Maharashtra Board Class 11 Maths Part 2 Chapter 7 Probability 7.1 Solutions

Question 1. State the sample space and n(S) for the following random experiments.
(i) A coin is tossed twice. If a second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.
Answer:
Solution:
(i) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown.
Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
Thus, the sample space S for (i) is:
\( \implies \) S = {HH, TH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
\( \implies \) n(S) = 14
(ii) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
Let A be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
\( \implies \) A = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event B.
\( \implies \) B = {HTT, HTH, TTT, TTH}
The sample space S of the experiment is A U B.
\( \implies \) S = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6, HTT, HTH, TTT, TTH}
\( \implies \) n(S) = 16
In simple words: This question asks for the total possible outcomes (sample space) and their count for two different scenarios involving coin tosses and die rolls, detailing conditions under which a die or another coin is used.

🎯 Exam Tip: Clearly define the conditions for each stage of the experiment and meticulously list all possible outcomes to construct the sample space correctly. Pay attention to "if-then" conditions.

Question 2. In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Answer:
Solution:
The bag contains 3 balls out of which one is black (B), one is red (R) and the other one is green (G).
Two balls are drawn one after the other, with replacement, from the bag.
\( \implies \) the sample space S is given by
S = {BB, BR, BG, RB, RR, RG, GB, GR, GG}
\( \implies \) n(S) = 9
In simple words: When drawing two balls with replacement from a set of three distinct balls, the sample space includes all combinations where the first and second draws can be any of the three balls, resulting in 3x3=9 total outcomes.

🎯 Exam Tip: For "with replacement" problems, remember that the number of choices remains constant for each draw, making it easier to calculate n(S) using multiplication (e.g., \( \text{choices}^{\text{draws}} \)).

Question 3. A coin and die are tossed. State sample space of following events.
(i) A: Getting a head and an even number.
(ii) B: Getting a prime number.
(iii) C: Getting a tail and perfect square.
Answer:
Solution:
When a coin and a die are tossed the sample space S is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) A: getting a head and an even number
\( \implies \) A = {H2, H4, H6}
(ii) B: getting a prime number
\( \implies \) B = {H2, H3, H5, T2, T3, T5}
(iii) C: getting a tail and a perfect square.
\( \implies \) C = {T1, T4}
In simple words: This question requires identifying specific subsets (events) from the complete sample space formed by tossing a coin and rolling a die, based on conditions like getting a head with an even number, just a prime number (regardless of coin side), or a tail with a perfect square.

🎯 Exam Tip: First, list the complete sample space clearly. Then, carefully filter elements that satisfy each event's condition. Remember that "prime numbers" are 2, 3, 5 for a die, and "perfect squares" are 1, 4.

Question 4. Find the total number of distinct possible outcomes n(S) for each of the following random experiments.
(i) From a box containing 25 lottery tickets and 3 tickets are drawn at random.
(ii) From a group of 4 boys and 3 girls, any two students are selected at random.
(iii) 5 balls are randomly placed into five cells, such that each cell will be occupied.
(iv) 6 students are arranged in a row for photographs.
Answer:
Solution:
(i) Let S be the event that 3 tickets are drawn at random from 25 tickets
\( \implies \) 3 tickets can be selected in \( {}^{25}\text{C}_3 \) ways
\( \implies \) n(S) = \( {}^{25}\text{C}_3 \)
= \[ \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \]
= 2300
(ii) There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in \( {}^{7}\text{C}_2 \) ways.
\( \implies \) n(S) = \( {}^{7}\text{C}_2 \)
= \[ \frac{7 \times 6}{2 \times 1} \]
= 21
(iii) 5 balls have to be placed in 5 cells in such a way that each cell is occupied.
\( \implies \) The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
\( \implies \) a total number of ways of filling 5 cells such that each cell is occupied = 5!
= \( 5 \times 4 \times 3 \times 2 \times 1 \)
= 120
\( \implies \) n(S) = 120
(iv) Six students can be arranged in a row for a photograph in \( {}^{6}\text{P}_6 \) = 6! ways.
= \( 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
= 720
\( \implies \) n(S) = 720
In simple words: This question asks to calculate the number of possible outcomes (n(S)) for various experiments, using combinations (\( {}^{n}\text{C}_r \)) for selection problems where order doesn't matter, and permutations (n! or \( {}^{n}\text{P}_r \)) for arrangement problems where order matters.

🎯 Exam Tip: Distinguish between permutations (arrangements, order matters) and combinations (selections, order doesn't matter). Use the correct formula (\( {}^{n}\text{P}_r \) or \( {}^{n}\text{C}_r \)) for each scenario, remembering that n! is a special case of \( {}^{n}\text{P}_n \).

Question 5. Two dice are thrown. Write favourable outcomes for the following events.
(i) P: The sum of the numbers on two dice is divisible by 3 or 4.
(ii) Q: sum of the numbers on two dice is 7.
(iii) R: sum of the numbers on two dice is a prime number.
Also, check whether
(a) events P and Q are mutually exclusive and exhaustive.
(b) events Q and R are mutually exclusive and exhaustive.
Answer:
Solution:
When two dice are thrown, all possible outcomes are
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)}
(i) P: sum of the numbers on two dice is divisible by 3 or 4.
\( \implies \) P = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}
(ii) Q: sum of the numbers on two dice is 7.
\( \implies \) Q = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
(iii) R: sum of the numbers on two dice is a prime number.
\( \implies \) R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
(a) P and Q are mutually exclusive events as P \( \cap \) Q = \( \phi \) and
P U Q = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} \( \ne \) S
\( \implies \) P and Q are not exhaustive events as P U Q \( \ne \) S.
(b) Q \( \cap \) R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
\( \implies \) Q \( \cap \) R \( \ne \phi \)
Also, Q U R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} \( \ne \) S
\( \implies \) Q and R are neither mutually exclusive nor exhaustive events.
In simple words: This problem involves listing outcomes for different events when two dice are rolled, specifically for sums divisible by 3 or 4, a sum of 7, or a prime sum. It also tests the understanding of mutually exclusive (no common outcomes) and exhaustive (union covers all possible outcomes) events.

🎯 Exam Tip: To check for mutually exclusive events, find the intersection (\( \cap \)). If it's empty (\( \phi \)), they are mutually exclusive. To check for exhaustive events, find the union (U). If the union equals the total sample space (S), they are exhaustive.

Question 6. A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
(i) is not taken into account.
(ii) is taken into account.
Answer:
Solution:
(i) If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in \( {}^{52}\text{C}_1 \) = 52 ways.
\( \implies \) n(S) = 52
(ii) If consideration of suits is taken into account, then one card can be drawn from each suit in \( {}^{13}\text{C}_1 \times {}^{4}\text{C}_1 \)
= 13 \( \times \) 4
= 52 ways.
\( \implies \) n(S) = 52
In simple words: This question demonstrates that when drawing one card from a standard 52-card deck, the total number of outcomes (sample space size) remains 52, regardless of whether you conceptualize it as choosing any card or choosing a rank from each suit.

🎯 Exam Tip: For single-item selections, \( {}^{n}\text{C}_1 \) is simply n. Understand how different interpretations of "consideration of suits" can lead to the same result for simple drawing. The phrasing might change, but the underlying counting principle should be applied correctly.

Question 7. Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Answer:
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box II contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
\( \implies \) the sample space is
S = {(H, R11), (H, R12), (H, R13), (H, B11), (H, B12), (T, R21), (T, R22), (T, B21), (T, B22), (T, B23), (T, B24)}
In simple words: The sample space for this experiment is a set of ordered pairs, where the first element is the coin toss outcome (Head or Tail) and the second element is the marble drawn, determined by which box is selected based on the coin toss.

🎯 Exam Tip: When an experiment has sequential stages or conditional choices, construct the sample space by pairing outcomes from each stage. Ensure all possible combinations are listed, clearly distinguishing outcomes from different conditional paths.

Question 8. Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn
(i) with replacement
(ii) without replacement.
Answer:
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8.
Two cards are to be drawn from this bag.
(i) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6,8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}
(ii) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}
In simple words: This problem asks for the sample space of drawing two cards from a set of four, first when the drawn card is put back (with replacement, allowing duplicates) and then when it is not (without replacement, disallowing duplicates and order matters for pairs).

🎯 Exam Tip: The key difference between "with replacement" and "without replacement" is whether an item can be selected more than once. For "without replacement" and ordered draws, ensure that (A, B) is included but (A, A) is not, and for distinct items, (A,B) and (B,A) are separate outcomes.