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Probability Exercise-14.1

Hey Guys!
Today we will discuss the term probability. What is probability? Probability is the likelihood of something occurring.Probability is like forecasting the likelihood of something happening. It helps us determine how likely an event is to occur. We define probability as a number between 0 (impossible) and 1 (certain). For example, a 10% chance of rain is considered a likelihood. Probabilities can be described using words such as unlikely, feasible, likely, or certain. 🎲🌦️

In mathematics, the probability of an event occurring is computed by dividing the number of positive outcomes by the total number of possible outcomes. Here's the formula: $$Probability\ of\ an\ event\ P(E)\ =\ \frac {Number\ of\ positive\ outcomes} {Total\ number\ of\ outcomes}$$ $$Here\ P(E)=Probability\ of\ occuring\ an\ event$$ $$and\ P(E̅̅̅̅̅̅)=Probability\ of\ not\ occuring\ an\ event$$ $$Remember\ always\ P(E)\ +\ P(E̅̅̅̅̅̅)\ =\ 1$$ For example:
1. If you have two yellow pillows out of six, the chances of picking a yellow pillow are 1/3.
2. If you pick bottles from a container, the chance of picking a green bottle is 0.45 when there are 450 green bottles among 1000.
2. If you throw a dice then chance of every outcome is 1/6.
Remember that probability allows us to make informed judgements and grasp uncertainty! $$Now\ let's\ start\ the\ fun$$ $$Exercise-14.1$$ 1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = 1
(ii) The probability of an event that cannot happen is 0 called impossible.
(iii) The probability of an event that is certain to happen is 1 is called sure event
(iv) The sum of the probabilities of all the elementary events of an experiment is 1
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1

 2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
Sol: Not equally likely.If engine is OK car will surely start
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Sol:Not equally likely.If player is master in shooting , he never miss the shot
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Sol: Equally likely
(iv) A baby is born. It is a boy or a girl.
Sol: Equally likely

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Sol:Because tossing a coin is an equally likely event and the chance for each outcome is 50% i.e. 1/2.

4. Which of the following cannot be the probability of an event? 
 (A) 2/3 (B)–1.5 (C) 15% (D) 0.7
Sol:(B) can not be the probablity of an event, because chance of happening an event could not be negative.

5. If P(E) = 0.05, what is the probability of ‘not E’?
Sol:$$Remember\ \ P(E)\ +\ P(E̅̅̅̅̅̅)\ =\ 1$$
$$\therefore\  0.05\ +\ P(E̅̅̅̅̅̅)\ =\ 1$$
$$\therefore\ \ P(E̅̅̅̅̅̅)\ =\ 0.95\ Ans.$$

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out 
 (i) an orange flavoured candy? (ii) a lemon flavoured candy?
Sol:
(i) Since the bag does not contains orange flavoured candy , so outcome of an orange flavoured candy is an impossible event, hence the probability will be ZERO.
(ii) Since the bag only have lemon flavoured candy , so outcome of a lemon flavoured candy is always ONE.
 
7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Sol:
$$\because\ \ P(E)\ +\ P(E̅̅̅̅̅̅)\ =\ 1$$
$$\therefore\  P(E)\ +\ 0.992\ =\ 1$$
$$\therefore\ \ P(E)\ =\ 0.008\ Ans.$$

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red?
Sol:A ball is drawn at random , it means all the outcomes are equally likely, hence
the total number of possible outcomes are : 3+5=8
Total number of favorable events are: 3
$$\because\ Probability\ of\ an\ event\ P(E)\ =\ \frac {Number\ of\ positive\ outcomes} {Total\ number\ of\ outcomes}$$
$$\therefore\ Probability\ of\ a\ red\ ball\ drawn\ is\ P(E)\ =\ \frac {3} {8}\ Ans.$$
$$\therefore\ Probability\ of\ a\ red\ ball\ not\ drawn\ is\ P(E̅̅̅̅̅̅)\ =\ 1-\frac {3} {8}$$
$$\therefore\ Probability\ of\ a\ red\ ball\ not\ drawn\ is\ P(E̅̅̅̅̅̅)\ =\frac {5} {8}\ Ans.$$

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green?
Sol:
A marble is taken out at random , it means all the outcomes are equally likely, hence
the total number of possible outcomes are : 5+8+4=17.
(i)Favourable event of outcome is red = 5
$$\therefore\ Probability\ of\ a\ red\ marble\ taken\ out\ is\ P(E)\ =\ \frac {5} {17}\ Ans.$$
(ii)Favourable event of outcome is white = 8
$$\therefore\ Probability\ of\ a\ white\ marble\ taken\ out\ is\ P(E)\ =\ \frac {8} {17}\ Ans.$$
(iii)Favourable event of outcome is not green = 5+8=13
$$\therefore\ Probability\ of\ a\ marble\ not\ green\ is\ P(E)\ =\ \frac {13} {17}\ Ans.$$

10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ` 5 coin?
Sol:
A coin fall out at random which is a equally likely event , hence
the total number of possible outcomes are : 100+50+20+10=180.
(i)Favourable event of outcome is 50p coin = 100
$$\therefore\ Probability\ of\ a\ 50p\ coin\ fallen\ out\ is\ P(E)\ =\ \frac {100} {180}.$$
$$\therefore\ Probability\ of\ a\ 50p\ coin\ fallen\ out\ is\ P(E)\ =\ \frac {5} {9}\ Ans.$$
(ii)Favourable event of outcome is 5 coin = 10
$$\therefore\ Probability\ of\ a\ 5\ coin\ is\ P(E)\ =\ \frac {10} {180}$$
$$\therefore\ Probability\ of\ a\ 5\ coin\ is\ P(E)\ =\ \frac {1} {18}$$
$$\therefore\ Probability\ of\ not\ a\ 5\ coin\ is\ P(E̅̅̅̅̅̅)\ =\ 1-\ \frac {1} {18}$$
$$\therefore\ Probability\ of\ not\ a\ 5\ coin\ is\ P(E̅̅̅̅̅̅)\ =\ \frac {17} {18}\ Ans$$

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish?
Solution:
$$Favourable\ events\ =\ 5$$
$$Total\ events\ =\ 5+8=\ 13$$
$$\therefore\ Probability\ of\ a\ male\ fish\ is\ P(E)\ =\ \frac {5} {13}\ Ans$$

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally likely outcomes. What is the probability that it will point at 
(i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Sol:
Total number of equally likely events = 8
(i) 
$$Favourable\ events\ =\ 1$$
$$Total\ events\ =\ 8$$
$$\therefore\ Probability\ of\ \ pointing\ 8\ is\ P(E)\ =\ \frac {1} {8}\ Ans$$
(ii) 
$$Favourable\ events\ of\ an\ odd\ number=\ 4$$
$$Total\ events\ =\ 8$$
$$\therefore\ Probability\ of\ an\ odd\ number\ is\ P(E)\ =\ \frac {4} {8}\ =\frac {1} {2}\  Ans$$
(iii) 
$$Favourable\ events\ of\ a\ number\ greater\ than\ 2=\ 6$$
$$Total\ events\ =\ 8$$
$$\therefore\ Probability\ of\ an\ odd\ number\ is\ P(E)\ =\ \frac {6} {8}\ =\frac {3} {4}\  Ans$$
(iv) 
$$Favourable\ events\ of\ a\ number\ less\ than\ 9=\ 8$$
$$Total\ events\ =\ 8$$
$$\therefore\ Probability\ of\ a\ number\ less\ than\ 9\ is\ P(E)\ =\ \frac {8} {8}\ =\ 1\  Ans$$

13. A die is thrown once. Find the probability of getting
 (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Sol:
Total number of equally likely events = 6
(i) 
$$Favourable\ events\ of\ prime\ number\ i.e.\ outcome\ is\ from\ 2,3,5=\ 3$$
$$Total\ events\ =\ 6$$
$$\therefore\ Probability\ of\ a\ prime\ number\ is\ P(E)\ =\ \frac {3} {6}\ =\ \frac {1} {2}\ Ans$$
(ii) 
$$Favourable\ events\ of\ a\ number\ lying\ between\ 2\ and\ 6\ =\ 3$$
$$Total\ events\ =\ 6$$
$$\therefore\ Probability\ of\ a\ number\ lying\ between\ 2\ and\ 6\ is\ P(E)\ =\ \frac {3} {6}\ =\frac {1} {2}\  Ans$$
(iii) 
$$Favourable\ events\ of\ an\ odd\ number\ =\ 3$$
$$Total\ events\ =\ 6$$
$$\therefore\ Probability\ of\ an\ odd\ number\ is\ P(E)\ =\ \frac {3} {6}\ =\frac {1} {2}\  Ans$$
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting 
 (i) a king of red colour      (ii) a face card      (iii) a red face card 
(iv) the jack of hearts         (v) a spade             (vi) the queen of diamonds
Sol:
Total number of equally likely events = 52
(i) 
$$Favourable\ events\ of\ a\ red\ color\ king\ is\ =\ 2$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ red\ color\ king\ is\ P(E)\ =\ \frac {2} {52}\ =\ \frac {1} {26}\ Ans$$
(ii) 
$$Favourable\ events\ of\ a\ face\ card\ =\ 12$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ face\ card\ is\ P(E)\ =\ \frac {12} {52}\ =\frac {3} {13}\  Ans$$
(iii) 
$$Favourable\ events\ of\ a\ red\ face\ card\ =\ 6$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ red\ face\ card\ is\ P(E)\ =\ \frac {6} {52}\ =\frac {3} {26}\  Ans$$
(iv) 
$$Favourable\ events\ of\ a\ jack\ of\ hearts\ =\ 1$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ jack\ of\ hearts\ is\ P(E)\ =\ \frac {1} {52}\  Ans$$
(v) 
$$Favourable\ events\ of\ a\ spade\ =\ 13$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ spade\ is\ P(E)\ =\ \frac {13} {52}\ =\frac {1} {4}\  Ans$$
(vi) 
$$Favourable\ events\ of\ a\ queen\ of\ diamond\ =\ 1$$
$$Total\ events\ =\ 52$$
$$\therefore\ Probability\ of\ a\ queen\ of\ diamond\ is\ P(E)\ =\ \frac {1} {52}\  Ans$$
15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. 
(i) What is the probability that the card is the queen? 
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Sol:
Total number of equally likely events = 5
(i) 
$$Favourable\ events\ of\ a\ queen\ king\ is\ =\ 1$$
$$Total\ events\ =\ 5$$
$$\therefore\ Probability\ of\ a\ queen\ is\ P(E)\ =\ \frac {1} {5}\ Ans$$
(ii) 
Total number of equally likely events = 4 , when queen is drawn and put aside.
(a)
$$Favourable\ events\ of\ an\ ace\ card\ =\ 1$$
$$Total\ events\ =\ 4$$
$$\therefore\ Probability\ of\ an\ ace\ card\ is\ P(E)\ =\ \frac {1} {4}\  Ans$$
(b)
$$Because\ in\ the\ ten\ of\ cards\ ,\ now\ only\ four\ cards\ are\ left\ after$$
$$queen\ drawn\ out\ and\ put\ aside,\ so\ again\ drawing\ a\ queen\ is\ an $$
$$ impossible\ event,\ hence\ the\ probability\ will\ be\ ZERO.\ Ans$$
16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Sol:
$$Favourable\ events\ of\ a\ good\ pen\ =\ 132$$
$$Total\ events\ =\ 144$$
$$\therefore\ Probability\ of\ a\ good\ pen\ taken\ out\ P(E)\ =\ \frac {132} {144}\ =\frac {11} {12}\  Ans$$
17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? 
 (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Sol:
(i)
$$Favourable\ events\ of\ a\ defective\ bulb\ =\ 4$$
$$Total\ events\ =\ 20$$
$$\therefore\ Probability\ of\ a\ defective\ bulb\ is\ P(E)\ =\ \frac {4} {20}\ =\frac {1} {5}\  Ans$$
(ii)
Since the bulb drawn is not defective and also not replaced ,therefore
$$Favourable\ events\ of\ a\ not\ defective\ bulb\ =\ 19-4=15$$
$$Total\ events\ =\ 19$$
$$\therefore\ Probability\ of\ a\ not\ defective\ bulb\ is\ P(E)\ =\ \frac {15} {19}\  Ans$$
18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears 
(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Sol:
From the range of numbers 1 to 90 , there are 9 numbers are of one digit and rest 81 are of two digits numbers.Hence:
(i)
$$Favourable\ events\ of\ a\ two-digit\ number\ =\ 81$$
$$Total\ events\ =\ 90$$
$$\therefore\ Probability\ of\ a\ two\ digit\ number\ is\ P(E)\ =\ \frac {81} {90}\ =\ \frac {9} {10}\  Ans$$
(ii)
perfect square numbers in the range are : 1,4,9,16,25,36,49,64,81, Hence:
$$Favourable\ events\ of\ a\ perfect\ square\ number\ =\ 9$$
$$Total\ events\ =\ 90$$
$$\therefore\ Probability\ of\ a\ perfect\ square\ number\ is\ P(E)\ =\ \frac {9} {90}\ =\ \frac {1} {10}\  Ans$$
(iii)
List of numbers divisible by 5 are: 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90.Hence:
$$Favourable\ events\ of\ a\ number\ divisible\ by\ 5\ =\ 18$$
$$Total\ events\ =\ 90$$
$$\therefore\ Probability\ of\ a\ number\ divisible\ by\ 5\ P(E)\ =\ \frac {18} {90}\ =\ \frac {1} {5}\  Ans$$
19. A child has a die whose six faces show the letters as given below:
[A] [B] [C] [D] [E] [A]
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Sol:
(i)
$$Favourable\ events\ of\ getting\ [A]\ =\ 2$$
$$Total\ events\ =\ 6$$
$$\therefore\ Probability\ of\ getting\ [A]\ is\ P(E)\ =\ \frac {2} {6}\ =\ \frac {1} {3}\  Ans$$
(ii)
$$Favourable\ events\ of\ getting\ [D]\ =\ 1$$
$$Total\ events\ =\ 6$$
$$\therefore\ Probability\ of\ getting\ [D]\ is\ P(E)\ =\ \frac {1} {6}\  Ans$$
20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?
Sol:
Dropping the ball is an equally likely event , which can drop anywhere within the rectangle. 
Area of the rectangle =3×2=6 m²
Area of the circle =π/2 × r²= π/2 m²

$$Favourable\ events\ when\ ball\ drop\ in\ the\ circle\ area\ =\ π/2$$ $$Total\ Area\ =\ 6$$ $$\therefore\ Probability\ of\ ball\ drop\ in\ the\ circle\ area\ is\ P(E)\ =\ \frac {\frac {π} {2}} {6}\ $$ $$\therefore\ Probability\ of\ ball\ drop\ in\ the\ circle\ area\ is\ P(E)\ =\ \frac {π} {12}\ Ans$$ 
21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it ? (ii) She will not buy it ?
Sol:
(i)Since Noori will only buy the pen, when it is not defective. Hence:
$$Favourable\ events\ of\ getting\ a\ not\ defective\ pen\ is\ =\ 124$$
$$Total\ events\ =\ 144$$
$$\therefore\ Probability\ of\ buying\ is\ P(E)\ =\ \frac {124} {144}$$
$$\therefore\ Probability\ of\ buying\ is\ P(E)\ =\ \frac {31} {36}\  Ans$$
(ii)$$Remember\ always\ P(E)\ +\ P(E̅̅̅̅̅̅)\ =\ 1$$
$$\therefore\ Probability\ of\ not\ buying\ P(E̅̅̅̅̅̅)\ =\ 1\ -\ \frac {31} {36}$$
$$\therefore\ Probability\ of\ not\ buying\ P(E̅̅̅̅̅̅)\ =\ \frac {5} {36}\ Ans.$$
22. Refer to Example 13. (i) Complete the following table:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11 . Do you agree with this argument? Justify your answer.
Sol:
(i)
Total numbers of equally likely events are: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) = 36 outcomes
Probability, when sum is 3 : = 2/36  Ans.Favourable events (1,2),(2,1)
Probability, when sum is 4 : = 3/36  Ans. Favourable events (1,3),(2,2),(3,1)
Probability, when sum is 5 : = 4/36  Ans. Favourable events (1,4),(2,3),(3,2),(4,1)
Probability, when sum is 6 : = 5/36  Ans. Favourable events (1,5),(2,4),(3,3),(4,2),(5,1)
Probability, when sum is 7 : = 6/36  Ans. Favourable events (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Probability, when sum is 9 : = 4/36  Ans. Favourable events (3,6),(4,5),(5,4),(6,3)
Probability, when sum is 10 : = 3/36 Ans. Favourable events (4,6),(5,5),(6,4)
Probability, when sum is 11 : = 2/36  Ans. Favourable events (5,6),(6,5)
(ii)
No, the student is wrong , because the sum is not equally likely.
 
23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. 
Sol:
Because tossing a coin 3 times is an equally likely events , so the possible outcomes in all three tosses are:(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,T,T),(T,T,H),(T,H,T),(T,H,H) = 8 outcomes
Favourable events : (H,H,T),(H,T,H),(H,T,T),(T,T,H),(T,H,T),(T,H,H) = 6 outcomes
Hence the probability = 6/8 =3/4 Ans. 
24. A die is thrown twice. What is the probability that 
 (i) 5 will not come up either time? (ii) 5 will come up at least once? 
Solution: 
Total numbers of equally likely events are: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) = 36 outcomes
(i)
Total number of favorable events: 25
Hence the probability is = 25/36

2nd way
Each time the probability of outcome 5 is = 5/6 
so total probability of outcome 5 in twice = 5/6 × 5/6 = 25/36

(ii)
Total number of favorable events: 11
Hence the probability is = 11/36

2nd way
We already know the P(E)\ +P(E̅̅̅)\ =1
Hence P(E̅̅̅) = 1 - 25/36 = 11/36

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. 
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

This is your practice question , write the solution in comment box.

I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏


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Real Numbers Exercise 1.1

 Hi Friends and Champs, Before coming to the exercise 1.1, we need to understand some points, which you have already read in class IX. Prime Number: Those numbers which could be divide by "1" and itself. i.e. there are only two multiple of the number, example 2,3,5,7...... etc. 2 is the smallest prime number we have. "1" is not a prime number, because for a prime number there should be exactly two multiples of the number, but "1" have only one multiple which is itself. Prime factors: Prime factors are the product of prime numbers for a given number. ex 2x2x3x5 are the prime factors for the number 60.   Exercise 1.1 Q-1 Express each number as a product of its prime factors: (i) 140        (ii) 156    (ii) 3825    (iv)  5005     (v) 7429 Note: Prime numbers are those numbers, which could be divide by 1 and by itself. Note: Do not try to skip the steps in prime factorization, start from the lowest prime number. Solution:...
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Introduction to Trigonometry Ex- 8.1

 Hi Friends and Champs Before we proceed to the next exercise let's understand some basic facts about trigonometry.  Trigonometry is the combination of three Greek words " tri", "gon"  and  " metron"  . " tri"  mean three " gon"  means sides and " metron"  mean measure. So, trigonometry means to study the relationship between sides and angles of a triangle.    Trigonometry Ratios:   Let  Δ ABC is a right angle triangle , right angle at B, then: Sin A = BC / CA           i.e. Side opposite to angle A / Hypotenuse Cos A = AB / CA             i.e. Side adjacent to angle A / Hypotenuse Tan A = BC / AB             i.e. Side opposite to angle A / Side adjacent to angle A  Cosec A = CA / BC      i.e Hypotenuse /. Side opposite to angle A Sec A = CA / AB          i.e. Hypotenuse / ...
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Learn Excel V-Lookup in Easy Way

Let We have three tables  names:  Table - 1 ( our Main Database Table ) and  Table - 2 Where we have only Name of customer and require Mob No  Table -3 Where we have customer name and require customer City.  VlookUP Function is: VLOOKUP(lookup_value , table_array, col_index_num ,[range_lookup])  lookup_value Attribute : what you want to search.   table_array attribute       : where vlookup funtion will search , Fix data range from database   col_index_num attribute : which column data result you require from selected data range .   range_lookup   attribute    : Either you require exact match for your lookup_value or partial match.                                                                     ...
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Surface Areas and Volumes Exercise - 12.2

Hi Friends and Champs!  In this chapter, we will learn to find the resulting volumes by joining or scooping of different shape bodies in the same way as in previous exercise. Let's start.  Exercise 12.2 1.A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.  Solution: radius of both cone and hemisphere r = 1 cm height of the cone h = r = 1 cm Volume "v" of the solid = volume of the hemisphere + volume of the cone v = 2/3  πr ³ + 1/3  πr ² h =2/3  π1 ³ +   1/3  π1 ² ×1 = 2/3  π +  1/3  π = π cm ³ Ans. 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contai...
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Statistics Exercise -13.2

Hi Friends and Champs! We will talk about the mode of some supplied grouped data today. Remember that you have already learnt the mode for ungrouped data from the last class. As you are previously aware, the mode for ungrouped data is the frequency that is repeated more frequently. As an illustration: Example - Run score by a batsman in 8 matches are 50,98,99,98,76,75,99,98.Find the mode ? Solution : 98 runs repeated 3 times , Hence mode is 98. It is evident from the example above that figure 98 is the mode because it is repeated three times.Thus, it is easy to find the mode for ungrouped data. However, it is a little more complicated with aggregated data. For ungrouped data, we select the class with the highest frequency and use the formula below to determine the mode. \[Mode=l + \frac{f1-f0} {2f1-f0-f2}×h \] where l = lower limit of the modal class, h = size of the class interval (assuming all class sizes to be equal), f1 = frequency of the modal class, f0 = frequency of t...
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