CBSE Class 11 Mathematics Probability Assignment Set 01

Probability MCQ Questions with Answers Class 11 Mathematics

Question 1. If a leap year is selected at random ,what is the chance that it will contain 53 Tuesdays?
Answer: A leap year has 366 days, which is equal to 52 weeks and 2 remaining days.
The 2 remaining days can be any consecutive pair of days of the week: (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), or (Sunday, Monday).
Total possible outcomes for the remaining 2 days = 7
Favorable outcomes for getting a Tuesday = 2 (namely, [Monday, Tuesday] and [Tuesday, Wednesday])
Probability = \(\frac{2}{7}\)
In simple words: A leap year has 52 full weeks and 2 extra days, giving a 2 out of 7 chance for any specific day to appear 53 times.

Exam Tip: Remember that a non-leap year has 365 days (52 weeks and 1 extra day, probability = \(\frac{1}{7}\)), while a leap year has 366 days (probability = \(\frac{2}{7}\)).

Question 2. If 12 persons are seated at round table ,what is the probability that two particular persons sit together?
Answer: Total number of persons = 12
Total number of ways to arrange 12 persons around a circular table = \((12 - 1)! = 11!\)
Let the two particular persons be considered as a single unit (entity).
Then the number of units to be arranged around the circular table is \(11 - 1 = 10\).
These 10 units can be arranged in \(10!\) ways.
The two particular persons can arrange themselves among each other in \(2!\) ways.
Favorable number of ways = \(10! \times 2!\)
Probability = \(\frac{10! \times 2!}{11!} = \frac{10! \times 2}{11 \times 10!} = \frac{2}{11}\)
In simple words: Treat the two specific people as one single block, arrange the blocks in a circle, and then multiply by the ways the two people can switch places inside their block.

Exam Tip: For circular permutations of \(n\) objects, the total arrangements are always \((n-1)!\). Always account for internal arrangements of grouped objects.

Question 3. from a group of 7 men and 4 ladies a committee of 6 persons is formed. What is the probability that the committee will consists of exactly 2 ladies?
Answer: Total persons = 7 men + 4 ladies = 11 persons.
A committee of 6 persons can be chosen from 11 persons in \(\binom{11}{6}\) ways.
Total number of ways = \(\frac{11!}{6!5!} = 462\)
For the committee to consist of exactly 2 ladies, we must choose 2 ladies out of 4 and 4 men out of 7.
Favorable number of ways = \(\binom{4}{2} \times \binom{7}{4} = 6 \times 35 = 210\)
Probability = \(\frac{210}{462} = \frac{5}{11}\)
In simple words: Use combinations to find total ways to form the committee, then find ways to pick exactly 2 ladies and 4 men, and divide.

Exam Tip: Use the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) and remember that choosing \(r\) items is the same as choosing \(n-r\) items (\(\binom{n}{r} = \binom{n}{n-r}\)).

Question 4. A box contains 3 white , 4 red and 5 blue balls. If three balls are drawn at random, find the probability that
(i) all of them are white
(ii) exactly one is white
(iii) each ball is of different colour.
Answer: Total number of balls = 3 (white) + 4 (red) + 5 (blue) = 12 balls.
Total number of ways to draw 3 balls out of 12 = \(\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\).

(i) Probability that all of them are white:
Favorable ways to choose 3 white balls out of 3 = \(\binom{3}{3} = 1\)
Probability = \(\frac{1}{220}\)

(ii) Probability that exactly one is white:
Favorable ways = Choose 1 white ball out of 3 and 2 non-white balls out of 9 (4 red + 5 blue)
\(\binom{3}{1} \times \binom{9}{2} = 3 \times \frac{9 \times 8}{2} = 3 \times 36 = 108\)
Probability = \(\frac{108}{220} = \frac{27}{55}\)

(iii) Probability that each ball is of different colour:
Favorable ways = Choose 1 white out of 3, 1 red out of 4, and 1 blue out of 5
\(\binom{3}{1} \times \binom{4}{1} \times \binom{5}{1} = 3 \times 4 \times 5 = 60\)
Probability = \(\frac{60}{220} = \frac{3}{11}\)
In simple words: Break down each sub-question into combinations of desired ball selections divided by total possible selections of 3 balls.

Exam Tip: "Different color" means one from each color group; "non-white" means combining all other color categories together.

Question 5. what is the probability of getting a total of less than 12 in throw of two dice?
Answer: Total number of outcomes when two dice are thrown = \(6 \times 6 = 36\).
The maximum possible sum on two dice is \(6 + 6 = 12\).
A total of 12 can only occur in 1 way: \((6, 6)\).
The number of outcomes where the total is less than 12 means all outcomes except \((6, 6)\).
Favorable outcomes = Total outcomes - Outcomes where sum is 12
Favorable outcomes = \(36 - 1 = 35\)
Probability = \(\frac{35}{36}\)
In simple words: Since a sum of 12 is only possible with (6,6), every other outcome gives a sum less than 12.

Exam Tip: Using the complement probability rule (\(1 - P(\text{sum is 12})\)) is a very fast way to solve this.

Question 6. Three dice are thrown together. Find the probability of getting a total of at least 6.
Answer: Total number of outcomes when three dice are thrown = \(6^3 = 216\).
We need a total of at least 6, which means the sum can be 6, 7, 8, ..., up to 18.
It is easier to find the complement: the probability of getting a total of less than 6 (i.e., sums of 3, 4, or 5 since the minimum sum on three dice is \(1+1+1 = 3\)).
Let us list the outcomes for sums less than 6:
- Sum = 3: \((1, 1, 1)\) -> 1 outcome
- Sum = 4: \((1, 1, 2), (1, 2, 1), (2, 1, 1)\) -> 3 outcomes
- Sum = 5: \((1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)\) -> 6 outcomes
Total unfavorable outcomes (sum < 6) = \(1 + 3 + 6 = 10\).
Favorable outcomes (sum \(\geq\) 6) = \(216 - 10 = 206\).
Probability = \(\frac{206}{216} = \frac{103}{108}\)
In simple words: Find the few cases where the sum is less than 6 (3, 4, or 5), subtract them from 216, and divide by the total.

Exam Tip: When the condition is "at least" or "at most", always check if using complementary probability (1 minus opposite) saves time.

Question 7. In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. Find the probability that:
(i) both the tickets drawn have prime number,
(ii)none of the tickets drawn has prime number
(iii)one tickets has prime number.
Answer: Total number of tickets = 50.
Total ways to draw 2 tickets = \(\binom{50}{2} = \frac{50 \times 49}{2} = 1225\).
Prime numbers between 1 and 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
Total prime numbers = 15.
Total non-prime numbers = \(50 - 15 = 35\).

(i) Both tickets drawn have prime numbers:
Favorable ways = \(\binom{15}{2} = \frac{15 \times 14}{2} = 105\)
Probability = \(\frac{105}{1225} = \frac{21}{245}\)

(ii) None of the tickets drawn has prime number:
Favorable ways = \(\binom{35}{2} = \frac{35 \times 34}{2} = 595\)
Probability = \(\frac{595}{1225} = \frac{17}{35}\)

(iii) One ticket has prime number:
Favorable ways = Choose 1 prime and 1 non-prime = \(\binom{15}{1} \times \binom{35}{1} = 15 \times 35 = 525\)
Probability = \(\frac{525}{1225} = \frac{3}{7}\)
In simple words: Count how many prime numbers exist between 1 and 50 (there are 15), then use combinations for both, none, or exactly one prime.

Exam Tip: Remember that 1 is neither prime nor composite. List primes carefully up to 50 to avoid counting errors.

Question 8. A word is consists of 9 letters; 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability that more than one vowel will be selected ?
Answer: Total letters = 9 (5 consonants and 4 vowels).
Total ways to choose 3 letters out of 9 = \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).
"More than one vowel" means either 2 vowels or 3 vowels.
Case 1: Exactly 2 vowels and 1 consonant.
Favorable ways = \(\binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30\).
Case 2: Exactly 3 vowels and 0 consonants.
Favorable ways = \(\binom{4}{3} \times \binom{5}{0} = 4 \times 1 = 4\).
Total favorable ways = \(30 + 4 = 34\).
Probability = \(\frac{34}{84} = \frac{17}{42}\)
In simple words: Calculate the number of ways to pick either (2 vowels and 1 consonant) or (3 vowels), add them up, and divide by total ways.

Exam Tip: "More than one" strictly means > 1, so 2 or 3 in this case since only 3 letters are drawn.

Question 9. Four persons are to be chosen at random from a group of 3 men, 2 women and 4 children . Find the probability of selecting :
(i) 1 man,1 woman and 2 children
(ii) exactly two children
Answer: Total persons = 3 (men) + 2 (women) + 4 (children) = 9 persons.
Total ways to choose 4 persons out of 9 = \(\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\).

(i) Probability of selecting 1 man, 1 woman, and 2 children:
Favorable ways = \(\binom{3}{1} \times \binom{2}{1} \times \binom{4}{2} = 3 \times 2 \times 6 = 36\)
Probability = \(\frac{36}{126} = \frac{2}{7}\)

(ii) Probability of selecting exactly two children:
Selecting exactly 2 children means the remaining 2 persons must be chosen from the remaining adults (3 men + 2 women = 5 adults).
Favorable ways = \(\binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60\)
Probability = \(\frac{60}{126} = \frac{10}{21}\)
In simple words: Group the people by category, apply combination rules for the specific criteria, and divide by total possible selections of 4 people.

Exam Tip: When picking "exactly two children", ensure the remaining slots are filled by combining all other categories into a single adult pool.

Question 10. Twelve balls are distributed among three boxes, find the probability that the first box will contain three balls.
Answer: Each of the 12 balls can be placed into any of the 3 boxes independently in 3 ways.
Total number of ways to distribute 12 balls into 3 boxes = \(3^{12}\).
For the first box to contain exactly 3 balls:
Choose 3 balls out of 12 to put in the first box in \(\binom{12}{3}\) ways.
The remaining 9 balls must be distributed between the other 2 boxes (each ball has 2 choices) in \(2^9\) ways.
Favorable number of ways = \(\binom{12}{3} \times 2^9\)
Probability = \(\frac{\binom{12}{3} \times 2^9}{3^{12}} = \frac{12! \times 2^9}{3! \times 9! \times 3^{12}}\)
In simple words: Pick 3 balls for the first box, and let the remaining 9 balls go into the other two boxes.

Exam Tip: This follows a binomial distribution model where success is landing in box 1 with probability \(p = \frac{1}{3}\) and failure with probability \(q = \frac{2}{3}\).

Question 11. In a single throw of three dice, determine the probability of getting
(i) a total of 5
(ii) a total of at most 5
(iii) a total of at least 5.
Answer: Total outcomes when three dice are thrown = \(6^3 = 216\).

(i) Probability of a total of 5:
Favorable outcomes for sum = 5:
\((1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)\) -> 6 outcomes.
Probability = \(\frac{6}{216} = \frac{1}{36}\)

(ii) Probability of a total of at most 5:
At most 5 means sum can be 3, 4, or 5.
- Sum = 3: \((1, 1, 1)\) -> 1 outcome
- Sum = 4: \((1, 1, 2), (1, 2, 1), (2, 1, 1)\) -> 3 outcomes
- Sum = 5: 6 outcomes (calculated above)
Total favorable outcomes = \(1 + 3 + 6 = 10\)
Probability = \(\frac{10}{216} = \frac{5}{108}\)

(iii) Probability of a total of at least 5:
Complement of "at least 5" is sum less than 5 (i.e., sum of 3 or 4).
Unfavorable outcomes (sum = 3 or 4) = \(1 + 3 = 4\)
Favorable outcomes = \(216 - 4 = 212\)
Probability = \(\frac{212}{216} = \frac{53}{54}\)
In simple words: Count outcomes for sums 3, 4, and 5 carefully, then use complementary subtraction for "at least" when needed.

Exam Tip: Systematically write down permutations for small sums on three dice to avoid missing combinations.

Question 12. In a lot of 12 microwave ovens, there are 3 defective units. A person has ordered 4 of these units and since each is identically packed, the selection will be random. what is the probability that
(i) all 4 units are good
(ii) exactly 3 units are good
(iii) at least 2 units are good.
Answer: Total microwave ovens = 12
Defective = 3, Good = \(12 - 3 = 9\).
Total ways to choose 4 ovens = \(\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495\).

(i) All 4 units are good:
Favorable ways = \(\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\)

Probability = \(\frac{126}{495} = \frac{14}{55}\)

(ii) Exactly 3 units are good:
Favorable ways = Choose 3 good and 1 defective = \(\binom{9}{3} \times \binom{3}{1} = 84 \times 3 = 252\)
Probability = \(\frac{252}{495} = \frac{28}{55}\)

(iii) At least 2 units are good:
At least 2 good means 2 good, 3 good, or 4 good ovens.
Let us find the complement: fewer than 2 good ovens (i.e., 0 good or 1 good oven).
- 0 good, 4 defective: \(\binom{9}{0} \times \binom{3}{4} = 0\) (since only 3 defective exist)
- 1 good, 3 defective: \(\binom{9}{1} \times \binom{3}{3} = 9 \times 1 = 9\)
Total unfavorable ways = \(0 + 9 = 9\).
Favorable ways = \(495 - 9 = 486\).
Probability = \(\frac{486}{495} = \frac{54}{55}\)
In simple words: Divide the good and defective batches, use combinations for each scenario, and use the complement for "at least 2".

Exam Tip: Checking "at least 2" via complement (0 or 1 good) is much faster than calculating 2, 3, and 4 good separately.

Question 13. A five digit number is formed by 1,2,3,4,5 without repetition. find the probability that the number is divisible by 4.
Answer: Total digits = 5 (1, 2, 3, 4, 5).
Total 5-digit numbers that can be formed without repetition = \(5! = 120\).
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
The possible last two digit combinations formed from {1, 2, 3, 4, 5} that are divisible by 4 are: 12, 24, 32, 52.
For each of these 2-digit endings, the remaining 3 digits can be arranged in \(3! = 6\) ways.
Total favorable numbers = \(4 \text{ endings} \times 6 = 24\).
Probability = \(\frac{24}{120} = \frac{1}{5}\)
In simple words: A number is divisible by 4 if its last two digits form a multiple of 4. Find all such endings and multiply by the ways to arrange the remaining digits.

Exam Tip: Divisibility rule for 4: check if the two-digit number at the end is divisible by 4.

Question 14. The letters of the word 'FORTUNATES' are arranged at random in a row. What is the chance that the two ' T ' come together.
Answer: The word FORTUNATES has 10 letters, where 'T' appears 2 times, and all other letters are distinct (F, O, R, U, N, A, E, S).
Total number of arrangements of the 10 letters = \(\frac{10!}{2!}\).
Treating the two 'T's as a single combined unit, we have 9 units in total.
These 9 units can be arranged in \(9!\) ways (since the two T's are identical, internal arrangement does not change the word).
Probability = \(\frac{9!}{\frac{10!}{2!}} = \frac{9! \times 2!}{10!} = \frac{2}{10} = \frac{1}{5}\)
In simple words: Group the two 'T's together as one letter, count total arrangements, and divide by the total unrestricted arrangements.

Exam Tip: Do not forget to account for repeating letters in the denominator when calculating permutations of words.

Question 15. A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has
(i) 1 boy and two girls?
(ii) At least one girl?
(iii) At most one girl?
Answer: Total students = 10 boys + 8 girls = 18 students.
Total ways to select 3 students = \(\binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816\).

(i) Probability that the group has 1 boy and 2 girls:
Favorable ways = \(\binom{10}{1} \times \binom{8}{2} = 10 \times \frac{8 \times 7}{2} = 10 \times 28 = 280\)
Probability = \(\frac{280}{816} = \frac{35}{102}\)

(ii) Probability of at least one girl:
Complement of "at least one girl" is "zero girls" (i.e., all 3 students are boys).
Unfavorable ways (3 boys, 0 girls) = \(\binom{10}{3} \times \binom{8}{0} = \frac{10 \times 9 \times 8}{6} = 120\)
Favorable ways = \(816 - 120 = 696\)
Probability = \(\frac{696}{816} = \frac{29}{34}\)

(iii) Probability of at most one girl:
"At most one girl" means either 0 girls or 1 girl.
- Case 1 (0 girls, 3 boys) = \(120\) ways.
- Case 2 (1 girl, 2 boys) = \(\binom{8}{1} \times \binom{10}{2} = 8 \times 45 = 360\) ways.
Total favorable ways = \(120 + 360 = 480\).
Probability = \(\frac{480}{816} = \frac{10}{17}\)
In simple words: Use combinations for each specific gender count, and use the complement rule for "at least one".
Exam Tip: "At most one" means 0 or 1, while "at least one" means 1, 2, or 3.