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Maths 3 Marks Question

Statistics and Probability · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 25 questions.

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Question 13 Marks
I toss three coins together. The possible outcomes are no heads, $1$ head, $2$ heads and $3$ heads. So, I say that probability of no heads is $\frac{1}{4}.$ What is wrong with this conclusion?
Answer
Three coins are tossed together.
Total outcomes $T(E) = 2^3 = 8$
$(TTH), (THH), (HTH), (HHT), (HTT), (THT)$ and $(HHH), (TTT)$, so, the number of favourable outcomes for event (getting no head) = 1
$\therefore$ Probability (getting no head) $=\frac{1}{8}$
Hence, the given statement is wrong $\Big(\because\frac{1}{8}\neq\frac{1}{4}\Big).$
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Question 23 Marks
Find the mean marks of students for the following distribution:
Marks
Number of students
0 and above
10 and above
20 and above
30 and above
40 and above
50 and above
60 and above
70 and above
80 and above
90 and above
100 and above
80
77
72
65
55
43
28
16
10
8
0
Answer
Marks
Class marks ($x_i$​​​​​​​)
Number of students (Cumulative frequency)
$f_i$
$f_i x_i$
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
100-110
5
15
25
35
45
55
65
75
85
95
105
80
77
72
65
55
43
28
16
10
8
0
3
5
7
10
12
15
12
6
2
8
0
15
75
175
350
540
825
780
450
170
760
0
 
 
 
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=4140$
$\text{mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}=\frac{4140}{80}=51.75$
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Question 33 Marks
A coin is tossed two times. Find the probability of getting at most one head.
Answer
The possible outcomes, if a coin is tossed 2 times is,
S = {(HH), (TT), (HT), (TH)}
$\therefore$ n(S) = 4
Let E = Event of getting atmost one head
= {(TT), (HT), (TH)}
$\therefore$ n(E) = 3
Hence, required probability $=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{4}$
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Question 43 Marks
A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.
Answer
We know that, in english alphabets, there are (5 vowels + 21 consonants) = 26 letters.
So, total number of outcomes in english alphabets are,
n(S) = 26
Let E = Event of choosing a english alphabet which is a consonent
= {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
$\therefore$ n(E) = 21
Hence, required porbability $=\frac{\text{n(E})}{\text{n(S)}}=\frac{21}{26}$
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Question 53 Marks
A carton of $24$ bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?
Answer
$\therefore$ Total number of bulbs, $n(S) = 24$
Let $E_1 =$ Event of selecting not defective bulb = Event of selecting good bulbs
$n(E_1) = 18$
$\therefore\ \ \text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{18}{24}=\frac{3}{4}$
Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton in a carton, n(S) = 23.
In them, 18 are good bulb and 5 are defective bulbs.
$\therefore$ P(selecting second defective bulb) $=\frac{5}{23}$
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Question 63 Marks
Two dice are thrown at the same time. Find the probability of getting:
  1. Same number on both dice.
  2. Different numbers on both dice.
Answer
Two dice are thrown at the same time. [given]
So, total number of possible outcomes = 36
  1. We have, same number on both dice.
So, possible outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

$\therefore$ Number of possible outcomes = 6

Now, required probability $=\frac{6}{36}=\frac{1}{6}$
  1. We have, different number on both dice.
So, number of possible outcomes

= 36 - Number of possible outcomes for same number on both dice

= 36 - 6 = 30

$\therefore\ \text{Required probability}=\frac{30}{36}=\frac{5}{6}$
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Question 73 Marks
What is the probability that the card is:
  1. A club.
  2. $10$ of hearts.
Answer
 
Let $E_3 =$ Event of gettiing a club
$n (E_3) = (13 - 3) = 10$
$\therefore\ \ \text{Required probability}=\frac{\text{n(E}_3)}{\text{n(S)}}=\frac{10}{49}$
Let $E_4 =$ Event of getting a king
$n(E_4) = 1$
[because in 52 playing cards only 13 are the heat cards and only one 10 in 13 heart cards]
$\therefore\ \ \text{Required probability}=\frac{\text{n(E}_4)}{\text{n(S)}}=\frac{1}{49}$
 
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Question 83 Marks
Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?
Answer
For Apoorv T(E) = 36
Favourable is only (6, 6) i. e., F(E) = 1
then P(F) by Apoorv $=\frac{\text{F(E)}}{\text{T(E)}}=\frac{1}{36}$
Now for Peehu T'(E) = 6
F'(E) = 1
$\text{p}'\text{(A)}=\frac{\text{F}'\text{(E)}}{\text{T}'\text{(E)}}=\frac{1}{6}$
$\frac{1}{6}>\frac{1}{36}$
$\therefore\ \ \text{P}'\text{(A)}>\text{P(A)}$
Hence, Peehu has the better chance.
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Question 93 Marks
A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is $\frac{1}{2}.$ Justify.
Answer
From 1 to 100 numbers, there are 50 even and 50 odd numbers.
Total number of outcomes T(E) = 100
Number of outcomes favourable for event E = F(E) = 50
So, $\text{P(F)}=\frac{50}{100}=\frac{1}{2}$
Similarly, the probability of getting odd numbers is $\frac{1}{2}.$ Hence the probability of getting odd and even each is,
Hence, the given statement is true.
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Question 103 Marks
An integer is chosen between $0$ and $100$. What is the probability that it is:
  1. Divisible by $7$?
  2. Not divisible by 7?
Answer
 
The number of integers between 0 and 100 is, $n(S)=99$
Let $E _1=$ Event of choosing an integer which is divisible by 7
$=$ Event of choosing an integer which is multiple of 7
$=\{7,14,21,28,35,42,49,56,63,70,77,84,91,98\}$
$\therefore$ $n(E_1) = 14$
$\therefore\ \ \text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{14}{99}$
Let $E_2 =$ Event of choosing an integer which is not divisible by $7$
$\therefore$ $n(E_2) = n(S) - n(E_1)$
$= 99 - 14 = 85$
$\therefore\ \ \text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{85}{99}$
 
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Question 113 Marks
The king, queen and jack of clubs are removed from a deck of $52$ playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is:
  1. A heart.
  2. A king.
Answer
 
If we remove one king, one queen and one jack of clubs from $52$ cards, then the remaining,
cards left, $n(S) = 49$
Let $E_1 = $Event of gettiing a heatt
$n (E_1) = 13$
$\therefore\ \ \text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{13}{49}$
Let $E_2 =$ Event of getting a king
$n(E_2) = 3$ [since, out of $4$ king, one club cards is already removed]
$\therefore\ \ \text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{3}{49}$
 
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Question 123 Marks
If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.
Answer
A coin is tossed 6 times so, T(E) = 6
In total six events, number of outcomes for getting head are 3 so F(E) = 3 again
P(F) getting head $=\frac{3}{6}=\frac{1}{2}$
Hence, the given statement is false.
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Question 133 Marks
The following table gives the number of pages written by Sarika for completing her own book for 30 days:
Number of pages written per day
6-18
19-21
22-24
25-27
28-30
Number of days
1
3
4
9
13
Find the mean number of pages written per day.
Answer
Since,
Class mark
mid-value ($x_i$​​​​​​​)
Number of days ($f_i​​​​​​​$)
$f_i x_i$
15.5-18.5
18.5-21.5
21.5-24.5
24.5-27.5
27.5-30.5
17
20
23
26
29
1
3
4
9
13
17
60
92
234
377
Total
 
30
780
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Therefore, mean $\bar{(\text{x})}=\frac{\sum\text{f}_\text{i}\text{ x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{780}{30}=26$
Hence, the mean of pages written per day is 26.
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Question 143 Marks
A lot consists of $48$ mobile phones of which $42$ are good, $3$ have only minor defects and $3$ have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is:
  1. acceptable to Varnika?
  2. acceptable to the trader?
Answer
Given, total number of mobile phones
$n(S) = 48$
Let $E _1=$ Event that Varnika will buy a mobile phone
$=$ Varnika buy only, it is good mobile
$\therefore n\left(E_1\right)=42$
$\therefore\ \ \text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{42}{48}=\frac{7}{8}$
Let $E _2=$ Event that trader will buy only when it has no major defects
$=$ Trader will buy only 45 mobiles
$\therefore$ $n(E_2) = 45$
$\therefore\ \ \text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{45}{48}=\frac{15}{16}$
 
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Question 153 Marks
Find the mean age of 100 residents of a town from the following data:
Age equal and above (in years)
0
10
20
30
40
50
60
70
Number of Persons
100
90
75
50
25
15
5
0
Answer
Here, we observe that, all 100 residents of a town have age equal and above 0. Since, 90 residents of a town have age equal and above 10.
So, 100 - 90 = 10 residents lies in the interval 0-10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table.
Class inteval
Number of persons $(f_i$​​​​​​​)
Class marks ($x_i$​​​​​​​)
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$
$f_i u_i$
0-10
10-20
20-30
30-40
40-50
50-60
60-70
100 - 90 = 10
90 - 75 = 15
75 - 50 = 25
50 - 25 = 25
25 - 15 = 10
15 - 5 = 10
5 - 0 = 5
5
15
25
35 = a
45
55
65
-3
-2
-1
0
1
2
3
-30
-30
-25
0
10
20
15
 
$\text{N}=\sum\text{f}_\text{i}=100$
 
 
$\sum\text{f}_\text{i}=\text{u}_\text{i}=-40$
Here, (assumed mean) a = 35
and (class width) h = 10
By step deviation method,
$\text{Mean }\bar{(\text{x})}=\text{a}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=35+\frac{(-40)}{100}\times10$
$=35-4=4=31.$
Hence, the required mean age is 31 year.
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Question 163 Marks
Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is:
  1. 7?
  2. A prime number?
  3. 1?
Answer
Two dice are thrown simultaneously. [given]
So, total number of possible outcomes = 36
  1. Sum of the number appearing on the dice is 7.
So, the possible ways are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).

Number of possible ways = 6

$\therefore\ \text{Required probability}=\frac{6}{36}=\frac{1}{6}$
  1. Sum of the number appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.
So, the possilble ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

Number of possible ways = 15

$\therefore\ \text{Required probability}=\frac{15}{36}=\frac{5}{12}$
  1. Sum of the number appearing on the dice is 1.
It is not possible, so its probaility is zero.
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Question 173 Marks
Weekly income of 600 families is tabulated below:
Weekly income (in Rs.)
Number of families
0-1000
1000-2000
2000-3000
3000-4000
4000-5000
5000-6000
250
190
100
40
15
5
Total
600
Compute the median income.
Answer
First we construct a cumulative frequency table,
Weekly income (in Rs.)
Number of familles ($f_i$​​​​​​​)
Cumulative frequency (cf)
0-1000
1000-2000 = mid class
2000-3000
3000-4000
4000-5000
5000-6000
250
190 = f
100
40
15
5
250
250 + 190 = 440
440 + 100 = 540
540 + 40 = 580
580 + 15 = 595
595 + 5 = 600
It is given that, n = 600
$\therefore\ \ \frac{\text{n}}{2}=\frac{600}{2}=300$
Since, cumulative frequency 400 lies in the interval 1000-2000
Here, (lover median class) l = 1000,
f = 190, cf = 250, (class width) h = 1000
and (total observation) n = 600
$\therefore\ \ \text{Median}=\text{l}+\frac{\Big\{\frac{\text{n}}{2}-\text{cf}\Big\}}{\text{f}}\times\text{h}$
$=1000+\frac{(300-250)}{190}\times1000$
$=1000+\frac{50}{190}\times1000$
$=1000+\frac{5000}{19}$
$=1000+236.15=1263.15$
Hence, the median income is Rs. 1263.15.
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Question 183 Marks
The daily income of a sample of 50 employees are tabulated as follows:
Income (in Rs.)
1-200
201-400
401-600
601-800
Number of employees
14
15
14
7
Find the mean daily income of employees.
Answer
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now, we first find class mark $x_i$​​​​​​​ of each class and then proceed as follows
Income (in Rs.)
Class marks $(x_i$)
Number of empolyees
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}=\frac{\text{x}_\text{i}-300.5}{200}$
$f_i u_i$
0.5-200.5
200.05-400.5
400.5-600.5
600.5-800.5
100.5
a = 300.5
500.5
700.5
14
15
14
7
-1
0
1
2
-14
0
14
14
 
 
$\text{N}=\sum\text{f}_\text{i}=50$
 
$\sum\text{f}_\text{i}\text{ u}_\text{i}=14$
$\therefore$ Assumed mean, a = 300.5
Class width, h = 200
and total obsevations, N = 50
By stop deviation method,
mean $=\text{a}+\text{h}\times\frac{1}{\text{N}}\sum_{\text{i}=1}^5\text{f}_\text{i}\text{ u}_\text{i}$
$=300.5+200\times\frac{1}{50}\times14$
$=300.5+56=356.5$
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Question 193 Marks
All the jacks, queens and kings are removed from a deck of $52$ playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value.
  1. $7$
  2. greater than $7$
  3. less than $7$
Answer
 
In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left, $n(S)=52-3 \times 4=40$.
Let $E_1=$ Event of getting a card whose value is 7
$E = Card$ value 7 may be of a spade, a diamond, a club or a heart
$\therefore n \left( E _1\right)=4$
$\therefore P \left( E _1\right)=\frac{ n \left( E _1\right)}{ n ( S )}=\frac{4}{40}=\frac{1}{10}$
Let $E_2=$ Event of getting a card whose value is greater than 7
$=$ Event of getting a card whose value is 8,9 or 10
$\therefore$ $n(E) = 3 × 4 = 12$
$\therefore\ \ \text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{12}{40}=\frac{3}{10}$
Let $E_3 =$ Event of getting a card whose value is less than$ 7$
= Event of getting a card whose value is $1, 2, 3, 4, 5 or 6$
$\therefore$ $n(E_3) = 6 \times 4 = 24$
$\therefore\ \ \text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}=\frac{24}{40}=\frac{3}{5}$
 
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Question 203 Marks
Cards with numbers $2$ to $101$ are placed in a box. A card is selected at random. Find the probability that the card has:
  1. An even number.
  2. A square number.
Answer
Total number of out comes with numbers 2 to 101,
$n(S) =100$
Let $E_1 =$ Event of selecting a card which is an even number,
$= {2, 4, 6, ...100}$
[in an AP, $l = a + (n - 1) d,$
here$ l = 100, a = 2$ and $d = 2 $
$\Rightarrow 100 = 2 + (n - 1)2 $
$\Rightarrow (n - 1) = 49 $
$\Rightarrow n = 50]$
$\therefore n(E_1) = 50$
$\therefore\ \ \text{Required probability}=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{50}{100}=\frac{1}{2}$
Let E_1 = Event of selecting a card which is an even number,
$= {4, 9, 16, 25, 36, 49, 64, 81, 100}$
$= {(2)^2, (3)^2, (4)^2, (5)^2, (6)^2, (7)^2, (8)^2, (9)^2, (10)^2}$
$\therefore n(E_2) = 9$
Hence, required porbability $=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{9}{100}$
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Question 213 Marks
A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a:
  1. triangle.
  2. square.
  3. square of blue colour.
  4. triangle of red colour.
Answer
Total number of figures,
n(S) = 8 triangles + 10 squares = 18
  1. P(lost piece is a triangle) $=\frac{8}{18}=\frac{4}{9}$
  2. P(lost piece is a squares) $=\frac{10}{18}=\frac{5}{9}$
  3. P(square of blue colour) $=\frac{6}{18}=\frac{1}{3}$
  4. P(triangle of red colour) $=\frac{5}{18}$
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Question 223 Marks
In a game, the entry fee is Rs. $5$. The game consists of a tossing a coin $3$ times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she:
  1. loses the entry fee.
  2. gets double entry fee.
  3. just gets her entry fee.
Answer
Total possible outcomes of tossing a coin 3 times,
$S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}$
$\therefore$ $n(S) = 8$
  1. Let $E_1 =$ Event that Sweta losses the entry fee
= She tosses tail on three times
$n(E_1) = {(TTT)}$
$\therefore\ \ \text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{1}{8}$
  1. Let $E_2 =$ Event that Sweta gets doluble entry fee
= She tosses heads on three times
= {(HHH)}
$\therefore$ $n(E_2) = 1$
$\therefore\ \ \text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{1}{8}$
  1. Let $E_3 =$ Event that Sweta gets her entry fee back
$= Sweta gets heads one or two times$
$= {(HTT), (THT), (TTH), (HHT), (HTH), (THH)}$
$\therefore$ $n(E_3) = 6$
$\therefore\ \ \text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}=\frac{6}{8}=\frac{3}{4}$
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Question 233 Marks
Determine the mean of the following distribution:
Marks
Number of students
Below 10
Below 20
Below 30
Below 40
Below 50
Below 60
Below 70
Below 80
Below 90
Below 100
5
9
17
29
45
60
70
78
83
85
Answer
Here, we observe that, 5 students have scored marks below 10, i.e. it lies between class interval 0-10 and 9 students have scored marks below 20,
So, (9 - 5) = 4 students lies in the class interval 10-20. Continuing in the same manner, we get the complete frequency distribution table for given data.
Marks
Number of students ($f_i$​​​​​​​)
Class marks ($x_i​​​​​​​$)
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}=\frac{\text{x}_\text{i}-45}{\text{h}}$
$f_i u_i$
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
5
9 - 5 = 4
17 - 9 = 8
29 - 17 = 12
45 - 29 = 16
60 - 45 = 45
70 - 60 = 10
78 - 70 = 8
83 - 78 = 5
85 - 83 = 2
5
15
25
35
a = 45
55
65
75
85
95
-4
-3
-2
-1
0
1
2
3
4
5
-20
-12
-16
-12
0
15.
20
24
20
10
 
$\text{N}=\sum\text{f}_\text{i}=85$
 
 
$\bar{\sum}\text{f}_\text{i}\text{u}_\text{i}=29$
Here, (assumed mean) a = 45
and (class width) h = 10
By step deviation method,
$\text{Mean }\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$ $=45+\frac{29}{85}\times10=45+\frac{58}{17}$
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Question 243 Marks
 A bag contains $10$ red, $5$ blue and $7$ green balls. A ball is drawn at random. Find the probability of this ball being a,
  1. red ball.
  2. green ball.
  3. not a blue ball.
Answer
If a ball is drawn out of 22 balls ( 5 blue +7 green +10 red), then the total number of outcomes are, $n(S)=22$
Let $E_1=$ Event of gettiing a red ball
$n\left(E_1\right)=10$
$\therefore$ Required probability $=\frac{ n \left( E _1\right)}{ n ( S )}=\frac{10}{22}=\frac{5}{11}$
Let $E_2=$ Event of getting a green ball
$n\left(E_2\right)=7$
$\therefore$ Required probability $=\frac{ n \left( E _2\right)}{ n ( S )}=\frac{7}{22}$
Let $E_3=$ Event of gettiing a red ball or green ball i.e., not a blue ball.
$n\left(E_3\right)=(10+7)=17$
$\therefore$ Required probability $=\frac{n\left(E_3\right)}{n(S)}=\frac{17}{22}$
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Question 253 Marks
Two dice are thrown at the same time. Determine the probabiity that the difference of the numbers on the two dice is 2.
Answer
The total number of sample space in two dice, n(S) = 6 × 6 = 36
Let E = Event of getting the number whose difference is 2.
= {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
$\therefore$ n(E) = 8
$\therefore\ \ \text{P(E})=\frac{\text{n(E})}{\text{n(S)}}=\frac{8}{36}=\frac{2}{9}$
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