Probability Solution of TS & AP Board Class 10 Mathematics
Question 1.
Complete the following statements:
Probability of an event E + Probability of the event ‘not E’=________________
Answer:
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
Question 2.
Complete the following statements:
The probability of an event that cannot happen is ____________ Such an event is called _____________
Answer:
0, Impossible event
⇒ Probability of an event which can not be happen is 0. And this type of event is called as impossible event.
Question 3.
Complete the following statements:
The probability of an event that is certain to happen is ________ Such an event is called___________
Answer:
1, Sure event
⇒ If an event is certain to happen then probability is always 1. And this type of event is called Sure event.
Question 4.
Complete the following statements:
The sum of the probabilities of all the elementary events of an experiment is __________
Answer:
1
⇒ The sum of probabilities of all the elementary events of an experiment is always 1.
Question 5.
Complete the following statements:
The probability of an event is greater than or equal to __________ and less than or equal to ____________
Answer:
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.
Question 6.
Which of the following experiments have equally likely outcomes? Explain.
A driver attempts to start a car. The car starts or does not starts.
Answer:
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.
Question 7.
Which of the following experiments have equally likely outcomes? Explain.
A player attempts to shoot a basketball. She/he shoots or misses the shot.
Answer:
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.
Question 8.
Which of the following experiments have equally likely outcomes? Explain.
A trail is made to answer a true –false question. The answer is right or wrong.
Answer:
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.
Question 9.
Which of the following experiments have equally likely outcomes? Explain.
A baby is born. It is a boy or a girl.
Answer:
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.
Question 10.
If P (E)=0.05. What is the probability of ‘notE’?
Answer:
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
Question 11.
A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out?
an orange flavored candy?
Answer:
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
Question 12.
A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out?
a lemon-flavored candy?
Answer:
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
Question 13.
Rahim removes all the hearts from the cards. What is the probability of
Getting an ace from the remaining pack.
Answer:
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 =
Question 14.
Rahim removes all the hearts from the cards. What is the probability of
Getting a diamond.
Answer:
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
Question 15.
Rahim removes all the hearts from the cards. What is the probability of
Getting a card that is not a heart.
Answer:
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.
Question 16.
Rahim removes all the hearts from the cards. What is the probability of
Getting the Ace of hearts.
Answer:
⇒ 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.
Question 17.
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?
Answer:
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
Question 18.
A die is rolled once. Find the probability of getting
a prime number:
Answer:
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 .
Question 19.
A die is rolled once. Find the probability of getting
a number lying between 2 and 6
Answer:
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 .
Question 20.
A die is rolled once. Find the probability of getting
an odd number.
Answer:
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 .
Question 21.
What is the probability of selection out a red king from a deck of cards?
Answer:
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 
Question 1.
A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is
red ?
Answer:
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 = .
Question 2.
A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is
not red ?
Answer:
As we know,
⇒ P(Not Red) = 1 – P(Red)
⇒ P(Not Red) = 
⇒ P(Not Red) = 
∴ Probability that the ball selected is not red = .
Question 3.
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
red?
Answer:
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 = .
Question 4.
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
white?
Answer:
⇒ 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 
Question 5.
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
not green?
Answer:
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 
Question 6.
A kiddy bank contains hundred 50p coins. Fifty DI coins, twenty D2 coins and ten D5 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.
will be a 50p coin?
Answer:
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 
Question 7.
A kiddy bank contains hundred 50p coins. Fifty DI coins, twenty D2 coins and ten D5 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.
will not be a D5 coin?
Answer:
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 .
Question 8.
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 figure). What is the probability that the fish taken out is a male fish?
Answer:
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 =
Question 9.
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 (Sec figure), and these are equally likely outcomes. What is the probability that it will point at
8?
Answer:
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 .
Question 10.
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 (Sec figure), and these are equally likely outcomes. What is the probability that it will point at
an odd number?
Answer:
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 
Question 11.
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 (Sec figure), and these are equally likely outcomes. What is the probability that it will point at
a number greater than 2?
Answer:
⇒ 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 .
Question 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 (Sec figure), and these are equally likely outcomes. What is the probability that it will point at
a number less than 9?
Answer:
⇒ 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
Question 13.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
a king of red colour
Answer:
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
Question 14.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
a face cards
Answer:
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 = 
Question 15.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
a red face card
Answer:
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 = 
Question 16.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
the jack of hearts
Answer:
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 = .
Question 17.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
a spade
Answer:
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 = .
Question 18.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
the queen of diamonds
Answer:
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 = 
Question 19.
Five cards- the ten jack. Queen. King and ace of diamonds. Are well- shuffled with their face downwards. One card is selected at random.
What is the probability that the card is the queen?
Answer:
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 = 
Question 20.
Five cards- the ten jack. Queen. King and ace of diamonds. Are well- shuffled with their face downwards. One card is selected at random.
If the queen is selected and put aside (without replacement). What is the probability that the second card selected is (a) an ace ? (b) a queen ?
Answer:
(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.
Question 21.
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 is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
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 
Question 22.
A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?
Answer:
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
Question 23.
A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears.
a two- digit number
Answer:
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 
Question 24.
A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears.
a perfect square number
Answer:
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 
Question 25.
A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears.
a number divisible by 5.
Answer:
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 .
Question 26.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m?
Answer:
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 = Ï€r2
⇒ Area of circle = Ï€
⇒ 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) = 
(∵ Ï€ =
⇒ P(die will lie inside circle) = 
⇒ P(die will lie inside circle) = 
∴ Probability die will lie inside circle is .
Question 27.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that
She will buy it?
Answer:
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 .
Question 28.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that
She will not buy it?
Answer:
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 
Question 29.
Two dice are rolled simultaneously and counts are added.
completes the table given below:
Answer:
when two dices are rolled.
⇒ Number of total outcomes = 6× 6.
⇒ Number of total outcomes = 36.
⇒ Probability = 
⇒ Complete table will be –
Question 30.
Two dice are rolled simultaneously and counts are added.
A student argues that ‘there are 11 possible outcomes 2.3.4.5.6.7.8.9.10.11 and12. Therefore, each of them has a probability 
Do you agree with this argument? Justify your answer.
Answer:
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.
Question 31.
A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Deskhitha wins if all the tosses give the same result i.e. three heads or three tails. And loses otherwise. Calculate the probability that she will lose the game.
Answer:
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 .
Question 32.
A die is thrown twice. What is the probability that
5 will not come up either time?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment ].
Answer:
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 .
Question 33.
A die is thrown twice. What is the probability that
5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment ].
Answer:
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 
