In this page, you can find the complete solutions of the first exercise of Probability chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, the derivatives is the 15th chapter and has only one exercise. Thus, this is the solution of the fir st and the final exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:
Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

 

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

1. A dice is thrown once. Determine the probability of getting


a) an even number

Solution: In a dice, there are three even numbers (2,4,6). So, probable events n(E) = 3

Total number of faces in a dice is 6. When it is thrown once, sample space n(S) = 6

So, probability of getting an even number (P) = $\dfrac{n(E)}{n(S)}$

$= \dfrac{3}{6}$

$= \dfrac{1}{2}$

b) a number $\geq$ 3

Solution: In a dice, there are four number greater than or equal to 3 (3,4,5,6). So probable events n(E) = 4

Total number of faces in a dice is 6. When it is thrown once, sample space n(S) = 6

So, probabilty of getting an even number (P) = $\dfrac{n(E)}{n(S)}$

$= \dfrac{4}{6}$
$= \dfrac{2}{3}$
c) a number $\leq$ 4 

Solution: In a dice, there are four number less than or equal to 4 (1,2,3,4). So probable events n(E) = 4

Total number of faces in a dice is 6. When it is thrown once, sample space n(S) = 6

So, probabilty of getting an even number (P) = $\dfrac{n(E)}{n(S)}$

$= \dfrac{4}{6}$
$= \dfrac{2}{3}$

2. A bag contains 9 red, 7 white and 4 black balls. A ball is drawn at random. What is the probability of drawing a) a red ball b) a red ball or a white ball c) not a red ball.

Solution:
In a bag,
number of red balls, n(R) = 9
number of white balls, n(W) = 7

number of black balls, n(B) = 4

total balls, n(S) = n(R) + n(W) + n(B) = 9 + 7 + 4 = 20

Now,
a) probability of drawing a red ball

$= \dfrac{n(R)}{n(S)}$

$= \dfrac{9}{20}$
b) probabilty of drawing a red ball or a white ball

$= \dfrac{n(R)}{n(S)} + \dfrac{n(W)}{n(S)}$

$= \dfrac{9}{20} + \dfrac{7}{20}$

$= \dfrac{16}{20}$
$= \dfrac{4}{5}$
It is because the question says we can either pick up a red ball or a white ball. If we pick a red ball, it counts. If we pick a white ball, it counts. So, we add the probabilities of both and find the result.

c) probability of not drawing a red ball

$= 1 - \text{probability of drawing a red ball}$

$= 1 - \dfrac{9}{20}$
$= \dfrac{20-9}{20}$
$= \dfrac{11}{20}$


3. A card is drawn from a well-shuffled deck of 52 cards. What is the probablity that it is a) a spade b) a red 8, a red 9 or a red 10 c) a king or a diamond d) a black or an ace

Solution:

Here, total number of events that can happen individually or sample space, n(S) = 52

a) probablity of drawing a spade

In a deck of cards, there are 13 spade cards. So, on drawing one at random from well-shuffled cards, probability of getting a spade is:

$\dfrac{\text{number of spade cards}}{n(S)}$

$= \dfrac{13}{52}$

$= \dfrac{1}{4}$

b) a red 8, a red 9 or a red 10

In a deck of cards, there are two red cards of the same number or character (one diamond and one hearts). So, there are two cards each of 8, 9 and 10. Thus, total number of required cards are n(C) = 6.

So, probablity is:

$= \dfrac{n(C)}{n(S)}$

$= \dfrac{6}{52}$

$= \dfrac{3}{26}$

c) a king or a diamond

In a deck of cards, there are only 4 king cards. And there are 13 diamond cards. But one of the diamond cards is the king itself. Therefore, total number of probable events n(D) = 4 + 13 - 1 = 16 $\implies n(D) = 16$

So, proability is:

$= \dfrac{n(D)}{n(S)}$

$= \dfrac{16}{52}$

$= \dfrac{4}{13}$

d) a black or an ace

There are 26 black cards in a deck of cards. There are four ace in the same deck. Out of which, 2 ace(s) are black themselves. So, number of probable events become n(E) = 26 + 4 - 2 = 28 $\implies n(E) = 28$

So, probabilty is:

$= \dfrac{n(E)}{n(S)}$

$= \dfrac{28}{52}$

$= \dfrac{7}{13}$



BELOW NOT SOLVED

11 a) $P(E_1) = 0.4, P(E_2) = 0.36 \ \text{and} \ P (E_1 \cup E_2) = 0.5$, find $P(E_1/E_2) \ \text{and} \ P(E_2/E_1)$.

Solution:

Given,
$P(E_1) = 0.4$
$P(E_2) = 0.36$
$P (E_1 \cup E_2) = 0.5$


$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$

$or, 0.5 = 0.4 + 0.36 - P(E_1 \cap E_2)$

$\therefore, P(E_1 \cap E_2) = 0.26$

Now,

$P(E_1/E_2) = \dfrac{ P(E_1 \cap E_2}{P(E_2)}$

$= \dfrac{0.26}{0.36}$

$= \dfrac{26}{100} * \dfrac{100}{36}$

$= \dfrac{13}{18}$


About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)