac{8-3}{20} \)
\( = \frac{5}{20} \)
\( = \frac{1}{4} \)
(iv) \( P(A' \cup B') = P(A \cap B)' \) .....[De Morgan's law]
\( = 1 - P(A \cap B) \)
\( = 1 - \frac{3}{20} \)
\( = \frac{17}{20} \)
(v) \( P(A' \cap B') = P(A \cup B)' \) .....[De Morgan's law]
\( = 1 - P(A \cup B) \)
\( = 1 - \frac{1}{2} \)
\( = \frac{1}{2} \)
In simple words: We used the given probabilities and applied standard set theory formulas to find the probabilities of intersection, complements, and combinations of events, including De Morgan's laws.

🎯 Exam Tip: Memorize the formulas for \( P(A \cap B) \), \( P(A \cap B') \), \( P(A' \cap B) \), and both forms of De Morgan's laws to quickly solve these types of questions.

Question 7. A computer software company is bidding for computer programs A and B. The probability that the company will get software A is \( \frac{3}{5} \), the probability that the company will get software B is \( \frac{1}{3} \) and the probability that company will get both A and B is \( \frac{1}{8} \). What is the probability that the company will get at least one software?
Answer: Solution:
Let A be the event that the company will get software A.
\( \therefore P(A) = \frac{3}{5} \)
Let B be the event that company will get software B.
\( \therefore P(B) = \frac{1}{3} \)
Also, \( P(A \cap B) = \frac{1}{8} \)
\( \therefore \) P(the company will get at least one software) = \( P(A \cup B) \)
\( = P(A) + P(B) - P(A \cap B) \)
\( = \frac{3}{5} + \frac{1}{3} - \frac{1}{8} \)
\( = \frac{72+40-15}{120} \)
\( = \frac{97}{120} \)
In simple words: Given the probabilities of getting software A, software B, and both, we calculated the probability of getting at least one by using the formula for the union of two events.

🎯 Exam Tip: For "at least one" scenarios, the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is essential. Ensure common denominators are used for fractions during addition/subtraction.

Question 8. A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Answer: Solution:
One card can be drawn from the pack of 52 cards in \( ^{52}C_1 = 52 \) ways
\( \therefore n(S) = 52 \)
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let A be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in \( ^{13}C_1 \) ways
\( \therefore n(A) = ^{13}C_1 = 13 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{13}{52} \)
Let B be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in \( ^{4}C_1 \) ways
\( \therefore n(B) = ^{4}C_1 = 4 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} \)
There is one queen card out of 4 which is also a heart card
\( \therefore n(A \cap B) = ^{1}C_1 = 1 \)
\( \therefore P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{52} \)
\( \therefore \) P(card is a heart or a queen) = \( P(A \cup B) \)
\( = P(A) + P(B) - P(A \cap B) \)
\( = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \)
\( = \frac{13+4-1}{52} \)
\( = \frac{16}{52} \)
\( \therefore P(A \cup B) = \frac{4}{13} \)
In simple words: We calculated the probability of drawing a heart, a queen, and the queen of hearts. Then we used the principle of inclusion-exclusion to find the probability of drawing either a heart or a queen.

🎯 Exam Tip: When dealing with overlapping categories like "heart" and "queen", correctly identify and subtract the intersection (queen of hearts) to avoid double-counting in the union probability.

Question 9. In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Answer: Solution:
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in \( ^{7}C_4 \) ways
\( = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \)
\( = 35 \) ways.
\( \therefore n(S) = 35 \)
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in \( ^{3}C_3 \) ways while a girl can be selected in \( ^{4}C_1 \) ways.
\( \therefore n(A) = ^{3}C_3 \times ^{4}C_1 = 1 \times 4 = 4 \)
\( \therefore P(A) = \frac{n(A)}{n(S)} = \frac{4}{35} \)
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in \( ^{4}C_3 \) ways and a boy can be selected in \( ^{3}C_1 \) ways.
\( \therefore n(B) = ^{4}C_3 \times ^{3}C_1 = 4 \times 3 = 12 \)
\( \therefore P(B) = \frac{n(B)}{n(S)} = \frac{12}{35} \)
Since A and B are mutually exclusive and exhaustive events
\( \therefore P(A \cap B) = 0 \)
\( \therefore \) Required probability = \( P(A \cup B) \)
\( = P(A) + P(B) \)
\( = \frac{4}{35} + \frac{12}{35} \)
\( = \frac{16}{35} \)
In simple words: We calculated the total ways to select 4 students. Then, we found the ways to select 3 boys and 1 girl (Event A) and 3 girls and 1 boy (Event B). Since these events are mutually exclusive, their probabilities are added to find the total probability.

🎯 Exam Tip: For "either-or" questions where events cannot happen simultaneously (mutually exclusive), simply add their probabilities. Use combinations for selection problems where the order doesn't matter.