Probability NCERT solutions Chapter 15 Exercise 15.1
Question 1
1. Complete the following statements.
(i) Probability
of an event E + Probability of the event 'not E'= 1
(ii) The probability
of event that cannot happen is zero. Such an event is called impossible event.
(iii) The
probability of an event that is certain to happen is 1. Such an event is called
sure event or certain 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 zero and less than or equal to 1.
Probability ncert solutions Chapter 15 Exercise 15.1
Question 2
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.
This experiment does
not have equally likely outcomes. The probability of outcomes depends on the
condition of car.
(ii) A player
attempts to shoot a basketball. She/he shoots or misses the shot.
This experiment does
not have equally likely outcomes. The probability of outcomes depends on the
skills of player.
(iii)
A trial is made to
answer a true-false question. The answer is right or wrong.
This experiment has
equally likely outcomes because answer is either true or false and any person
would answer either true or false. The probability of giving right answer is
0.5 and also, the probability of giving wrong answer is 0.5.
(iv)
A baby is born. It is
a boy or a girl.
It has equally likely
outcomes. The probability that baby is going to be boy is 0.5 and the
probability that baby is going to be girl is also 0.5.
Probability ncert solutions Chapter 15 Exercise 15.1
Question 3
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?
Answer:
It is because there
are equal chances of winning a toss for both of the teams. The probability of
winning a toss for a particular team is 50 % or 0.5.
Probability ncert solutions Chapter 15 Exercise 15.1
Question 4
4. Which of the
following cannot be the probability of an event?
(A) 2/3
(B) -1.5 (C) 15 %
(D) 0.7
Answer is
(B) because probability of happening any event is always greater than or equal
to 0 and less than equal to 1.
Probability ncert solutions Chapter 15 Exercise 15.1
Question 5
5. If
P(E)=0.05, what is the probability of 'not E'?
Solution:
We know that P(E) +
P(not E)=1
⇒ 0.05+P(not E)=1
⇒ P(not E)=1-0.05=0.95
Probability ncert solutions Chapter 15 Exercise 15.1
Question 6
6. 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
(i) an orange flavored candy
(ii) a lemon flavored candy
Solution:
(i) There are no
orange flavored candies in the bag. Therefore, the probability of taking out
orange flavored candy is zero.
(ii) All the candies
present in the bag are lemon flavored. Therefore, Malini takes out lemon flavored
candy for sure which means probability of taking out lemon flavored candy is 1.
Probability ncert solutions Chapter 15 Exercise 15.1
Question 7
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?
Solution:
Let E be the event
that 2 students do not have the same birthday.
It is given that
P(E)=0.992
We have P(E)+P(E¯¯¯)=1 {where P(E¯¯¯) is probability of not happening of event E or
we can say that it is the probability of 2 students having the same birthday.}
⇒0.992+P(E¯¯¯)=1
⇒P(E¯¯¯)=1−0.992=0.008
Therefore, Probability of 2 students having the same
birthday=0.008
Chapter 15 Exercise 15.1 Question 8
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?
Solution:
(i)
Let E be the event of
drawing a red ball from the bag.
The number of possible
outcomes favorable to E=3 because there are 3 red balls.
Number of all possible
outcomes of experiment =3+5=8
Therefore, P(E)=
Number of outcomes favorable to E/Number of all possible outcomes of experiment =38
It means that
Probability of drawing a red ball =P(E)=3/8
(ii)
Continuing from part
(i)
We know that P(E)+P(E¯¯¯)=1 {where P(E¯¯¯) is probability of not happening of event E. In
other words it is the probability of not drawing a red ball}
⇒3/8+P(E¯¯¯)=1
⇒P(E¯¯¯)=1−3/8=5/8
Therefore, Probability
of not drawing a Red ball =P(E¯¯¯)=5/8
Probability ncert solutions Chapter 15 Exercise 15.1
Question 9
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?
Solution
(i)
Let E be the Event
that selected marble is red.
Number of possible
outcomes favorable to event E=5 {Because there are 5 red marbles}
Total number of
possible outcomes=5+8+4=17 {Total number of balls}
Probability that the
selected marble is red=(Number of outcomes favorable to Event E)/Total number
of possible outcomes=5/17
(ii)
Let E1 be the event
that selected marble is white.
Number
of possible outcomes favorable to event E1=8 {Because there are 8
white marbles}
Total number of
possible outcomes=5+8+4=17 {Total number of balls}
Probability that the
selected marble is white=(Number of outcomes favorable to Event E1)/Total
number of possible outcomes=8/17
(iii)
Continuing from part (i) and (ii)
Let E2 be the Event
that selected marble is not green.
The ball cannot be
green if it is either red or white. Therefore, Event E2 will happen if there is
either Event E or event E1.
The number of possible
outcomes favorable to Event E2=5+8=13 (Number of red balls + number
of white balls)
Total number of
possible outcomes=5+8+4=17 {Total number of balls}
Probability that
the selected marble is not green=(Number of outcomes favorable to Event
E2)/Total number of possible outcomes=13/17
Probability ncert solutions Chapter 15 Exercise 15.1
Question 10
10. A piggy bank
contains hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 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 50p
coin. (ii) will not be a Rs 5 coin
Solution:
(i)
Let E be the event
that 50p coin falls out.
Number of outcomes
favorable to event E =Total number of 50p coins=100
Total number of
possible outcomes=total number of coins=100+50+20+10=180
P(E)=(Number of outcomes
favorable to event E)/Total number of possible outcomes =100180=59
(ii)
Let F be the event
that the coin falls out is not Rs 5.
Number of possible
outcomes favorable to event F=Total number of coins except Rs 5
coins=100+50+20=170
Total number of
possible outcomes=Number of coins=100+50+20+10=180
P(F)=Number of possible
outcomes favorable to event F/Total number of possible outcomes =170/180=17/18
Probability ncert solutions Chapter 15 Exercise 15.1
Question 11
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. What is the
probability that the fish taken out is a male fish?
Solution:
Let E be the event
that the fish taken out is male.
Number of possible
outcomes favorable to E=Number of male fish=5
Total number of
possible outcomes=total number of fish in the tank=5+8=13
P(E)=(Number of
possible outcomes favorable to event E)/Total number of possible outcomes =5/13
Probability NCERT solutions Chapter 15 Exercise 15.1
Question 12
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, 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?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution (i)
Let E be the event that arrow rests at number 8.
Let E be the event that arrow rests at number 8.
Number of outcomes
favorable to Event E=1
Number of total possible outcomes=8
Number of total possible outcomes=8
P(E)=(Number of
outcomes favorable to Event E)/(Number of total possible outcomes)=1/8
Solution(ii)
Let F be the event that arrow rests at an odd number.
Let F be the event that arrow rests at an odd number.
Number of outcomes
favorable to Event F=4 (There are 4 odd numbers from numbers 1 to 8)
Total number of possible outcomes=8 (There are total of 8 numbers in the game).
Total number of possible outcomes=8 (There are total of 8 numbers in the game).
P(F)=(Number of
outcomes favorable to Event F)/(Total number of possible outcomes)=4/8=1/2
Solution (iii)
Let A be the event that arrow stops at number greater than 2.
Number of outcomes favorable to Event A=6 (There are 6 numbers greater than 2 in a game)
Total number of possible outcomes=8
Let A be the event that arrow stops at number greater than 2.
Number of outcomes favorable to Event A=6 (There are 6 numbers greater than 2 in a game)
Total number of possible outcomes=8
P(A)=(Number of
outcomes favorable to Event A)/(Total number of possible outcomes)=6/8=3/4
Solution(iv)
Let B be the event that arrow stops at number less than 9.
Let B be the event that arrow stops at number less than 9.
Each number in the
game is less than 9. Therefore, Event B will certainly happen.
Therefore, P(B)=1
Probability ncert solutions Chapter 15 Exercise 15.1
Question 13
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
(i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number
Solution (i)
Let E be the Event of
getting a prime number with a single throw of dice.
Number of outcomes
favorable to Event E=3 (There are three prime numbers from numbers 1 to 7 which
are 2, 3 and 5.)
Total number of possible outcomes=6 {There are six possible outcomes in a dice.}
Total number of possible outcomes=6 {There are six possible outcomes in a dice.}
P(E)=(Number of
outcomes favorable to Event E)/(Total number of possible outcomes)=3/6=1/2
Solution (ii)
Let F be the Event of
getting a number between 2 and 6 with a single throw of dice.
Number of outcomes
favorable to Event F=3 (There are 3 numbers between 2 and 6).
Total number of possible outcomes=6 {There are six possible outcomes in a dice.}
Total number of possible outcomes=6 {There are six possible outcomes in a dice.}
P(F)=(Number of
outcomes favorable to Event F)/(Total number of possible outcomes)=3/6=1/2
Solution (iii)
Let A be the Event of
getting an odd number with a single throw of dice.
Number of outcomes
favorable to Event A=3 {There are 3 odd numbers from 1 to 6}
Total number of possible outcomes=6 {There are six possible outcomes in a dice}
Total number of possible outcomes=6 {There are six possible outcomes in a dice}
P(A)=(Number of
outcomes favorable to Event A)/Total number of possible outcomes=3/6=1/2
Probability ncert solutions Chapter 15
Exercise 15.1 Question 14
14. A card is drawn from a well-shuffled deck
of 52 cards. Find the probability of getting
(i) a king of red color (ii) a face card (iii) a red face card
(iv) the jack of hearts (v) a spade (vi) the queen of diamonds
(i) a king of red color (ii) a face card (iii) a red face card
(iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Solution (i)
Let E be the
event of drawing a king of red color
Number of outcomes favorable to event E=2 {There are two red kings in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
Number of outcomes favorable to event E=2 {There are two red kings in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(E)=(Number of
outcomes favorable to event E)/(Total number of possible outcomes)=2/52=1/26
(ii)
Let F be the
event of drawing a face card.
Number of outcomes favorable to event F=12 {There are 12 face cards in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(F)=(Number of outcomes favorable to event F)/(Total number of possible outcomes)=12/52=3/13
Number of outcomes favorable to event F=12 {There are 12 face cards in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(F)=(Number of outcomes favorable to event F)/(Total number of possible outcomes)=12/52=3/13
(iii)
Let A be the event of drawing a red face card.
Number of outcomes favorable to event F=6 {There are 6 red face cards in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(A)=(Number of outcomes favorable to event A)/(Total number of possible outcomes)=6/52=3/26
(iv)
Let B be the event of drawing jack of hearts.
Number of outcomes favorable to event B=1 {There is 1 jack of hearts in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(B)=(Number of outcomes favorable to event B)/(Total number of possible outcomes)=1/52
(v)
Let C be the event of drawing a spade.
Number of outcomes favorable to event C=13
{There are 13 spade cards in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(C)=(Number of outcomes favorable to event C)/(Total number of possible outcomes)=13/52=1/4
(vi)
Let D be the event of drawing a queen of diamonds.
Number of outcomes favorable to event D=1 {There is 1 queen of diamonds in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(D)=(Number of outcomes favorable to event D)/(Total number of possible outcomes)=1/52
Let D be the event of drawing a queen of diamonds.
Number of outcomes favorable to event D=1 {There is 1 queen of diamonds in a deck of 52 cards}
Total number of possible outcomes=52 {There are total of 52 cards}
P(D)=(Number of outcomes favorable to event D)/(Total number of possible outcomes)=1/52
Probability ncert solutions Chapter 15
Exercise 15.1 Question 15
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?
Solution:
(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?
Solution:
(i)
There are five cards.
There are five cards.
Let E be the
event that card drawn is queen.
Total number of outcomes favorable to event E=1 (Because there is just 1 queen)
Total number of possible outcomes=5 (We have 5 cards)
P(E)=(Total number of possible outcomes favorable to event E)/Total number of possible outcomes=1/5
Total number of outcomes favorable to event E=1 (Because there is just 1 queen)
Total number of possible outcomes=5 (We have 5 cards)
P(E)=(Total number of possible outcomes favorable to event E)/Total number of possible outcomes=1/5
(ii)
Queen has already been drawn and put aside. Therefore, we now have just 4 cards left.
(a)
Let A be the event that card drawn is an ace.
Total number of outcomes favorable to Event A=1 (We have 1 ace out of 4 cards)
Total number of possible outcomes=4 (We have 4 cards left)
P(A)=(Total
number of outcomes favorable to event A)/(Total number of possible
outcomes)=1/4
(b)
Let B be the event that card drawn is queen.
Total number of outcomes favorable to Event B=0 (We have no queen left in 4 cards.)
Total number of possible outcomes=4 (We have 4 cards left)
P(B)=(Total number of outcomes favorable to event B)/(Total number of possible outcomes)=0/4=0
Probability ncert solutions Chapter 15
Exercise 15.1 Question 16
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.
Solution:
Solution:
Let E be the
event that pen taken out is a good pen.
Number of
outcomes favorable to event E=132 (There are 132 good pens)
Total number of
possible outcomes=12+132=144 (There are total of 144 pens)
P(E)=(Number of
outcomes favorable to event E)/(Total number of possible outcomes) =132/144=11/12
Probability ncert solutions Chapter 15
Exercise 15.1 Question 17
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?
(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?
Solution
(i)
Let E be the event of drawing a defective bulb.
Total number of outcomes favorable to event E=4 (There are 4 defective bulbs)
Total number of possible outcomes=20 (There are total of 20 bulbs)
P(E)=(Total number of outcomes favorable to event E)/(Total number of possible outcomes)=4/20=1/5
Let E be the event of drawing a defective bulb.
Total number of outcomes favorable to event E=4 (There are 4 defective bulbs)
Total number of possible outcomes=20 (There are total of 20 bulbs)
P(E)=(Total number of outcomes favorable to event E)/(Total number of possible outcomes)
(ii)
One non-defected bulb has been taken out already. Therefore, we now have just 19 bulbs out of which 4 are defective.
Let A be the
event of drawing a non-defective bulb.
Total number of outcomes favorable to event A=19-4=15
(There are 15 non-defective bulbs out of 19 bulbs)
Total number of possible outcomes=19 (There are total of 19 bulbs)
P(A)=(Number of outcomes favorable to event A)/(Total number of possible outcomes)=15/19
Total number of outcomes favorable to event A=19-4=15
(There are 15 non-defective bulbs out of 19 bulbs)
Total number of possible outcomes=19 (There are total of 19 bulbs)
P(A)=(Number of outcomes favorable to event A)/(Total number of possible outcomes)
Probability ncert solutions Chapter 15
Exercise 15.1 Question 18
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
Solution:
(i)
Let E be the event of drawing a disc bearing two-digit number.
Let E be the event of drawing a disc bearing two-digit number.
Number of
possible outcomes favorable to event E=81
(There are 81 two digit numbers from 1 to 90)
Total number of possible outcomes=90 (There are 90 discs)
P(E)=(Number of possible outcomes favorable to event E)/(Total number of possible outcomes)=81/90=9/10
(There are 81 two digit numbers from 1 to 90)
Total number of possible outcomes=90 (There are 90 discs)
P(E)=(Number of possible outcomes favorable to event E)/(Total number of possible outcomes)
(ii)
Let A be the event of drawing a disc bearing a perfect square number.
Let A be the event of drawing a disc bearing a perfect square number.
The perfect
squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
Therefore, there are 9 perfect square numbers from 1 to 90.
Number of possible outcomes favorable to event A=9
(There are 9 perfect square numbers from 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(A)=(Number of possible outcomes favorable to event A)/(Total number of possible outcomes)=9/90=1/10
Therefore, there are 9 perfect square numbers from 1 to 90.
Number of possible outcomes favorable to event A=9
(There are 9 perfect square numbers from 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(A)=(Number of possible outcomes favorable to event A)/(Total number of possible outcomes)
(iii)
Let B be the event of drawing a disc bearing a number divisible by 5.
The numbers
divisible by 5 from numbers 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50,
55, 60, 65, 70, 75, 80, 85 and 90.
Therefore, there are 18 numbers divisible by 5 from numbers 1 to 90.
Therefore, there are 18 numbers divisible by 5 from numbers 1 to 90.
Number of
outcomes favorable to event B=18
(There are 18 numbers divisible by 5 from numbers 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(B)=(Number of possible outcomes favorable to event B)/(Total number of possible outcomes)=18/90=1/5
(There are 18 numbers divisible by 5 from numbers 1 to 90)
Total number of possible outcomes=90 (There are total of 90 discs)
P(B)=(Number of possible outcomes favorable to event B)/(Total number of possible outcomes)
Probability ncert solutions Chapter 15
Exercise 15.1 Question 19
19. A child has a die whose six faces show
the letters A, B, C, D, E and A.
The die is thrown once. What is the probability of getting (i) A? (ii) D?
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution
(i)
(i)
Let E be the
event of getting A with a single throw of dice.
Number of
possible outcomes favorable to event E=2
(There are 2 faces on a dice with A written on them)
Total number of possible outcomes=6 (There are 6 faces on a dice)
P(E)=(Number of outcomes favorable to event E)/(Total number of possible outcomes)=2/6=1/3
(There are 2 faces on a dice with A written on them)
Total number of possible outcomes=6 (There are 6 faces on a dice)
P(E)=(Number of outcomes favorable to event E)/(Total number of possible outcomes)
(ii)
Let F be the event of getting D with a single throw of dice.
Let F be the event of getting D with a single throw of dice.
Number of
possible outcomes favorable to event F=1
(There is 1 face with letter D on it)
(There is 1 face with letter D on it)
Total number of
possible outcomes=6 (There are 6 faces on a dice)
P(F)=(Number of outcomes favorable to event F)/(Total number of possible outcomes)=1/6
P(F)=(Number of outcomes favorable to event F)/(Total number of possible outcomes)
Probability ncert solutions Chapter 15
Exercise 15.1 Question 20
20. Suppose you drop a die at random on the
rectangular region shown in fig. What is the probability that it will land
inside the circle with diameter 1 m?
Solution:
Let E be the
event that dice lands inside the circle.
Radius of circle
=Diameter/2==1/2=0.5 m
Area of circle =π.r2=π.(0.5)2=π.( 0.25) m2
Area of rectangle
=length x breadth=3 x 2 =6 m2
P(E)=(Area of
circle)/(Area of rectangle) =π.(0.25)/ 6=π/24
Probability ncert solutions Chapter 15
Exercise 15.1 Question 21
21. A lot
consists of 144 balls 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?
Solution:
(i)
Let E be the
event that she buys pen.
Number of
outcomes favorable to Event E=144-20=124
(There are 124
good pens)
Total number of
possible outcomes=144
(There are total
of 144 pens)
P(E)=(Number of
outcomes favorable to E)/(Total number of possible outcomes)
(ii)
Let F be the
event that she does not buy a pen.
Number of
outcomes favorable to Event F=20
(There are 20
defective pens)
Total number of
possible outcomes=144
(There are total
of 144 pens)
P(F)=(Number of
outcomes favorable to F)/(Total number of possible outcomes)
Probability ncert solutions Chapter 15
Exercise 15.1 Question 22
22. Two dice, one blue and one grey are
thrown at the same time. What is the probability of different sums that can
appear on the top of different dice.
Also, 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 .
Do you agree with this argument? Justify your answer.
Solution:
Sum equal to 2
can appear in just 1 way which is (1, 1) (We have 1 appearing on both of
the dice).
Sum equal to 3
can appear in 2 ways which are (1, 2) and (2, 1)
Sum equal to 4
can appear in 3 ways which are (1, 3), (3, 1) and (2, 2)
Sum equal to 5
can appear in 4 ways which are (2, 3), (3, 2), (1, 4) and (4, 1)
Sum equal to 6
can appear in 5 ways which are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3)
Sum equal to 7
can appear in 6 ways which are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4,
3)
Sum equal to 8
can appear in 5 ways which are (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4)
Sum equal to 9
can appear in 4 ways which are (3, 6), (6, 3), (4, 5) and (5, 4)
Sum equal to 10
can appear in 3 ways which are (4, 6), (6, 4) and (5, 5)
Sum equal to 11
can appear in 2 ways which are (5, 6) and (6, 5)
Sum equal to 12
can appear in 1 way which is (6, 6)
Therefore, we can
find probabilities of all the sums which can appear on the top of two dice.
Total number of
possible outcomes =1+2+3+4+5+6+5+4+3+2+1=36
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
We do not agree
with this argument because there are different number of possible outcomes for
each sum. According to the above table, we can see that each sum has different
probability.
Probability ncert solutions Chapter 15 Exercise 15.1 Question 23
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.
Solution:
Let E be the event
that Hanif loses the game.
The different possible
outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT and TTT.
Therefore, there are
total of 8 different possible outcomes.
Number of possible
outcomes favorable to event E=6
Total number of
possible outcomes =8
P(E)=(Number of
outcomes favorable to event E)/(Total number of possible outcomes)
=6/8=3/4
Probability ncert solutions Chapter 15
Exercise 15.1 Question 24
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?
[Hint: Throwing
a die twice and throwing two dice simultaneously are treated as the same
experiment]
Solution:
The different
outcomes possible when throwing two dice 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)
(i)
Let E be the
event that 5 does not appear either time.
Number of
outcomes favorable to event E=25
Total number of
possible outcomes=36
P(E)=(Number of
outcomes favorable to Event E)/(Total number of possible outcomes) =25/36
(ii)
Let F be the even
that 5 appear at least once.
Number of
outcomes favorable to event E=11
Total number of
possible outcomes=36
P(F)=(Number of
outcomes favorable to Event F)/(Total number of possible outcomes) =11/36
Probability ncert solutions Chapter 15
Exercise 15.1 Question 25
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
(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
Solution:
(i) Incorrect
If two coins are
tossed simultaneously there are 4 possible outcomes.
HH
(We get heads on both the coins)
HT
(Heads on the first coin and tails on the second coin)
TH
(Tails on the first coin and heads on the second coin)
TT
(Tails on both the coins)
Therefore,
Probability of each outcomes is equal to 1
(ii) Correct
With a single
throw of a die, we have 6 possible outcomes.
1, 2, 3, 4, 5, or
6.
There are 3 even
numbers and 3 odd numbers from 1 to 6.
Therefore, the
probability of getting an even or odd number is equal to 1
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