In the following situations, the outcomes do not have equal chance:
(i) In drawing a red marble from a box having 3 red marbles and 5 white marbles.
(ii) In drawing a black ball from a box having 6 black balls and 4 yellow balls.
(iii) Choosing an odd number from first 11 natural numbers.
(iv) Choosing a girl as class monitor from a group of 10 girls and 15 boys.
(v) In drawing a red ace card from a well shuffled pack of 52 cards

From the bar graph, it is clear that number of books in the years 1989 is 180 (approx.). We can see in the bar graph that the number of book sold in the year 1989 is little less than 200, but greater than 175.

The frequency table is shwn below.

It is dear from the table that, the lowest number = 1

Out of 6 marbles, one can be drawn in 6 ways
\(\therefore\) Total number of possible outcomes = 6
A probability of drawing marble with number 5, P(5) = \(\frac{1}{6}\)

5

We have, 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
On arranging the data in ascending order, we get 2,10,12,14,14,14,14,14,14,16,16,18 Now, we will arrange it in a table form

Since, 14 occurs the maximum number of times i.e. 6. Hence, mode of the data is 14.

On arranging the data in ascending order, we get 128,132,135,139,141,143,146,149,150,151 From ascending order of hights, we observe that first term is minimum.
\(\therefore\) A height of the shortest girl = 128 cm

23.42

From the given table, age of children in group A
=7yr,7yr,9yr,8yr, 10yr, 10yr
Age of children in group 8 
= 7yr , 9yr, 11 yr, 12yr, 12yr, 12yr
(a) Mode in group A = 7 yr and 10 yr
( \(\because\)  7 y rand 10 yr occurs 2 times)
Range in group A = Maximum value - Minimum value
=10-7=3
Mode in group 8= 12 yr
Range in group 8=12 -7 = 5
(b) If both groups are combined together 7, 7, 7, 9, 9,11,8, 12, 10, 12, 10, 12
Mode = 7 and 12 [\(\because\)  7 and 12 occurs 3 times]
Range = Maximum value - Minimum value
=12-7=5

As per the given information, mean of 10 observations is 40.
\(\therefore\)   Mean​​​​​​ ​\(=\frac { sum\ of\ all\ observations }{ number\ of\ observations } \)
\(\Rightarrow\)   40 = \(\frac { sum\ of\ all\ observations }{ 10 } \)
Sum of all observations = 40 x 10 = 400
But this is incorrect sum. Since, one observation was copied wrongly.
Correct sum = Incorrect sum - Incorrect observation + Correct observation
= 400 - 15 + 45 = 430
correct mean ​​​​​​​\(=\frac { Correct\ sum }{ number\ of\ observations } =\frac { 430 }{ 10 } =43\)

(a) Possible situations to happen are as follows:
(i) On tossing a coin, getting either a head or a tail
(ii) On drawing one card from a pack of 52 cards one side will appear.
(iii) Getting a number from 1 to 6 by throwing a die.
(b) Impossible to happen are as follows:
(i) A girl in the boy's school.
(ii) Getting a number 8 by throwing a die.
(iii) A person of height 3 m.
(c) Mayor may not happen situations are as follows:
(i) To toss a coin and get tail.
(ii) Probably it may rain.
(iii) An ant rising to 4 m height

A double bar graph can be used to compare two collections of data.

number less than 3 are 1 and 2
p (getting a number less thn 3) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

The representation of data with bars of uniform width is called bar graph

For median, arrangement of given data in ascending order
14,18,24,27,29
\(\therefore\) Median = middle number in data = 24

\(\underset { \overset { \downarrow }{ minimum\quad value } }{ 1 } ,2,4,\underset { \overset { \downarrow }{ median } }{ 6 } ,8,9,\underset { \overset { \downarrow }{ maximum\quad value } }{ 11 } \)