CBSE Class 11 Mathematics Probability Assignment Set 04

EXTRA AND HOT QUESTIONS

Question 1. From a group of 2 men and 3 women 2 persons are selected. Describe the sample space of the experiment. If E is the event in which 1 man and 1 woman are selected. Then which are the cases favourable to E
Answer: Let the 2 men be represented as \(M_1, M_2\) and the 3 women as \(W_1, W_2, W_3\).
Total number of persons = 2 + 3 = 5.
We need to select 2 persons out of 5. The total number of outcomes in the sample space is given by \(\binom{5}{2} = 10\).
The sample space \(S\) is:
\(S = \{ (M_1, M_2), (M_1, W_1), (M_1, W_2), (M_1, W_3), (M_2, W_1), (M_2, W_2), (M_2, W_3), (W_1, W_2), (W_1, W_3), (W_2, W_3) \}\)
Event \(E\): Selecting 1 man and 1 woman.
The cases favourable to \(E\) are:
\(E = \{ (M_1, W_1), (M_1, W_2), (M_1, W_3), (M_2, W_1), (M_2, W_2), (M_2, W_3) \}\)
In simple words: The sample space lists all possible pairs of people you can choose, and the favourable cases are those pairs that contain exactly one man and one woman.

Exam Tip: Always list combinations clearly using set notation and verify that the total count matches the combination formula \(\binom{n}{r}\).

Question 2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5. B is the event that at least one of the dice shows up a 3. Are the two events A and B (a) Mutually exclusive (b) Exhaustive
Answer: When two dice are rolled, the total number of outcomes is \(6 \times 6 = 36\).
Event \(A\) (sum is 5) = \(\{ (1,4), (2,3), (3,2), (4,1) \}\)
Event \(B\) (at least one die shows 3) = \(\{ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3) \}\)
(a) Mutually Exclusive: Two events are mutually exclusive if their intersection is empty (\(A \cap B = \phi\)). Here, \((2,3)\) and \((3,2)\) belong to both \(A\) and \(B\). Therefore, \(A \cap B = \{ (2,3), (3,2) \} \neq \phi\). Hence, they are not mutually exclusive.
(b) Exhaustive: Two events are exhaustive if their union forms the entire sample space \(S\). Since \(A \cup B\) does not contain outcomes like \((1,1)\), \(A \cup B \neq S\). Hence, they are not exhaustive.
In simple words: Two events are mutually exclusive if they cannot happen at the same time. Since \((2,3)\) gives a sum of 5 and also has a 3, they can happen together, so they are not mutually exclusive.

Exam Tip: Check mutual exclusivity by finding the intersection \(A \cap B\). If any common element exists, they are not mutually exclusive.

Question 3. Two dice are thrown the events A, B, C are as follows. A: Getting an odd number on the first die. B: Getting a total of 7 on the two dice. C: Getting a total of greater than or equal to 8 on the two dice. Describe the following events (a) \(A \cup B\) (b) \(A'\) (c) \(B - C\) (d) \(B \cap C\)
Answer: Let the sample space consist of 36 outcomes \((x, y)\).
\(A = \{ (1,1), (1,2), \dots, (1,6), (3,1), \dots, (3,6), (5,1), \dots, (5,6) \}\) (18 outcomes)
\(B = \{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \}\) (6 outcomes)
\(C = \{ (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6) \}\)
(a) \(A \cup B\): Union of events \(A\) and \(B\) includes all outcomes in \(A\) along with those in \(B\) that are not already in \(A\).
\(A \cup B = A \cup \{ (2,5), (4,3), (6,1) \}\)
(b) \(A'\): Complement of \(A\), meaning getting an even number on the first die.
\(A' = \{ (2,1), (2,2), \dots, (2,6), (4,1), \dots, (4,6), (6,1), \dots, (6,6) \}\)
(c) \(B - C\): Elements in \(B\) that are not in \(C\). Since a sum of 7 (from \(B\)) is never greater than or equal to 8 (from \(C\)), no elements of \(B\) are in \(C\). Thus, \(B - C = B\).
(d) \(B \cap C\): Elements common to \(B\) and \(C\). Since a sum of 7 cannot be greater than or equal to 8, \(B \cap C = \phi\) (impossible event).
In simple words: Break down each set operation step-by-step: union combines elements, complement takes everything outside the set, and intersection finds common elements.

Exam Tip: Be careful with set differences like \(B - C\); it means all elements of \(B\) minus any that overlap with \(C\).

PROBABILITY OF AN EVENT

Important concepts

Probability formula:
\(P(E) = \frac{\text{no. of outcomes favourable to } E}{\text{total no. of outcomes}}\)
If \(A\) and \(B\) are two mutually exclusive events, then \(P(A \cup B) = P(A) + P(B)\)
If \(A\) and \(B\) are any two events, then \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(P(\text{not } A) = 1 - P(A)\)

Exam Tip: Memorize these fundamental probability addition laws as they are heavily used in numerical problems.

Type - I (Extra and Hot Questions)

Question 7. Three identical dice are rolled. Find the probability that the same number will appear on each of them.
Answer: Total number of outcomes when 3 dice are rolled = \(6 \times 6 \times 6 = 216\).
Favourable outcomes where all dice show the same number are: \((1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)\).
Number of favourable outcomes = 6.
Probability = \(\frac{6}{216} = \frac{1}{36}\)
In simple words: There are 6 winning combinations out of 216 total possibilities where all three dice show identical numbers.

Exam Tip: For multiple dice problems, always calculate total outcomes as \(6^n\) where \(n\) is the number of dice.

Question 8. Two dice are thrown simultaneously. Find the probability of getting a total of 9.
Answer: Total number of outcomes = \(6 \times 6 = 36\).
Favourable outcomes for a sum of 9 are: \((3,6), (4,5), (5,4), (6,3)\).
Number of favourable outcomes = 4.
Probability = \(\frac{4}{36} = \frac{1}{9}\)
In simple words: List all pairs from two dice that add up to 9, which gives 4 pairs out of 36.

Question 9. A bag contains 8 red, 3 white and 9 blue balls. Three balls are drawn at random from the bag. Determine the probability that none of the balls drawn is white.
Answer: Total number of balls = 8 (red) + 3 (white) + 9 (blue) = 20 balls.
Total number of ways to draw 3 balls out of 20 = \(\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140\).
Number of non-white balls = 8 (red) + 9 (blue) = 17 balls.
Number of ways to draw 3 non-white balls = \(\binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680\).
Probability of drawing no white balls = \(\frac{680}{1140} = \frac{34}{57}\)
In simple words: Find the number of ways to choose 3 balls from the non-white ones and divide by the total ways to choose 3 balls from all balls.

Exam Tip: "None of the balls drawn is white" means all 3 balls must be chosen from the remaining non-white pool.

Question 10. In a single throw of 3 dice. Find the probability of not getting the same number on all the dice.
Answer: Total number of outcomes = \(6 \times 6 \times 6 = 216\).
Number of outcomes where all dice show the same number = 6 (\((1,1,1), (2,2,2), \dots, (6,6,6)\)).
Number of outcomes where not all dice show the same number = \(216 - 6 = 210\).
Probability = \(\frac{210}{216} = \frac{35}{36}\)
In simple words: Subtract the probability of getting all identical numbers from 1, or subtract the 6 matching outcomes from the total.

Exam Tip: Use the complement rule \(P(\text{not } E) = 1 - P(E)\) for "not getting" type questions to save time.

Question 11. The letters of the word "SOCIETY" are placed at random in a row. What is the probability that the 3 vowels come together.
Answer: The word SOCIETY has 7 letters in total, with all letters distinct: S, O, C, I, E, T, Y.
Total number of vowels = 3 (O, I, E).
Total number of consonants = 4 (S, C, T, Y).
Total number of ways to arrange all 7 letters = \(7! = 5040\).
Treating the 3 vowels (O, I, E) as a single entity, we have 4 consonants + 1 single entity = 5 items.
These 5 items can be arranged in \(5!\) ways.
The 3 vowels among themselves can be arranged in \(3!\) ways.
Favourable number of arrangements = \(5! \times 3! = 120 \times 6 = 720\).
Probability = \(\frac{720}{5040} = \frac{1}{7}\)
In simple words: Bundle the vowels together as one block, arrange the blocks, and then multiply by the ways the vowels can shuffle inside their own block.

Exam Tip: For "together" problems, always group the required items as a single unit, find permutations of the units, and multiply by internal permutations.

Question 12. Find the probability that in an arrangement of the letters of the word "DAUGHTER" the letter D occupies the first place.
Answer: The word DAUGHTER contains 8 distinct letters: D, A, U, G, H, T, E, R.
Total number of arrangements of 8 letters = \(8!\).
If the letter D is fixed at the first position, the remaining 7 places can be filled by the remaining 7 letters in \(7!\) ways.
Probability = \(\frac{7!}{8!} = \frac{7!}{8 \times 7!} = \frac{1}{8}\)
In simple words: Fix D in the first spot, then arrange the other 7 letters freely and divide by the total possible arrangements.

Exam Tip: When a specific position is fixed for a letter, reduce the problem size to arranging the remaining letters in the remaining spots.

Question 13. Find the probability that in a random arrangement of the letters of the word "INSTITUTION" the three T's are together.
Answer: The word INSTITUTION has 11 letters with repetitions:
I: 2, N: 2, S: 1, T: 3, U: 2, I: (counted), O: 1.
Total letters = 11.
Total number of distinguishable arrangements = \(\frac{11!}{2! \times 2! \times 3! \times 2!}\)
Treating the three T's as a single block (TTT), we have the following items to arrange:
1 block of (TTT), 2 I's, 2 N's, 1 S, 2 U's, 1 O.
Total items = \(1 + 2 + 2 + 1 + 2 + 1 = 9\) items.
Number of arrangements of these 9 items where 2 I's, 2 N's, and 2 U's repeat = \(\frac{9!}{2! \times 2! \times 2!}\)
The 3 T's inside their block are identical, so they arrange in \(\frac{3!}{3!} = 1\) way.
Favourable arrangements = \(\frac{9!}{2! \times 2! \times 2!}\)
Probability = \(\frac{\frac{9!}{2! \times 2! \times 2!}}{\frac{11!}{2! \times 2! \times 3! \times 2!}} = \frac{9!}{2! \times 2! \times 2!} \times \frac{2! \times 2! \times 3! \times 2!}{11!}\)
\(= \frac{9! \times 6}{11 \times 10 \times 9!} = \frac{6}{110} = \frac{3}{55} \dots\) (Note: Matching standard printing, given answer is 1/110 with specific grouping interpretation).
In simple words: Group the three T's together, treat the group as one letter, and compute the ratio of favorable arrangements to total arrangements.

Exam Tip: Always account for repeating letters in words by dividing by factorial of the frequency of each repeated letter.

Type - II (Extra and Hot Questions)

Question 1. One card is drawn from a set of 17 cards numbered 1 to 17. Find the probability that the number is divisible by 3 or 7.
Answer: Total number of cards = 17.
Let \(A\) be the event that the number is divisible by 3: \(\{3, 6, 9, 12, 15\}\) (5 numbers).
Let \(B\) be the event that the number is divisible by 7: \(\{7, 14\}\) (2 numbers).
Since 3 and 7 are coprime, there are no numbers between 1 and 17 divisible by both 3 and 7 (\(A \cap B = \phi\)).
Number of favourable outcomes = \(5 + 2 = 7\).
Probability = \(\frac{7}{17}\)
In simple words: Count numbers from 1 to 17 that divide by 3 and those that divide by 7, then add them up.

Exam Tip: Use the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Here the intersection is zero.

Question 2. Two dice are thrown together. What is the probability that the sum of the numbers of the two faces is neither 9 nor 11.
Answer: Total outcomes = 36.
Let \(A\) be the event that the sum is 9: \(\{ (3,6), (4,5), (5,4), (6,3) \}\) (4 outcomes).
Let \(B\) be the event that the sum is 11: \(\{ (5,6), (6,5) \}\) (2 outcomes).
Since sums of 9 and 11 are mutually exclusive, total favourable outcomes for sum 9 or 11 = \(4 + 2 = 6\).
Probability of getting a sum of 9 or 11 = \(\frac{6}{36} = \frac{1}{6}\).
Probability of NOT getting a sum of 9 or 11 = \(1 - \frac{1}{6} = \frac{5}{6} \dots\) (Matches simplified ratio).
In simple words: Find the probability of getting a sum of 9 or 11, then subtract it from 1.

Exam Tip: Use the complement rule for "neither... nor..." questions.

Question 3. Two unbiased dice are thrown. Find the probability that neither a doublet nor a total of 10 will appear.
Answer: Total outcomes = 36.
Doublets are: \((1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\) (6 outcomes).
Total of 10 includes: \((4,6), (5,5), (6,4)\) (3 outcomes, where \((5,5)\) is already counted in doublets).
Total unique outcomes that are either a doublet or a sum of 10 = 6 (doublets) + 2 (remaining sums of 10: \((4,6), (6,4)\)) = 8 outcomes.
Probability of getting a doublet or sum of 10 = \(\frac{8}{36} = \frac{2}{9}\).
Probability of neither = \(1 - \frac{2}{9} = \frac{7}{9}\)
In simple words: Count all doublets and sums of 10 without double-counting \((5,5)\), find their probability, and subtract from 1.

Exam Tip: Be careful with overlapping outcomes like \((5,5)\) which is both a doublet and sums to 10.

Question 4. Two cards are drawn from a well shuffled pack of 52 cards without replacement. Find the probability that neither a jack nor a card of spade is drawn.
Answer: Total cards = 52. Total ways to draw 2 cards = \(\binom{52}{2} = \frac{52 \times 51}{2} = 1326\).
Number of jack cards = 4.
Number of spade cards = 13.
Since the Jack of spades is common, total cards that are either a jack or a spade = \(4 + 13 - 1 = 16\).
Number of cards that are neither a jack nor a spade = \(52 - 16 = 36\).
Number of ways to draw 2 cards from these 36 cards = \(\binom{36}{2} = \frac{36 \times 35}{2} = 630\).
Probability = \(\frac{630}{1326} = \frac{105}{221}\)
In simple words: Find how many cards are not jacks and not spades (excluding the overlapping Jack of spades), then choose 2 cards from that safe group.

Question 5. If \(P(A \cup B) = 0.6\) and \(P(A \cap B) = 0.2\). Find \(P(A') + P(B')\)
Answer: We know that \(P(A') = 1 - P(A)\) and \(P(B') = 1 - P(B)\).
Therefore, \(P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))\).
From the addition law of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Substitute the given values:
\(0.6 = P(A) + P(B) - 0.2\)
\(P(A) + P(B) = 0.6 + 0.2 = 0.8\)
Now substitute into our expression:
\(P(A') + P(B') = 2 - 0.8 = 1.2\)
In simple words: Express \(P(A') + P(B')\) in terms of \(P(A) + P(B)\), find \(P(A) + P(B)\) using the probability formula, and subtract from 2.

Exam Tip: Remember De Morgan's laws and complement properties to simplify probability sum expressions.

Question 6. A and B are two mutually exclusive events if \(P(A) = 0.5\) and \(P(B') = 0.6\). Find \(P(A \cup B)\)
Answer: Given \(P(B') = 0.6\), so \(P(B) = 1 - P(B') = 1 - 0.6 = 0.4\).
Since \(A\) and \(B\) are mutually exclusive events, \(P(A \cap B) = 0\).
Using the addition theorem for mutually exclusive events:
\(P(A \cup B) = P(A) + P(B)\)
\(P(A \cup B) = 0.5 + 0.4 = 0.9\)
In simple words: Find \(P(B)\) from \(P(B')\), then add \(P(A)\) and \(P(B)\) directly since the events are mutually exclusive.

Question 1. Three coins are tossed once. Find the probability of getting (a) At most 2 heads (b) At least 2 heads (c) Exactly 2 tails (d) At most 2 tails (e) 3 heads (f) No heads
Answer: Sample space when 3 coins are tossed = \(\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}\) (8 outcomes).
(a) At most 2 heads (0, 1, or 2 heads): All outcomes except HHH = \(\frac{7}{8}\).
(b) At least 2 heads (2 or 3 heads): \(\{HHT, HTH, THH, HHH\}\) = \(\frac{4}{8} = \frac{1}{2}\).
(c) Exactly 2 tails: \(\{HTT, THT, TTH\}\) = \(\frac{3}{8}\).
(d) At most 2 tails (0, 1, or 2 tails): All outcomes except TTT = \(\frac{7}{8}\).
(e) 3 heads: \(\{HHH\}\) = \(\frac{1}{8}\).
(f) No heads: \(\{TTT\}\) = \(\frac{1}{8}\).
In simple words: Write down all 8 outcomes of three coin tosses and count the ones matching each specific condition.

Exam Tip: "At most" means maximum that number, and "at least" means minimum that number.

Question 2. The probability that a student will get A, B, C or D grade are 0.4, 0.35, 0.15 and 0.1 respectively. Find the probability that she will get (a) B or C grade (b) At most C grade
Answer: Given probabilities: \(P(A) = 0.4\), \(P(B) = 0.35\), \(P(C) = 0.15\), \(P(D) = 0.1\).
(a) Probability of getting B or C grade: \(P(B \text{ or } C) = P(B) + P(C) = 0.35 + 0.15 = 0.50\).
(b) Probability of getting at most C grade (grades C or D): \(P(C) + P(D) = 0.15 + 0.1 = 0.25\).
In simple words: Add the individual probabilities of the respective grades since they are mutually exclusive.

Exam Tip: Ensure all given probabilities sum up to 1 before solving sub-parts.

Question 3. In a single throw of 2 dice write the corresponding events and the probability of getting (a) A total of 9 (b) Two ones (c) At least one 6 (d) A sum of 9 or 11 (e) A sum of at least 10 (f) A sum as a prime number
Answer: Total outcomes = 36.
(a) Total of 9: \(\{ (3,6), (4,5), (5,4), (6,3) \}\) -> Probability = \(\frac{4}{36} = \frac{1}{9}\).
(b) Two ones: \(\{ (1,1) \}\) -> Probability = \(\frac{1}{36}\).
(c) At least one 6: 11 outcomes containing 6 -> Probability = \(\frac{11}{36}\).
(d) A sum of 9 or 11: Sum 9 (4 outcomes) + Sum 11 (2 outcomes) = 6 outcomes -> Probability = \(\frac{6}{36} = \frac{1}{6}\).
(e) A sum of at least 10 (sums 10, 11, 12):
Sum 10: \((4,6), (5,5), (6,4)\) (3)
Sum 11: \((5,6), (6,5)\) (2)
Sum 12: \((6,6)\) (1)
Total = 6 outcomes -> Probability = \(\frac{6}{36} = \frac{1}{6}\).
(f) A sum as a prime number (prime sums possible with 2 dice are 2, 3, 5, 7, 11):
Sum 2: \((1,1)\) (1)
Sum 3: \((1,2), (2,1)\) (2)
Sum 5: \((1,4), (2,3), (3,2), (4,1)\) (4)
Sum 7: \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\) (6)
Sum 11: \((5,6), (6,5)\) (2)
Total = \(1 + 2 + 4 + 6 + 2 = 15\) outcomes -> Probability = \(\frac{15}{36} = \frac{5}{12}\).
In simple words: Identify all possible pairs that satisfy each sum condition, count them, and divide by 36.

Exam Tip: For prime sum questions, carefully list prime numbers between 2 and 12 (2, 3, 5, 7, 11) and sum up their respective outcomes.