Probability Important Questions

Probability

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Very Short Answer Questions (1 Mark Questions)

Ques. One letter is chosen randomly from “ASSASSINATION”. What is the probability that the letter is a vowel? 

Ans. Vowel = a, a, i, a, i, o = 6 vowels

 P (Vowel) = 6/ 13 

Ques. In the city council, there are around 6 men and 4 women. If a council member is selected for a committee randomly, then find the probability that it is men? 

Ans. P (men member is selected)= 6/ 10 = 3/ 5.

Ques. Two dice are thrown simultaneously. What is the probability of getting doublet? 

Ans. n = (S) = 36 

Let A be the event of getting doublet

{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

P (A) = 6/ 36 =\(1 \over 6\)

Read More: Types of Events in Probability

Ques. If 4/ 10 is the probability that an event will occur, Find the probability that it will not occur. 

Ans. P (E) = 4/ 10

As we know that,

= 1 – P (A) = 1- 4/ 10 = 6/ 10 = 3/ 5. 

Ques. If A and B are events such that P (A) = 1/ 4, P (B) = 1/2, P (A and B) = 1/8. Find the probability of getting no A and no B. 

Ans. P (A’ B’) = P (A B)’

= 1 - P (A B) { P (A B) = 1/ 4 + 1/ 2+ 1/ 8 = 5/ 8}

= 1 – 5/ 8 = 3/ 8.

Ques. A coin is tossed and a die is thrown, what is the sample space?

Ans. {H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6}

Ques. If E and F are two mutually exclusive events such as P (E) = 1/2, P (F) = 1/3. Find P (E or F).

Ans. As we know, 

P (E or F) = P (E) + P (F) – P (E ∩B) = 1/2 + 1/3 - ∅= 5/ 6

Read More: Difference Between Mutually Exclusive and Independent Events

Ques. Four cards are drawn from a deck of 52 cards. Find the probability of obtaining three diamonds and 1 spade. 

Ans. The number of ways of drawing four cards from 52 cards = 52C4

In a well-shuffled deck of 52 cards, there are 13 diamonds and 13 spades so, the possibility of getting 3 diamonds and 1 spade = 13C₃ X 13C₁

So, the probability of obtaining 3 diamonds and 1 spade = \(13C_3X13C_1 \over52C_4\)

Ques. What is the possibility that a leap year that is selected randomly will contain 53 Sundays? 

Ans. Total days in a leap year = 366 and complete weeks = 52 and 2 days over. Two days can be any two from Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. 

P (53 Sundays in a leap year)= 2/ 7

Ques. usually, the probability of any event lies between?

Ans. the probability of any event lies between 0 and 1. 

Read More: Sample Space

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Short Answer Questions (2 Marks Questions) 

Ques. In a throw of two dice, what is the probability that there is neither a doublet nor a total of 10 that appears? 

Ans. Let us assume S as the sample space and the event of getting doublet be E1 and the event of getting a total of 10 is E2. 

E1 = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

E2 = {(4, 6), (5, 5), (6, 4)}

n (S) = 36

then, P (E1) = 6/ 36 = 1/6 and P (E2) – 3/ 36 = 1/ 12

P (E1 E2) = 1 and P (E1 E2) = 2/ 9

P (E1’ E2’) = P (E1 E2)’

1 – P (E1 E2) = 1- 2/9 = 7/ 9. 

Ques. The probability of a person that will get an electrification contract is 2/ 5 and the probability of not getting a plumbing contractor is 4/ 7. If the probability that he will get not less than 1 contract is 2/ 3. Find the probability of the event that he will get both contracts. 

Ans. Let E be an event of getting an electrification contract and F be an event of getting a plumbing contract. 

P (E) = 2/ 5 and P (not F) = 4/ 7

P (F) = 1 – 4/ 7 = 3/ 7

P (E F) = 2/ 3 and P (E F) = P (E) + P (F) - P (E F)

= 2/ 5 + 3/ 7 – 2/ 3 = 17/ 105. 

Read More: Independent Events in Probability

Ques. What is the probability that in a random arrangement of letters of UNIVERSITY the two I’s come together? 

Ans. The word UNIVERSITY, the total words that can be formed 

= 10!/ 2!.

If we consider 2 I’s together, then the number of ways of arranging 

= 9!/ (10!/ 2!) = 1/ 5

Probability Formula

Ques. Let W = {W 1, W 2, W 3, W 4, W 5, W 6} be the sample space. Is the valid probability outcome? 

Ans. Yes, 1/ 6 + 1/ 6+ 1/6 + 1/6 + 1/6 + 1/6 = 1

Ques. If odds against any event are 6: 9, what is the probability of non-occurrence of this event? 

Ans. 1- 9/15 = (15 - 9)/ 15 = 6/ 15

Ques. In a random sampling, three items are selected from a lot and each item is tested. They are classified as defective (A) and non-defective (B). Write the sample space. 

Ans. Let the sample space be S

S= {AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB}

Read More: Theoretical Probability

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Long Answer Questions (3 Marks Questions)

Ques. One coin is tossed thrice. Consider the following event A: all tails appear, B: only one head comes and C: not less than two heads come, do they form a set of mutual exclusive & exhaustive events? 

Ans. Let the sample events be S

S= {HHH, HHT, HTT THT, HTH, THH, TTT, TTH}

A = {TTT}, B = {HTT, TTH, THT} and C = {HHH, HTH, THH, HHT}

Considering all events, A B C = S

So, all are exhaustive events and A\( \cap\) B = ∅, A \(\cap\) C = ∅ C\(\cap\)C = ∅, they are mutually exclusive. 

Ques. A book consists of 100 pages. One page is chosen randomly. Find the probability that the sum of the digit on the page is 9. 

Ans. Let the event be E.

E= {9, 18, 27, 36, 45, 54, 63, 72, 81, 90} and total number of pages is 100. 

P (E) = 10/100 = 1/10.

Ques. A hockey match timings are 3 pm to 5 pm. One man arrives late for the match. Find the probability that he misses the only goal of the match that was scored in the 20th minute of the match? 

Ans.  He can arrive between 3 pm to 5 pm. So, according to it, the time is 2 hours which means 120 minutes. 

He can see the goal of the match when he arrives within the first 20 minutes. 

P (see the goal) = 20/ 120 = 1/6.

P (does not see the goal) = 1 – 1/6 = 5/6. 

Ques. E and F are events such that P (E) = 0.42, P (F) = 0.48, and P (E and F) = 0.16. Determine 

(i) P (not E) 

(ii) P (not F)

 (iii) P (E or F)

Ans. According to the given equation, 

(i) P (not E) = 1- P (E) 

= 1 – 0.42 = 0.58

(ii) P (not F) = 1- P (F) 

= 1 – 0.48 = 0.52

 (iii) P (E or F) = P (E) + P (F) - P (A B)

= 0.42 + 0.48 – 0.16

= 0.74 

Ques. There is a group of 2 girls and 3 boys, two children are selected randomly. What will be the sample space of the following?

(i) E1 = both are girls

(ii) E2 = both are boys

(iii) E3 = one girl and one boy

(iv) E4 = at least one is girl 

Ans. Let the sample space be S

S = {G1 G2, G1 B1, G1 B2, G1 B3, G2 B1, G2 B2, G2 B3, B1 B2, B1 B3, B2 B3}

E1 ={G1 G2}

E2 ={B1 B2, B1 B3, B2 B3}

E3 ={ G1 B1, G1 B2, G1 B3, G2 B1, G2 B2, G2 B3}

E4 ={G1 G2, G1 B1, G1 B2, G1 B3, G2 B1, G2 B2, G2 B3}

Read More: Bayes' Theorem: Introduction, Proof, Formula and Derivation

Ques. In a town, there are 6000 people of which 1200 are over 50 years old and 2000 are females. It is said that 30% of females are over 50 years. Find the probability that an individual chosen randomly from the town is either female or over 50 years. 

Ans. Let the event of person being a female is E1, and the event of person being over 50 years old is E2. 

n (E1) = 2,000, n (E2) = 1,200

n (E1 E2) = 30% of 2,000 = 30/100 2,000 = 600

n (E1 E2) = n (E1) + n (E2) - n (E1 E2)

= 2000 + 1200 – 600 = 2,600 

n (E1 E2) = 2600/ 6000 = 13/30. 

Ques. What is the probability when 7 cards are drawn randomly from a deck of 52 cards, it contains

(i) All kings 

(ii) Three kings

 (iii) At least three kings

Ans. (i) P (all king) =\( 4C_4 X 48C_3\over52C_7\)=\(1 \over 7735\)

(ii) P (3 kings)= \(4C_3X48C_4\over 52C_7\)=91547

 (iii) P (at least three king) = P (three king) + P (4 king) = 9/ 1547 + 1/ 7735 = 46/ 7735

Ques. In class XI of a school, 40% of students study mathematics and 30% study biology. 10% of the class study both mathematics and biology. If a student is selected at random from the class, what is the probability that she will be studying mathematics or biology? 

Ans. Let M be mathematics and B be biology. 

P (M) = 40/ 100 and P (B) = 30/ 100

P (M B) = 10/ 100

P (M B) = P (M) + P (B) – P (M B)

= 40/ 100 + 30/ 100 – 10/ 100 = 0.6

Read More: Probability Distribution

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Very Long Answer Questions (5 Marks Question)

Ques. A card has been drawn from a well-shuffled deck of 52 cards. What will be the probability that a card will be an 

(i) Diamond

(ii) Black card

(iii) Not an ace

(iv) Not a diamond

Ans. (i) The probability of a diamond card

As we know that in a deck of 52 cards, there are 13 diamond cards. So, the required probability is: 

P( a diamond card) = 13/52 = 1/4

(ii) The probability of a black card

As we know that in a deck of 52 cards, there are 26 black cards. So, the required probability is: 

P( a black card) = 

26/52 = 1/2

(iii) The probability of not having an ace card

As we know that in a deck of 52 cards, there are 4 ace cards. So, the required probability is: 

P( not an ace card) = 

= 1- 4/52 = 1- 1/ 13

= (13 - 1)/ 13 = 12/ 13

(iv) The probability of not a diamond card

As we know that in a deck of 52 cards, there are 13 diamond cards. So, the required probability is: 

P( not a diamond card) = 

= 1- 13/52 = 1- 1/4

= (4 - 1)/ 4 = 3/4 

Ques. There are 20 cards that are numbered from 1 to 20. If a card is withdrawn randomly, then find the probability that a number on the card will be:

(i) Multiple of 4

(ii) Even number

(iii) Not divided by 5

(iv) Prime Number 

Ans. Let us assume that the sample space is S, which gives

S= { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

(i) Multiple of 4

The probability of a card drawn for a multiple of 4 

Let us assume that, the event of getting a multiple of 4 numbers is E1. 

Then E1={4, 8, 12, 16, 20}

Hence, P(E1) = 5/20 = 1/4 

(ii) Even number

The probability of a card drawn for an even number 

Let us assume that, the event of getting an even number is E2. 

Then E2={2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

Hence, P(E2) = 10/20 = 1/2

(iii) Not divided by 5

The probability of a card drawn for not divided by 5 

Let us assume that, the event of getting not divided by 5 numbers is E3. 

Then E3={1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19}

Hence, P(E3) = 16/20 = 8/10 = 4/5 

(iv) Prime Number 

The probability of a card drawn for a prime number

Let us assume that, the event of getting a prime number is E4. 

Then E4={2, 3, 5, 7, 11, 13, 17, 19}

Hence, P(E1) = 8/20 = 4/ 10 = 2/ 5.

Read More: Even and Odd Numbers

Ques. One die has two faces each with number 1, three faces each with number 2 and one face with number 3, if the die rolled only once, the find

(i) P (2)

(ii) P (1 or 3) 

(iii) P (not 3)

Ans. Let E be the event getting a face with number 1

Let F be the event getting a face with number 2

Let G be the event getting a face with number 3

P (E) = 2/ 6 = 1/ 3

P (F) = 3/ 6 = 1/ 2

P (G) = 1/ 6

(i) P (2) = 1/ 3

(ii) P (1 or 3) = P (1) + P (3)

= 1/ 3+ 1/ 6 = 1/ 2

(iii) P (not 3) = 1 – 1/6 = 5/ 6

Ques. Two students Anil and Sheena appeared in an exam. The probability that Anil will clear the exam is 0.05 and that Sheena will clear the exam is 0. 10. The probability of both will clear the exam is 0.02. What is the probability that

(a) Both Anil and Sheena will clear the exam

(b) At least one of them will not clear the exam

(c) only one of them will clear the exam

Ans. Let the events A and B denote that Anil and Sheena will pass the exam

P (A) = 0. 05, P (B) = 0. 10, P (A B) = 0. 02

(a) P (Both Anil and Sheena will clear the exam)

P (A’ B’) = P (A B)’ = 1 - P (A B) = 1- [P (A) + P (B) - P (A B)]

= 1 – 0.13 = 0.87

(b) P (At least one of them will not clear the exam) = 

1 – P (both will clear) = 1- 0. 02 = 0. 98

(c) P (only one of them will clear the exam) = 

P (A\(\cap\) B’) + P (A’ \(\cap\) B) = P (A) - P (A \(\cap\) B) + P (B) - P (A \(\cap\) B)

= 0. 05 – 0.02 + 0. 10 – 0. 02

= 0. 11 

Ques. If an entrance exam that is graded based on two exams, the probability of chosen at random, students clearing the 1st exam is 0.8 and the probability of passing the 2nd exam is 0.7. The probability of clearing at least one of them is 0. 95. Find the probability of clearing both. 

Ans. Let E be the students cleared first exam and F be the students passes 2nd exam

P (E) = 0. 8, P (F) = 0. 7, P (A B) = 0. 95

P (E F) = ?

P (A B) = P (E) + P (F) – P (E \(\cap\) F)

0.95 = 0.8 + 0.7 - P (E \(\cap\) F)

0.55 = P (E \(\cap\) F)

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