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CBSE 11th Standard Maths Subject Probability Ncert Exemplar 4 Marks Questions 2021

Published on: 27/05/2021

QB365 Provides the NCERT Exemplar Question Papers for Class 11 Maths , and also provide the detail solution for each and every ncert exemplar papers. Ncert Exemplar papers will help to get more idea about question pattern in every exams and also will help to get more marks in Exams

Download CBSE Class 11th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 11th Standard CBSE Mathematics

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Std: 11th Standard CBSE - Mathematics

chapter: - NCERT Exemplers

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Questions:

Generic 3

5 x 3 = 15

One card is drawn from a well-shuffled deck of 52 cards.Calculate the probability that the card will be
an ace

One card is drawn from a well-shuffled deck of 52 cards.Calculate the probability that the card will be
a black card

One card is drawn from a well-shuffled deck of 52 cards.Calculate the probability that the card will be
not an ace

Suppose an integer from 1 through 1000 is chosen at random.Find the probability that the integer is a multiple of 2 or a multiple of 9.

A die is rolled in such a way that each odd number is twice as likely to occur as likely to occur as each even number.Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

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Answers Key:

Generic 3

5 x 3 = 15

\(\frac { 1 }{ 13 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 12 }{ 13 } \)

0.556

When a die is rolled , then we get 1 or 2 or 3 or 4 or 5 or 6.
S = {1,2,3,4,5,6}
It is given that
P(each odd number = 2 x P (each even number)
P(1) = 2x, P(2) = x, P(3) = 2x ,P(4) = x, P(5) =2x and P(6) = x
We know that , P(5) = 1
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
\(\Rightarrow \) 2x + x + 2x + x + 2x + x = 1 \(\Rightarrow \) \(x=\frac { 1 }{ 9 }\)
Thus P(1) = P(3) = P(5) = \(\frac { 2 }{ 9 } \)
and P(2) = P(4) = P(6) = \(\frac { 1 }{ 9 } \)
Now, P(G) = P (getting 4 or 5 or 6)
= P(4) + P(5) +P(6) = \(\frac { 4 }{ 9 } \)