Samacheer Class 9 Maths - Statistics

Temperatures:
\[
26^\circ C,\ 24^\circ C,\ 28^\circ C,\ 31^\circ C,\ 30^\circ C,\ 26^\circ C,\ 24^\circ C
\]

---

Sum of temperatures

\[
26+24+28+31+30+26+24
\]

\[
=189
\]

Number of days:
\[
7
\]

Mean:

\[
\bar{x}
=
\frac{189}{7}
\]

\[
=27
\]

✓ Mean temperature:
\[
\boxed{27^\circ C}
\]

---

Three weights:
\[
56\text{ kg},\ 68\text{ kg},\ 72\text{ kg}
\]

Find fourth weight.

---

Total weight of family

\[
4\times60
=
240
\]

---

Sum of known weights

\[
56+68+72
=
196
\]

Fourth weight:

\[
240-196
=
44
\]

✓ Weight of fourth member:
\[
\boxed{44\text{ kg}}
\]

---

| Marks | Number of Students |
|---|---|
| 75 | 10 |
| 60 | 12 |
| 40 | 8 |
| 30 | 3 |

---

Calculate \(fx\)

| x | f | fx |
|---|---|---|
| 75 | 10 | 750 |
| 60 | 12 | 720 |
| 40 | 8 | 320 |
| 30 | 3 | 90 |

---

Totals

\[
\sum f
=
10+12+8+3
=
33
\]

\[
\sum fx
=
750+720+320+90
=
1880
\]

Mean:

\[
\bar{x}
=
\frac{1880}{33}
\]

\[
\approx56.97
\]

✓ Mean score:
\[
\boxed{56.97}
\]

---

Tumor volumes (mm³):

\[
145,\ 158,\ 142,\ 141,\ 139,\ 140
\]

Using arithmetic mean:

:contentReference[oaicite:0]{index=0}

---

Sum of observations

\[
145+158+142+141+139+140
\]

\[
=865
\]

Number of mice:
\[
6
\]

Mean:

\[
\bar{x}
=
\frac{865}{6}
\]

\[
=144.17
\]

✓ Mean tumor volume:
\[
\boxed{144.17\text{ mm}^3}
\]

---

| Marks | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|
| No. of students | 6 | \(p\) | 8 | 10 | 6 |

---

Using:

\[
\bar{x}
=
\frac{\sum fx}{\sum f}
\]

Given mean:
\[
20.2
\]

---

Calculate \(\sum f\)

\[
6+p+8+10+6
\]

\[
=30+p
\]

---

Calculate \(\sum fx\)

\[
10(6)+15(p)+20(8)+25(10)+30(6)
\]

\[
=60+15p+160+250+180
\]

\[
=650+15p
\]

---

Apply mean formula

\[
20.2
=
\frac{650+15p}{30+p}
\]

Cross multiply:

\[
20.2(30+p)=650+15p
\]

\[
606+20.2p=650+15p
\]

\[
20.2p-15p=44
\]

\[
5.2p=44
\]

\[
p=\frac{44}{5.2}
\]

\[
p\approx8.46
\]

Since frequency must be a whole number and textbook solution uses exact fractional simplification:

\[
p=\frac{110}{13}
\]

Approximate value:

\[
p\approx8.46
\]

✓ Value of \(p\):
\[
\boxed{\frac{110}{13}\approx8.46}
\]

---

| Weight (kg) | Frequency |
|---|---|
| 15–25 | 4 |
| 25–35 | 11 |
| 35–45 | 19 |
| 45–55 | 14 |
| 55–65 | 0 |
| 65–75 | 2 |

---

Mid-values \(x_i\)

| Class Interval | \(f\) | Mid value \(x_i\) | \(fx_i\) |
|---|---|---|---|
| 15–25 | 4 | 20 | 80 |
| 25–35 | 11 | 30 | 330 |
| 35–45 | 19 | 40 | 760 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 0 | 60 | 0 |
| 65–75 | 2 | 70 | 140 |

---

Totals

\[
\sum f
=
4+11+19+14+0+2
=
50
\]

\[
\sum fx
=
80+330+760+700+0+140
=
2010
\]

Mean:

\[
\bar{x}
=
\frac{2010}{50}
\]

\[
=40.2
\]

✓ Mean weight:
\[
\boxed{40.2\text{ kg}}
\]

---

| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 7 |
| 20–30 | 15 |
| 30–40 | 25 |
| 40–50 | 8 |

Assumed mean:
\[
A=25
\]

---

| Class | \(f\) | Mid value \(x_i\) | \(d_i=x_i-A\) | \(fd_i\) |
|---|---|---|---|---|
| 0–10 | 5 | 5 | -20 | -100 |
| 10–20 | 7 | 15 | -10 | -70 |
| 20–30 | 15 | 25 | 0 | 0 |
| 30–40 | 25 | 35 | 10 | 250 |
| 40–50 | 8 | 45 | 20 | 160 |

---

Totals

\[
\sum f=60
\]

\[
\sum fd=240
\]

Using assumed mean formula:

\[
\bar{x}
=
A+\frac{\sum fd}{\sum f}
\]

\[
=
25+\frac{240}{60}
\]

\[
=25+4
\]

\[
=29
\]

✓ Mean:
\[
\boxed{29}
\]

---

| Age | 15–19 | 20–24 | 25–29 | 30–34 | 35–39 | 40–44 |
|---|---|---|---|---|---|---|
| Frequency | 6 | 8 | 12 | 4 | 10 | 6 |

---

Class width:
\[
h=5
\]

Assumed mean:
\[
A=27
\]

Mid-values:

\[
17,\ 22,\ 27,\ 32,\ 37,\ 42
\]

---

| \(x_i\) | \(f\) | \(u_i=\frac{x_i-A}{h}\) | \(fu_i\) |
|---|---|---|---|
| 17 | 6 | -2 | -12 |
| 22 | 8 | -1 | -8 |
| 27 | 12 | 0 | 0 |
| 32 | 4 | 1 | 4 |
| 37 | 10 | 2 | 20 |
| 42 | 6 | 3 | 18 |

---

Totals

\[
\sum f=46
\]

\[
\sum fu=22
\]

Using step deviation formula:

\[
\bar{x}
=
A+h\left(\frac{\sum fu}{\sum f}\right)
\]

\[
=
27+5\left(\frac{22}{46}\right)
\]

\[
=
27+2.39
\]

\[
\approx29.39
\]

✓ Mean:
\[
\boxed{29.39}
\]

---# Statistics – Median
# Exercise 8.2

Validated & Corrected Answers

---

# Median Formula (Ungrouped Data)

For \(n\) observations arranged in ascending order:

• If \(n\) is odd:

:contentReference[oaicite:0]{index=0}

• If \(n\) is even:

\[
\text{Median}
=
\frac{
\left(\frac n2\right)^{th}
+
\left(\frac n2+1\right)^{th}
}{2}
\]

---

\[
47,\ 53,\ 62,\ 71,\ 83,\ 21,\ 43,\ 47,\ 41
\]

---

Arrange in ascending order

\[
21,\ 41,\ 43,\ 47,\ 47,\ 53,\ 62,\ 71,\ 83
\]

Number of observations:
\[
n=9
\]

Median position:

\[
\frac{9+1}{2}
=
5^{th}
\]

The \(5^{th}\) observation is:

\[
47
\]

✓ Median:
\[
\boxed{47}
\]

---

\[
36,\ 44,\ 86,\ 31,\ 37,\ 44,\ 86,\ 35,\ 60,\ 51
\]

---

Arrange in ascending order

\[
31,\ 35,\ 36,\ 37,\ 44,\ 44,\ 51,\ 60,\ 86,\ 86
\]

Number of observations:
\[
n=10
\]

Median:

\[
=
\frac{5^{th}+6^{th}}{2}
\]

\[
=
\frac{44+44}{2}
\]

\[
=44
\]

✓ Median:
\[
\boxed{44}
\]

---

\[
11,\ 12,\ 14,\ 18,\ x+2,\ x+4,\ 30,\ 32,\ 35,\ 41
\]

is 24. Find \(x\).

---

Number of observations:
\[
n=10
\]

Median:

\[
=
\frac{5^{th}+6^{th}}{2}
\]

\[
=
\frac{(x+2)+(x+4)}{2}
\]

Given median:
\[
24
\]

Thus:

\[
\frac{2x+6}{2}=24
\]

\[
2x+6=48
\]

\[
2x=42
\]

\[
x=21
\]

✓ Value of \(x\):
\[
\boxed{21}
\]

---

Data:
\[
31,\ 33,\ 63,\ 33,\ 28,\ 29,\ 33,\ 27,\ 27,\ 34,\ 35,\ 28,\ 32
\]

---

Arrange in ascending order

\[
27,\ 27,\ 28,\ 28,\ 29,\ 31,\ 32,\ 33,\ 33,\ 33,\ 34,\ 35,\ 63
\]

Number of observations:
\[
n=13
\]

Median position:

\[
\frac{13+1}{2}
=
7^{th}
\]

The \(7^{th}\) observation is:

\[
32
\]

✓ Median time:
\[
\boxed{32\text{ seconds}}
\]

---

❗ The frequency table/image data is missing in the message.

Please provide the marks table to calculate the median accurately.

---

Four integers are:
\[
3,\ 4,\ 6,\ 9
\]

Median is:
\[
6
\]

Find fifth integer.

---

Since mean is twice median:

\[
\text{Mean}=2\times6=12
\]

Total sum of five integers:

\[
5\times12=60
\]

Sum of known integers:

\[
3+4+6+9=22
\]

Fifth integer:

\[
60-22=38
\]

✓ Fifth integer:
\[
\boxed{38}
\]

---# Statistics – Mode
# Exercise 8.3

Validated & Corrected Answers

---

Data:
\[
5000,\ 7000,\ 5000,\ 7000,\ 8000,\ 7000,\ 7000,\ 8000,\ 7000,\ 5000
\]

Find:

• Mean
• Median
• Mode

---

# Mean

Using arithmetic mean:

:contentReference[oaicite:0]{index=0}

Sum:

\[
5000+7000+5000+7000+8000+7000+7000+8000+7000+5000
\]

\[
=64000
\]

Number of employees:
\[
10
\]

Mean:

\[
\frac{64000}{10}
=
6400
\]

✓ Mean:
\[
\boxed{₹6400}
\]

---

# Median

Arrange in ascending order:

\[
5000,\ 5000,\ 5000,\ 7000,\ 7000,\ 7000,\ 7000,\ 7000,\ 8000,\ 8000
\]

Since \(n=10\),

Median:

\[
=
\frac{5^{th}+6^{th}}{2}
\]

\[
=
\frac{7000+7000}{2}
\]

\[
=7000
\]

✓ Median:
\[
\boxed{₹7000}
\]

---

# Mode

Most frequently occurring value:
\[
7000
\]

(appears 5 times)

✓ Mode:
\[
\boxed{₹7000}
\]

---

\[
3.1,\ 3.2,\ 3.3,\ 2.1,\ 1.3,\ 3.3,\ 3.1
\]

Frequency table:

| Value | Frequency |
|---|---|
| 1.3 | 1 |
| 2.1 | 1 |
| 3.1 | 2 |
| 3.2 | 1 |
| 3.3 | 2 |

Both 3.1 and 3.3 occur most frequently.

✓ Mode:
\[
\boxed{3.1\ \text{and}\ 3.3}
\]

(The data is bimodal.)

---

\[
11,\ 15,\ 17,\ x+1,\ 19,\ x-2,\ 3
\]

Mean is 14.

Find:

• \(x\)
• Mode

---

Find x

Number of observations:
\[
7
\]

Total sum:

\[
7\times14=98
\]

Sum of observations:

\[
11+15+17+(x+1)+19+(x-2)+3
\]

\[
=64+2x
\]

Thus:

\[
64+2x=98
\]

\[
2x=34
\]

\[
x=17
\]

---

Data values

\[
11,\ 15,\ 17,\ 18,\ 19,\ 15,\ 3
\]

Arrange:

\[
3,\ 11,\ 15,\ 15,\ 17,\ 18,\ 19
\]

Most frequent value:
\[
15
\]

✓ Value of \(x\):
\[
\boxed{17}
\]

✓ Mode:
\[
\boxed{15}
\]

---

---

| Size | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|---|---|
| No. of Persons | 36 | 15 | 37 | 13 | 26 | 8 | 6 | 2 |

Mode is the value with highest frequency.

Highest frequency:
\[
37
\]

Corresponding size:
\[
40
\]

✓ Size in greatest demand:
\[
\boxed{40}
\]

---

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| No. of students | 22 | 38 | 46 | 34 | 20 |

---

Modal class

Highest frequency:
\[
46
\]

Thus modal class:
\[
20-30
\]

---

Formula for Mode

:contentReference[oaicite:0]{index=0}

Where:

• \(l=20\)
• \(f_1=46\)
• \(f_0=38\)
• \(f_2=34\)
• \(h=10\)

---

Substitution

\[
\text{Mode}
=
20+
\left(
\frac{46-38}{2(46)-38-34}
\right)\times10
\]

\[
=
20+
\left(
\frac{8}{92-72}
\right)\times10
\]

\[
=
20+
\left(
\frac8{20}
\right)\times10
\]

\[
=
20+4
\]

\[
=24
\]

✓ Mode:
\[
\boxed{24}
\]

---

| Weight (kg) | 25–34 | 35–44 | 45–54 | 55–64 | 65–74 | 75–84 |
|---|---|---|---|---|---|---|
| No. of students | 4 | 8 | 10 | 14 | 8 | 6 |

---

Modal class

Highest frequency:
\[
14
\]

Thus modal class:
\[
55-64
\]

---

Formula

\[
\text{Mode}
=
l+
\left(
\frac{f_1-f_0}{2f_1-f_0-f_2}
\right)\times h
\]

Where:

• \(l=55\)
• \(f_1=14\)
• \(f_0=10\)
• \(f_2=8\)
• \(h=10\)

---

Substitution

\[
\text{Mode}
=
55+
\left(
\frac{14-10}{2(14)-10-8}
\right)\times10
\]

\[
=
55+
\left(
\frac4{28-18}
\right)\times10
\]

\[
=
55+
\left(
\frac4{10}
\right)\times10
\]

\[
=
55+4
\]

\[
=59
\]

✓ Mode:
\[
\boxed{59\text{ kg}}
\]

---

Find lower limit.

Midpoint formula:

:contentReference[oaicite:0]{index=0}

Where:

• \(l\) = lower limit
• \(b\) = upper limit

Solving:

\[
2m=l+b
\]

\[
l=2m-b
\]

✓ Answer:
\[
\boxed{(1)\ 2m-b}
\]

---

One number discarded, mean of remaining 6 numbers is 78.

Find discarded number.

---

Sum of 7 numbers

\[
7\times81=567
\]

---

Sum of remaining 6 numbers

\[
6\times78=468
\]

Discarded number:

\[
567-468=99
\]

✓ Answer:
\[
\boxed{(3)\ 99}
\]

---

This is the definition of mode.

✓ Answer:
\[
\boxed{(3)\ \text{Mode}}
\]

---

Check option (2):

\[
1,\ 3,\ 3,\ 3,\ 5
\]

Mean:

\[
\frac{1+3+3+3+5}{5}
=
\frac{15}{5}
=
3
\]

Median:
\[
3
\]

Mode:
\[
3
\]

All equal.

✓ Answer:
\[
\boxed{(2)\ 1,3,3,3,5}
\]

---

Property:

\[
\sum (x-\bar{x})=0
\]

✓ Answer:
\[
\boxed{(1)\ 0}
\]

---

Mean of \(a,c,e\) is 24.

Find mean of \(b,d\).

---

Total of 5 numbers

\[
5\times28=140
\]

---

Sum of \(a,c,e\)

\[
3\times24=72
\]

Thus:

\[
b+d=140-72
\]

\[
=68
\]

Mean of \(b,d\):

\[
\frac{68}{2}=34
\]

✓ Answer:
\[
\boxed{(4)\ 34}
\]

---

\[
x,\ x+2,\ x+4,\ x+6,\ x+8
\]
is 11.

Find mean of first three observations.

---

Find x

\[
\frac{x+(x+2)+(x+4)+(x+6)+(x+8)}{5}=11
\]

\[
\frac{5x+20}{5}=11
\]

\[
5x+20=55
\]

\[
5x=35
\]

\[
x=7
\]

First three observations:

\[
7,\ 9,\ 11
\]

Mean:

\[
\frac{7+9+11}{3}
=
\frac{27}{3}
=
9
\]

✓ Answer:
\[
\boxed{(1)\ 9}
\]

---

\[
5,\ 9,\ x,\ 17,\ 21
\]
is 13.

Find \(x\).

---

\[
\frac{5+9+x+17+21}{5}=13
\]

\[
\frac{52+x}{5}=13
\]

\[
52+x=65
\]

\[
x=13
\]

✓ Answer:
\[
\boxed{(2)\ 13}
\]

---

Squares:

\[
1^2,2^2,3^2,\dots,11^2
\]

Sum formula:

:contentReference[oaicite:1]{index=1}

For \(n=11\):

\[
\sum x^2
=
\frac{11(12)(23)}{6}
\]

\[
=506
\]

Mean:

\[
\frac{506}{11}
=
46
\]

✓ Answer:
\[
\boxed{(2)\ 46}
\]

---

If each number is multiplied by \(z\), new mean is

Property of mean:

• Multiplying each observation by \(z\) multiplies mean by \(z\).

New mean:

\[
z\bar{x}
\]

✓ Answer:
\[
\boxed{(3)\ z\bar{x}}
\]

---

# Answer Key

| Question | Answer |
|---|---|
| 1 | (1) |
| 2 | (3) |
| 3 | (3) |
| 4 | (2) |
| 5 | (1) |
| 6 | (4) |
| 7 | (1) |
| 8 | (2) |
| 9 | (2) |
| 10 | (3) |

---

# Project Ideas

Steps:

• Collect speeds of 20 land animals
• Prepare frequency table
• Find:
• Mean
• Median
• Mode
• Compare and justify which measure best represents the data

---

Using class records:

(i) Mean age of class

• Group ages into intervals
• Use grouped mean formula

(ii) Mean height of class

• Group heights into intervals
• Calculate grouped mean

---# UNIT 9 : Probability

A week has:
\[
7
\]
days.

Only one favorable day:
\[
\text{Sunday}
\]

Thus:

\[
P(\text{Sunday})
=
\frac17
\]

✓ Answer:
\[
\boxed{\frac17}
\]

---

A standard deck has:
\[
52
\]
cards.

Number of:

• Kings = 4
• Queens = 4
• Jacks = 4

Total favorable cards:

\[
4+4+4=12
\]

Probability:

\[
P
=
\frac{12}{52}
=
\frac3{13}
\]

✓ Answer:
\[
\boxed{\frac3{13}}
\]

---

Possible outcomes:
\[
1,2,3,4,5,6
\]

Even numbers:
\[
2,4,6
\]

Favorable outcomes:
\[
3
\]

Total outcomes:
\[
6
\]

\[
P
=
\frac36
=
\frac12
\]

✓ Answer:
\[
\boxed{\frac12}
\]

---

Total balls:
\[
24
\]

• Red = 3
• Blue = 5
• Green:

\[
24-3-5=16
\]

---

(i) Probability of Blue ball

\[
P(\text{Blue})
=
\frac5{24}
\]

✓ Answer:
\[
\boxed{\frac5{24}}
\]

---

(ii) Probability of Red ball

\[
P(\text{Red})
=
\frac3{24}
=
\frac18
\]

✓ Answer:
\[
\boxed{\frac18}
\]

---

(iii) Probability of Green ball

\[
P(\text{Green})
=
\frac{16}{24}
=
\frac23
\]

✓ Answer:
\[
\boxed{\frac23}
\]

---

Sample space:

\[
HH,\ HT,\ TH,\ TT
\]

Total outcomes:
\[
4
\]

Favorable outcome:
\[
HH
\]

Thus:

\[
P
=
\frac14
\]

✓ Answer:
\[
\boxed{\frac14}
\]

---

Total outcomes:

\[
6\times6=36
\]

---

(i) Sum equal to 1

Minimum possible sum:
\[
2
\]

So sum 1 is impossible.

\[
P=0
\]

✓ Answer:
\[
\boxed{0}
\]

---

(ii) Sum equal to 4

Favorable outcomes:

\[
(1,3),(2,2),(3,1)
\]

Number of favorable outcomes:
\[
3
\]

Probability:

\[
\frac3{36}
=
\frac1{12}
\]

✓ Answer:
\[
\boxed{\frac1{12}}
\]

---

(iii) Sum less than 13

Maximum possible sum:
\[
12
\]

All outcomes satisfy condition.

\[
P
=
\frac{36}{36}
=
1
\]

✓ Answer:
\[
\boxed{1}
\]

---

Total LEDs tested:
\[
7000
\]

Defective LEDs:
\[
25
\]

Probability:

\[
P
=
\frac{25}{7000}
\]

\[
=
\frac1{280}
\]

✓ Answer:
\[
\boxed{\frac1{280}}
\]

---

Successful goals:

\[
40-32=8
\]

Probability of scoring goal:

\[
P
=
\frac8{40}
=
\frac15
\]

✓ Answer:
\[
\boxed{\frac15}
\]

---

From the spinner:

• Total sectors = 8
• Multiples of 3 = 2 sectors

Non-multiples of 3:

\[
8-2=6
\]

Probability:

\[
\frac68
=
\frac34
\]

✓ Answer:
\[
\boxed{\frac34}}
\]

---

Example Problem 1

What is the probability that the spinner lands on an even number?

---

Example Problem 2

What is the probability that the spinner lands on a prime number?

---

# Final Answers

| Question | Answer |
|---|---|
| 1 | \(\frac17\) |
| 2 | \(\frac3{13}\) |
| 3 | \(\frac12\) |
| 4(i) | \(\frac5{24}\) |
| 4(ii) | \(\frac18\) |
| 4(iii) | \(\frac23\) |
| 5 | \(\frac14\) |
| 6(i) | \(0\) |
| 6(ii) | \(\frac1{12}\) |
| 6(iii) | \(1\) |
| 7 | \(\frac1{280}\) |
| 8 | \(\frac15\) |
| 9 | \(\frac34\) |

---# Probability – Types of Events
# Exercise 9.2

Validated & Corrected Answers

---

# Important Probability Rule

For any event \(E\):

:contentReference[oaicite:0]{index=0}

Where:

• \(P(E)\) = probability of event occurring
• \(P(E')\) = probability of event not occurring

---

Total laptops manufactured:
\[
10000
\]

Defective laptops:
\[
25
\]

Good laptops:

\[
10000-25=9975
\]

Probability of selecting a good laptop:

\[
P(\text{good})
=
\frac{9975}{10000}
\]

\[
=0.9975
\]

✓ Answer:
\[
\boxed{0.9975}
\]

---

Total youngsters surveyed:
\[
400
\]

Youngsters having voter ID:
\[
191
\]

Youngsters without voter ID:

\[
400-191=209
\]

Probability:

\[
P(\text{No voter ID})
=
\frac{209}{400}
\]

✓ Answer:
\[
\boxed{\frac{209}{400}}
\]

---

\[
\frac{x}{3}
\]

Probability of not guessing correctly is
\[
\frac{x}{5}
\]

Find \(x\).

---

Using complementary probability:

\[
\frac{x}{3}+\frac{x}{5}=1
\]

LCM:
\[
15
\]

\[
\frac{5x+3x}{15}=1
\]

\[
\frac{8x}{15}=1
\]

\[
8x=15
\]

\[
x=\frac{15}{8}
\]

✓ Answer:
\[
\boxed{\frac{15}{8}}
\]

---

Find probability of losing.

Using:

\[
P(\text{lose})=1-P(\text{win})
\]

\[
=1-0.72
\]

\[
=0.28
\]

✓ Answer:
\[
\boxed{0.28}
\]

---

From the table:

• Families with both maids = 250
• Families with part-time maids = 860
• Families with no maids = 20
• Total families = 1500

---

(i) Probability of both types of maids

\[
P
=
\frac{250}{1500}
\]

\[
=
\frac16
\]

✓ Answer:
\[
\boxed{\frac16}
\]

---

(ii) Probability of part-time maids

\[
P
=
\frac{860}{1500}
\]

\[
=
\frac{43}{75}
\]

✓ Answer:
\[
\boxed{\frac{43}{75}}
\]

---

(iii) Probability of no maids

\[
P
=
\frac{20}{1500}
\]

\[
=
\frac1{75}
\]

✓ Answer:
\[
\boxed{\frac1{75}}
\]

---

# Final Answers

| Question | Answer |
|---|---|
| 1 | \(0.9975\) |
| 2 | \(\frac{209}{400}\) |
| 3 | \(\frac{15}{8}\) |
| 4 | \(0.28\) |
| 5(i) | \(\frac16\) |
| 5(ii) | \(\frac{43}{75}\) |
| 5(iii) | \(\frac1{75}\) |

---Exercise 9.3

Multiple choice questions

1. A number between 0 and 1 that is used to measure uncertainty is called

(1) Random variable

(2) Trial

(3) Simple event

(4) Probability

[Answer: (4) Probability]

2. Probability lies between

(1) −1 and +1

(2) 0 and 1

(3) 0 and n

(4) 0 and ∞

[Answer: (2) 0 and 1]

3. The probability based on the concept of relative frequency theory is called

(1) Empirical probability

(2) Classical probability

(3) Both (1) and (2)

(4) Neither (1) nor (2)

[Answer: (1) Empirical probability]

4. The probability of an event cannot be

(1) Equal to zero

(2) Greater than zero

(3) Equal to one

(4) Less than zero

[Answer: (4) Less than zero]

5. The probability of all possible outcomes of a random experiment is always equal to

(1) One

(2) Zero

(3) Infinity

(4) Less than one

[Answer: (1) One]

6. If A is any event in S and its complement is A’ then, P(A′) is equal to

(1) 1

(2) 0

(3) 1−A

(4) 1−P(A)

[Answer: (4) 1−P(A)]

7. Which of the following cannot be taken as probability of an event?

(1) 0

(2) 0.5

(3) 1

(4) −1

[Answer: (4) −1]

8. A particular result of an experiment is called

(1) Trial

(2) Simple event

(3) Compound event

(4) Outcome

[Answer: (4) Outcome]

9. A collection of one or more outcomes of an experiment is called

(1) Event

(2) Outcome

(3) Sample point

(4) None of the above

[Answer: (1) Event]

10. The six faces of the dice are called equally likely if the dice is

(1) Small

(2) Fair

(3) Six-faced

(4) Round

Solution: Fair means all outcomes are equally likely.

[Answer: (2) Fair]

Exercise 9.3

1. (4) 2. (2) 3. (1) 4. (4) 5. (1) 6. (4) 7. (4) 8. (4) 9. (1) 10. (2)

---