Probability

1

As we know that,

Probability of Event ‘E’=P(E)

Probability of an event ‘not E’ = P(E̅ )

⇒ P(not E) = 1 – P(E)

⇒ P(not E) + P(E) = 1

0, Impossible event

⇒ Probability of an event which can not be happen is 0. And this type of event is called as impossible event.

1, Sure event

⇒ If an event is certain to happen then probability is always 1. And this type of event is called Sure event.

1

⇒ The sum of probabilities of all the elementary events of an experiment is always 1.

0, 1

⇒ Probability of an event lies between 0 and 1 i.e. Probability can’t be less than 1 or more than unity.

⇒ 0 ≤ P(E) ≥ 1

∴ The probability of an event is greater than or equal to 0 and less than or equal 1.

No

⇒ If a driver is trying to start a car, The Car may start or may not be start it depends on fuel in car engine. Starting also depends on how old car is and there is no accident happen with car. Now we can conclude that the factors on which car will start or not start is not same. Thus, we can say that it is not an equally likely event.

No

⇒ When a player shoot basketball the number of shots player shoots can go in or misses it depends on players ability. Players ability is not defined here. Therefore, it is not equally likely event.

Yes

⇒ In the trial there is only two cases – first answer may be right or it may be wrong because it is true false question.

⇒ 50% probability answer is right or 50% probability question is wrong. Therefore, it is equally likely event.

Yes

⇒ When a baby child is born. Born baby may be boy or girl. Probability for baby is boy is 50% while baby is girl also 50%. Therefore, it is also an equally likely event.

If a probability of an event to be happen is P. Then event not to happen will (1 – P).

Here, probability of event ‘E’ is given as –

⇒ P(E)=0.05

⇒ P(not E)= P(E̅ ) = 1 – P(E)

⇒ P(E̅ )= 1 – 0.05

⇒ P(E̅ )=0.95

∴ probability of ‘notE’ = 0.95

let total number of candies be ‘x’

All candies are lemon flavored.

∴ Total number of orange flavored candy is = 0.

⇒ Probability of Malini takes out an orange flavored candy (P) =

⇒ P(Taken out orange flavor) =

⇒ P(Taken out orange flavor) = 0

∴ Probability that Malini takes out an orange flavored candy = 0

Let Total number of candies be ‘x’ in bag

all candies in the bag are lemon flavored only.

∴ lemon flavored candies will also ‘x’.

∵ there is no other candies in the bag.

⇒ If Malini takes out lemon flavored candy from bag.

∴ P(lemon flavored candies) =

⇒ P(lemon flavored candies) =

⇒ P(lemon flavored candies) = 1.

∴ Probability that Malini takes out a lemon-flavored candy = 1

There are 52 cards in a pack of cards.

A pack of 52 cards have 4 aces.

Total number of hearts in pack of cards = 13

⇒ Rahim removes all the hearts. 1 ace will also removed.

∴ Remaining cards = 52 – 13 = 39

⇒ remaining ace cards = 4 – 1 = 3

⇒ Number of favourable cases = 3

⇒ P(getting an ace) =

⇒ P(getting an ace) =

⇒ P(getting an ace) =

∴ Probability of getting an ace =

∴ Probability of getting ace from remaining cards =

As we know there are always 52 cards in a pack of card.

⇒ out of 26 red cards there are 13 diamonds cards.

⇒ Rahim removes all the hearts. 1 ace will also removed.

∴ Remaining cards = 52 – 13 = 39

⇒ P(Getting diamond) =

⇒ P(Getting diamond) =

⇒ P(Getting diamond) =

∴ Probability of getting diamond is

Rahim removes all the heart cards.

∴ remaining cards = 52 – 13 = 39

⇒ There is no heart cards in this 39 cards.

⇒ P(Getting a card that is not heart) =

⇒ P(Getting a card that is not heart) =

⇒ P(Getting a card that is not heart) = 1.

∴ Probability of getting card that is not heart is 1.

⇒ Rahim removes all the hearts.

⇒ remaining heart cards = 0.

⇒ P(Getting the ace of hearts) =

⇒ P(Getting the ace of hearts) =

⇒ P(Getting the ace of hearts) = 0.

∴ Probability of getting the ace of heart is zero.

Given,

⇒ Probability of 2 students not having same birthday = 0.992

∴ Probability of 2 students having same birthday = 1 – Probability of 2 students not having same birthday.

⇒ P(2 students having same birthday) = 1 – P(2 students having same birthday)

⇒ P(2 students having same birthday) = 1 – 0.992

⇒ P(2 students having same birthday) = 0.008

⇒ Probability that the 2 students have the same birthday is 0.008

When a dice is thrown total outcomes can be (1,2,3,4,5,6).

⇒ Possible outcomes = 6

⇒ Prime numbers = (2,3,5)

⇒ number of prime numbers = 3

⇒ Probability of getting prime number –

⇒ P(a prime number) =

⇒ P(a prime number) =

⇒ P(a prime number) =

∴ Probability of getting prime number is .

When a dice is thrown total outcomes can be (1,2,3,4,5,6).

⇒ Possible outcomes = 6

⇒ numbers lying between 2 and 6 = (3,4,5)

⇒ Possible outcomes = 3

⇒ P(number lying between 2 and 6) =

⇒ P(number lying between 2 and 6) =

⇒ P(number lying between 2 and 6) =

∴ Probability for number lying between 2 and 6 is .

When a dice is thrown total outcomes can be (1,2,3,4,5,6).

⇒ Possible outcomes = 6

⇒ Odd numbers = (1,3,5)

⇒ Possible outcomes = 3

⇒ P(an odd number) =

⇒ P(an odd number) =

⇒ P(an odd number) =

∴ Probability of getting odd number is .

There are total 52 cards in a pack of cards.

⇒ There are 4 kings in which 2 kings are red color.

⇒ Total outcomes = 52

⇒ Favorable outcomes = 2 (∵ only 2 kings are of red color)

⇒ Probability of selection out a red king from a deck of cards –

⇒ P(a red king) =

⇒ P(a red king) =

⇒ P(a red king) =

∴ Probability of getting red king from deck of cards is

From given data,

⇒ Total number of balls in bag = 5 + 3

⇒ Total number of balls in bag = 8

⇒ Total number of Red balls in bag = 3

⇒ Probability of selected ball is red =

⇒ P(Red) =

∴ Probability that the ball selected is red = .

As we know,

⇒ P(Not Red) = 1 – P(Red)

⇒ P(Not Red) =

⇒ P(Not Red) =

∴ Probability that the ball selected is not red = .

Total number of marbles = 5 + 8 + 4 = 17

⇒ Number of red marbles = 5

⇒ P(Marbles taken out is red) =

⇒ P(Marbles taken out is red) =

∴ Probability that marble taken out is Red = .

⇒ Total number of marbles = 5 + 8 + 4 = 17

⇒ Number of White marbles = 8

⇒ P(Marbles taken out is White) =

⇒ P(Marbles taken out is White) =

∴ Probability that marble taken out is white is

Total number of marbles = 5 + 8 + 4 = 17

⇒ Number of green marbles = 4

⇒ P(Marbles taken out is green) =

⇒ P(Marbles taken out is green) =

⇒ P(Marbles taken out is not green) = 1 – P(marbles taken out is green)

⇒ P(Marbles taken out is not green) =

⇒ P(Marbles taken out is not green) =

⇒ P(Marbles taken out is not green) =

∴ Probability that marble taken out is not green is

There are hundred 50p coins, 50 DI coins, 20 D2 coins and 10 D5 coins.

⇒ Total number of coins in kiddy bank = 100 + 50 + 20 + 10

⇒ Total number of coins in kiddy bank = 180

⇒ Probability that the coin will be 50p coins –

⇒ P(coin will be 50p coin) =

⇒ P(coin will be 50p coin) =

⇒ P(coin will be 50p coin) =

∴ Probability that the coin will be 50p coin is

Probability the coin will be a D5 coin –

⇒ P(Coin will be a D5 coin) =

⇒ P(Coin will be a D5 coin) =

⇒ P(Coin will be a D5 coin) =

∴ P(coin will not be a D5 coin) = 1 – P(coin will be a D5 coin)

⇒ P(coin will not be a D5 coin) =

⇒ P(Coin will not be a D5 coin) =

⇒ P(Coin will not be a D5 coin) =

∴ Probability that the coin will not be D5 coin is .

Total number of fish in tank = 5 male + 8 female

⇒ Total number of fish in tank = 13

⇒ number of male fish = 5

⇒ Probability the shopkeeper takes out a male fish –

⇒ P(Male fish) =

⇒ P(male fish) =

∴ Probability that the fish taken out is a male fish =

There are total 8 numbers.

⇒ We can get 8, one times only.

∴ number of favourable cases = 1

⇒ Probability that arrow will point 8 –

⇒ P(arrow will point 8) =

⇒ P(arrow will point 8) =

∴ Probability that arrow will point 8 will be .

Total numbers = 8

⇒ Total odd numbers = (1,3,5,7)

⇒ Number of favourable cases = 4

⇒ P( an odd number) =

⇒ P( an odd number) =

⇒ P( an odd number) =

∴ Probability arrow will point an odd number will be

⇒ Probability of numbers greater than 2 –

⇒ Total numbers = 8

⇒ Numbers greater than 2 = (3,4,5,6,7,8)

⇒ Total numbers greater than 2 = 6

⇒ P(>2) =

⇒ P(>2) =

⇒ P(>2) =

∴ Probability of numbers greater than 2 is .

⇒ total numbers = 8

⇒ total numbers which are less than 9 = (1,2,3,4,5,6,7,8)

i.e. all 8 numbers are less than 9.

∴ P(Numbers less than 9) =

⇒ P(Numbers less than 9) =

⇒ P(Numbers less than 9) = 1.

∴ Probability of arrow will point number less than 9 is 1

There are total 52 cards in a pack of cards.

⇒ There are 4 kings in which 2 kings are red color.

⇒ Total outcomes = 52

⇒ Favourable outcomes = 2 (∵ only 2 kings are of red color)

⇒ Probability of selection out a red king from a deck of cards –

⇒ P(a red king) =

⇒ P(a red king) =

⇒ P(a red king) =

∴ Probability of getting a red king is

There are total 52 cards in a pack of cards.

In which face cards are (King, Queen and Jack).

⇒ Each type has four cards.

⇒ Total face cards = 3× 4

⇒ Face cards = 12

⇒ Probability of getting a face card –

⇒ P(face card) =

⇒ P(face cards) =

⇒ P(face cards) =

∴ Probability of getting face card =

There are total 52 cards in a pack of cards.

⇒ Red face card = 2× 3 =6

⇒ P(a red face card) =

⇒ P(a red face card) =

⇒ P(a red face card) =

∴ Probability of getting a red face card =

There are total 52 cards in a pack of cards. There is only one jack of hearts.

⇒ Number of favourable outcome = 1

⇒ P(Jack of hearts) =

⇒ P(Jack of hearts) =

∴ Probability of getting Jack of hearts = .

There are total 52 cards in a deck of cards.

⇒ Number of spade cards = 13

⇒ Favourable cases = 13

⇒ Probability of getting a spade card –

⇒ P(a spade) =

⇒ P(a spade) =

⇒ P(a spade) =

∴ Probability of getting a spade = .

There are total 52 cards in a pack of cards. In which only one queen of diamonds.

⇒ Number of favourable cases = 1

⇒ Probability of getting a queen of diamonds –

⇒ P(queen of diamonds) =

⇒ P(queen of diamonds) =

∴ Probability of getting a queen of diamonds =

total cards = 5

⇒ Total cases while drawing a card at random = 5

⇒ There is one queen.

⇒ So favourable case for getting a queen = 1

⇒ probability that the card is the queen –

⇒ P( card is queen ) =

⇒ P( card is queen ) =

∴ Probability the card is queen =

(a)

queen is selected and put aside.

Remaining cards = 5 – 1

⇒ Remaining cards = 4

⇒ Total cases drawing a card at random = 4

⇒ One ace card.

∴ favourable cases to select ace = 1

⇒ probability that the second card selected is an ace –

⇒ P( card selected is an ace) =

⇒ P( card selected is an ace) =

∴ Probability selected card is an ace = .

(b) 0

queen is selected and put aside.

∴ number of favourable cases to select queen = 0

∴ probability that the second card selected is a queen will bw zero.

Given,

⇒ Total defective pens = 12

⇒ Total good pens = 132

∴ number of favourable cases = 132

⇒ Total number of pens = 12 + 132

⇒ Total number of pens = 144

⇒ Number of total outcomes will 144.

⇒ Probability that the pen taken out is a good one =

⇒ P(pen taken out is good one) =

⇒ P(pen taken out is good one) =

∴ Probability of pen taken out is good one is

Given,

Total number of bulbs = 20

Defective bulbs = 4

Probability selected this bulb is defective –

P(selected bulb is defective) =

P(selected bulb is defective) =

P(selected bulb is defective) =

∴ Probability the selected bulb is defective is

Given,

⇒ There are 90 discs in a box.

⇒ Number of outcomes in which discs is selected = 90.

⇒ Discs are numbered from 1 to 90.

∴ total 2 digits numbers = 90 – 9 (∵ 1 to 9 will be one digit numbers)

⇒ Total 2 digits numbers = 81

⇒ favourable outcomes = 81

⇒ Probability that discs bears a two- digit number –

⇒ P(two digit numbers) =

⇒ P(two digit numbers) =

⇒ P(two digit numbers) =

∴ Probability that disc bears two digit number is

There are 90 discs in a box.

⇒ Number of outcomes in which discs is selected = 90.

⇒ Total perfect square in the box –

⇒ 1² = 1

⇒ 2² = 4

⇒ 3² = 9

⇒ 4² = 16

⇒ 5² = 25

⇒ 6² = 36

⇒ 7² = 49

⇒ 8² = 64

⇒ 9² = 81

⇒ Number of favourable cases = 9

⇒ Probability that discs bears a perfect square number –

⇒ P(Perfect square numbers) =

⇒ P(Perfect square numbers) =

⇒ P(Perfect square numbers) =

∴ Probability that discs bears a perfect square number is

There are 90 discs in a box.

⇒ Number of outcomes in which discs is selected = 90.

⇒ Number of multiples of 5 from 1 to 90.

⇒ (5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90)

⇒ Number of favourable cases = 18

⇒ Probability that disc bears number divisible by 5 –

⇒ P(number divisible by 5) =

⇒ P(number divisible by 5) =

⇒ P(number divisible by 5) =

∴ Probability that disc bears a number divisible by 5 is .

Given,

⇒ Length of rectangular region = 3m

⇒ Breadth of rectangular region = 2m

⇒ Area of rectangular region = length× breadth

⇒ Area of rectangular region = 3× 2

⇒ Area of rectangular region = 6m²

⇒ Radius of circle =

⇒ Radius of circle = m

⇒ Area of circle = πr²

⇒ Area of circle = π

⇒ Area of circle = m²

⇒ Probability die will lie inside circle –

⇒ P(die will lie inside circle) =

⇒ P(die will lie inside circle) =

⇒ P(die will lie inside circle) =

⇒ P(die will lie inside circle) = (∵ π =

⇒ P(die will lie inside circle) =

⇒ P(die will lie inside circle) =

∴ Probability die will lie inside circle is .

Given,

⇒ Lot consists of total 144 ball pens.

⇒ Defective pens = 20

⇒ Non – defective pens = 144 – 20

⇒ Non – defective pens = 124

⇒ Sudha will buy pens if pen is good.

⇒ Probability that Sudha will buy pens –

⇒ P(Sudha will buy pen) =

⇒ P(Sudha will buy pen) =

⇒ P(Sudha will buy pen) =

∴ Probability of Sudha will buy a pen will .

Given,

⇒ Lot consists of total 144 ball pens.

⇒ Defective pens = 20

⇒ Sudha will not buy pens if pen is defective.

⇒ P(Sudha will not buy pen) =

⇒ P(Sudha will not buy pen) =

⇒ P(Sudha will not buy pen) =

∴ Probability Sudha will not buy a pen is

when two dices are rolled.

⇒ Number of total outcomes = 6× 6.

⇒ Number of total outcomes = 36.

⇒ Probability =

⇒ Complete table will be –

Sum 2,3,4,5,6,7,8,9,10,11,12 have different number of favourable outcomes. As we know total number of outcomes when two dice are thrown is 36. Therefore students argument is wrong.

We know that when a coin is tossed ‘n’ times.

⇒ Total number of outcome = 2ⁿ

∴ when a coin is tossed 3 times.

⇒ Total number of outcomes = 2³

⇒ Total number of outcomes = 8

⇒ Deskhitha wins if all the tosses give same result i.e. three heads and three tails.

⇒ When 3 coins tossed following cases will appear –

T T T

T T H

T H T

H T T

H H T

H T H

T H H

H H H

⇒ From the above analysis we conclude that there are 1 case where 3 heads and 1 case where three tails are appears.

⇒ And other 6 outcomes are of different results.

⇒ Probability of Deskhitha loose the game –

⇒ P(Deskhitha loose the game) =

⇒ P(Deskhitha loose the game) =

⇒ P(Deskhitha loose the game) =

∴ Probability of Deskhitha loose the game is .

If a dice thrown ‘n’ times then total number of outcome = 6ⁿ

⇒ Dice is thrown 2 times then total outcomes = 6²

⇒ Dice is thrown 2 times then total outcomes = 36

⇒ Favourable outcomes where 5 will not come up either time = (1,1)(1,2)(1,3)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,6)(3,1)(3,2)(3,3)(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,6)(6,1)(6,2)(6,3)(6,4)(6,6).

⇒ Number of favourable events = 25

⇒ Probability 5 will not come up either time –

⇒ P(5 will not come up either time)=

⇒ P(5 will not come up either time)=

∴ Probability 5 will not come up either time is .

If a dice thrown ‘n’ times then total number of outcome = 6ⁿ

⇒ Dice is thrown 2 times then total outcomes = 6²

⇒ Dice is thrown 2 times then total outcomes = 36

⇒ Favourable outcomes where 5 will come up at least once.

⇒ (1,5)(2,5)(3,5)(4,5)(5,5)(5,1)(5,2)(5,3)(5,4)(5,6)(6,5).

⇒ Favourable events = 11

⇒ Total events = 36

⇒ P(5 will come at least once) =

⇒ P(5 will come at least once) =

∴ Probability that 5 will come up at least once will