NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | MATHS IMPORTANT QUESTIONS | CHAPTER – 16 | PROBABILITY | EDUGROWN |

In This Post we are  providing Chapter-16 PROBABILITY NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PROBABILITY

Question 1.
A coin is tossed twice, what is the probability that at least one tail occurs?

Solution:
Let S is the sample space, then
S = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let E be the event of getting at least one tail, then
E = {(H, T), (T, H), (T, T)}
∴ n(E) = 3
Hence, required probability P(E) = n(E)n(S) = 34

Question 2.
There are four men and six women on the city council. If one council member is selected for a committee at random, then how likely is it that it is a woman?

Solution:
Let S is the sample space, then
n (S) = 10
Let E be the event that woman is selected, then
n(E) = 6
Hence, required probability P(E) = n(E)n(S)
P(E) = 610 = 35

Question 3.
If 211 is the probability of an event A, then find what is the probability of the event is ‘not A’.

Solution:
Given that :
P(A) = 211
We know that,
P(A) + P(A¯) = 1
211 + P(A¯) = 1
P(A¯) = 1 – 211 = 11–211
P(A¯) = 911.

Question 4.
A box contain 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that:

1. All marbles will be blue.
2. At least one marble will be green.

Solution:
In the box, there are 10 red, 20 blue and 30 green marbles.
The No. of marbles = 10 + 20 + 30 = 60
1. Total ways of choosing 5 marbles out of 60 marbles,
n(S) = ⁶⁰C₅
Let E is the event of choosing blue marbles, then
n(E) = ²⁰C₅
Probability = n(E)n(S) = 20C560C5

2. P(At least one green marble)
= 1 – P(No green)
= 1 – 30C560C5

Question 5.
4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining in card of 3 diamond and one spade?

Solution:
Let S is the sample space.
Total number of selecting 4 cards out of 52 cards, n(S) = ⁵²C₄
If E is the event obtaining card of 3 diamond and 2 spade, then
n(E) = ¹³C₃ x ¹³C₁
Probability P(E) = n(E)n(S) = 13C3×13C152C4

Question 6.
Given : P(A) = 35 and P(B) = 15. If A and B are mutually exclusive events, then find P(A or B). (NCERT)

Solution:
A and B are mutually exclusive events.
∴ P (A or B) = P(A ∪ B) = P(A) + P(B)
P(A ∪ B) = 35 + 15 = 3+15 = 45

Question 7.
In a lottery, a person choose six different natural numbers at random from 1 to 20 and if there six numbers match with the six numbers already fixed by lottery committee, he wins the prize, what is the probability of winning the prize in the game? (Hint: Order of the numbers is not important)

Solution:
Let S is the sample space, then
n(S) = ²⁰C₆
= 20×19×18×17×16×156×5×4×3×2×1 = 38760
Only one prize can be win.
∴ n(E) = 1
Hence, required probability P(E) = n(E)n(S) = 13876

Question 8.
Two dice are thrown simultaneously. Find the probability of getting a sum 9 in a single throw.

Solution:
Total number of ways in which two dice may be thrown
= 6 x 6 = 36
∴ n(S) = 36
Event of getting sum 9 is A = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(A) = 4
∴ Required probability P(A) = n(A)n(S) = 19

Question 9.
A bag contains 8 black, 6 white and 5 red balls. Find the probability of drawing a black or a white ball from it

Solution:
Total number of balls = 8 + 6 + 5 = 19
∴ n(S) = 19
Event A of drawing 1 black or 1 white ball
n(A) = 8 + 6 = 14
∴ n(A) = 14
∴ Required probability P(A) = n(A)n(S) = 1419

Question 10.
From a pack of well shuffled cards two cards drawn simultaneously. Find the probability that both the cards are ace.

Solution:
Total number of ways of drawing two cards out of 52
= ⁵²C₂
Number of ways drawing two ace out of 4 ace
= ⁴C₂
∴ Required Probability

Question 11.
A pair of dice are thrown. Find the probability that the sum is 9 or 11.

Solution:
Let the sample space be S, then
∴ n(S) = 36
Let E be the event that sum is 9 or 11, then
E = {(5, 4), (4, 5), (6, 3), (3, 6), (6, 5), (5, 6)}
∴ n (E) = 6
The probability of getting sum 9 or 11 is
P(E) = n(E)n(S) = 636 = 16
Hence, probability that the sum is not 9 or 11 is
P(E¯) = 1 – P(E)
= 1 – 16 = 56

Question 12

A letter is selected at random from the word ‘ASSASSINATION’. Find the probability that letter is

1. a vowel
2. a consonant

Solution:
Number of letters is 13 in which there are 6 vowels and 7 consonants.
1. Let sample space is S, then
∴ n(S) = 13
E₁ is the event of choosing a vowel, then

n(E₁) = 6
Hence, required probability P(E₁) = n(E1)n(S)
P(E₁) = 613

2. E₂ is the event of choosing a consonant, then
n(E₁) = 7
Hence, required probability P(E₂) = n(E2)n(S)

Question 13.
In class XI of a school, 40% of the 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, find the probability that he will be studying Mathematics or Biology.

Solution:
Let M and B denote the students of Mathematics and Biology respectively. Then, as given:
P(M) = 40% = 40100
= P(B) = 30% = 30100
P(M ∩ 5) = 10% = 10100
∴ P(M ∪ B) = P(M) + P(B) – P(M ∩ B)
40100 + 30100 – 10100
= 60100 = 60% = 0.6

Question 14.
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both? (NCERT)

Solution:
Let probability of passing in first examination is A and passing in the second examination is B.
P(A) = 0.8, P(B) = 0.7, P(A ∪ B) = 0.95, P(A ∩ B) = ?
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.95 = 0.8 + 0.7 – P(A ∩ B)
⇒ P(A ∩ B) = 1.5 – 0.95
∴ P(A ∩ B) = 0.55.

Question 15.
Check whether the following probabilities P{A) and P(B) are consistently defined:

1. P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
2. P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8.

Solution:
1. Given : P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
If P(A ∩ B) ≤ P(A) and P(A ∩ B) ≤ P(B)
Then, P(A) and P(B) are consistent.
Here

P(A) and P(B) are not consistent.