Example-1 Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
sol) In the experiment of tossing a coin once, the number of possible outcomes is
two- head(H) and Tail(T).
Let E be the event ' getting a head'. The number of outcomes favourable to E, (i.e. of getting a head) is 1. Therefore.
P(E) = P(head)
= No.of outcomes favourable to E = 1
No. of all possible outcomes 2
Similarly, if F is the event ' getting a tail', then
P(F) = P(Tail) = 1
2
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Example-2. A bag contains a red ball, a blue ball and an yellow ball, all the balls beings of the same size. Manasa take out a ball from the bag without looking into it. What is the probability that she takes a
(!) yellow ball (!!) red ball (!!!) blue ball
sol) Manasa takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let Y be the event ' the ball taken out is yellow',
B be the event ' the ball taken out is blue',
and R be the event ' the ball taken out is red'.
Now, the number of possible outcomes = 3.
(!) The number of outcomes favourable to the event Y = 1
So,
P(Y) = 1/3
P(R) = 1/3
P(B) = 1/3
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Remarks
1) An event having only one outcome in an experiment is called an elementary event.
In example 1, both the event E and F are elementary events,
Similarly, in Example 2, all the three events, Y,B and R are elementary events.
2) In Example 1, we note that : P(E) + P(F) =1
In Example 2, we note that : P(Y)+P(R)+P(B) = 1.
If we find the probability of all the elementary events and add them, we would get the total as 1.
3) In events like a throw of dice, probability of getting less than 3 and of getting a 3 or more than three are not elementary events of the possible outcomes.
In tossing two coins(HH), (HT) and (TT) are elementary events.
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Example-3. Suppose we throw a dice once
(!) What is the probability of getting a number greater than 4?
sol) In rolling an unbaised dice
Sample space S = {1,2,3,4,5,6}
No.of outcomes n(S) = 6
Favourable outcomes for number greater than 4
E = {5,6}
No.of favourable outcomes n(E) = 2
Probability P(E) = 2/6 = 1/3
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(!!) What is the probability of getting a number less than or equal to 4?
sol) let F be the event ' getting a number less than or equal to 4'
Sample space S = { 1,2,3,4,5,6}
No.of outcomes n(S) = 6
Favourable outcomes for F = {1,2,3,4}
number less or equal to 4
No.of favourable outcomes n(F) = 4
Probability P(F) = 4/6 = 2/3
Note : Are the events E and F in the above example elementary events?
No, they are not elementary events. The event E has 2 outcomes and the event F has 4 outcomes.
(*) Complementary Events and probability
In example-3, we calculated probability of events which are not elementary. We saw,
P(E) + P(F) = 1/3 + 2/3 = 1
Here F is the same as ' not E' by E. This is called the complement event of event E.
So, P(E) + P (not E) = 1
i.e., P(E) + P(E) = 1 , which gives us
P(E) = 1 - P(E)
In general, it is true that for an event E,
P(E) = 1 - P(E)
(*) Impossible and certain events
consider the following about the throws of a die with sides marked as 1,2,3,4,5,6
(!) What is the probability of getting a number 7 in a single throw of a die?
We know that there are only six possible outcomes in a single throw of this die. These outcomes are 1,2,3,4,5 and 6.
Since no face of the die is marked 7, there is no outcome favourable to 7, i.e., the number of such outcomes is zero.
In other words, getting 7 in a single throw of a die, is impossible.
So P(getting 7) = 0/6 = 0
That is, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.
(!!) What is the probability of getting 6 or a number less than 6 in a single throw of a die?
Since every face of a die is marked with 6 or a number less than 6, it is sure that we will always get one of these when the dice is thrown once.
So, the number of favourable outcomes is the same as the number of all possible outcomes, which is 6.
Therefore, P(E) = P(getting 6 or a number less than 6) = 6/6= 1
So, the probability of an event which is sure( or certain) to occur is 1.Such an event is called a sure event or a certain event.
Note: From the definition of probability P(E), we see that the numerator (number of outcomes favourable to the event E) is always less than or equal to the denominator(the number of all possible outcomes).
Therefore, 0< P(E) < 1.
(*) Deck of cards and Probability
Have you seen a deck of playing cards?
A deck of playing cards consists of 52 cards which are divided into 4 suits of 13 cards each.
They are
black spades(♠),
red hearts (💟),
Red diamonds(◆) and
black clubs(♣)
The cards in each suit are Ace, King, Queen, Jack, 10,9,8,7,6,5,4,3, and 2.
Kings, Queens and jacks are called face cards.
many games are played with this deck of cards, some games are played with part of the deck and some with two decks even.
The study of probability has a lot to do with card and dice games as it helps players to estimate possibilities and predict how the cards could be distributed players.
Example-4. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will
(!) be an ace
sol) Well-shuffling ensures equally likely outcomes.
There are 4 aces in a deck.
Let E be the event ' the card is an ace'.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52
Therefore, P(E) = 4/52 = 1/13
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(!!) Let F be the event ' card drawn is not an ace'.
sol) The number of outcomes favourable to the event F = 52-4=48
The number of possible outcomes = 52
Therefore, P(F) = 48/52= 12/13
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Alternate Method : Note that F is nothing but
_
E
Therefore, we can also calculate P(F) as follows.
_
P(F) = P (E) = 1 - 1/13 = 12/13
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(*) Use of Probability
We know that in sports some countries are strong and other are not so strong. We also know that when two players are playing it is not that they win equal times..
The probability of winning of the player or team that wins more often is more than the probability of the other player or team.
We also discuss and keep track of birthdays.Sometimes happens it that people we know have the same birthdays.
Can we find out whether this is a common event or would it only happen occasionally . Classical probability helps us do this.
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Example-5. Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Sol) Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of Sangeeta's winning chances = P(S) =0.62 (given)
The probability of Reshma's winning chances = P(R)= 1- P(S)
= 1- 0.62 = 0.38 [ R & S are complementary]
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Example-6. Sarada and Hamida are friends. What is the probability that both will have
(!) different birthdays?
sol) Out of the two friends, one girl,say, Sarada's birthday can be any day of the year.
Now , Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely.
If Hamid's birthday is different from Sarada's, the number of favourable outcomes for her birthday is
365-1 = 364
So,
P(Hamida's birthday is different from Sarada's birthday) = 364/365
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(!!) P(Saradan and hamida have the same birthday) = 1 - P( both have different birthdays)
_
= 1 - 364/365 [ using P(E) = 1-P(E)]
= 1
365
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Example-7. There are 40 students in class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative . She writes the name of each student on a separate cards, the cards being identical.
Then she put cards in a box and stirs them thoroughly. She then draws one card from the box.
What is the probability that the name written on the card is the name of
(!) a girl?
sol) There are 40 students , and only one name card has to be chosen.
The number of all possible outcomes is 40
(!) The number of outcomes favourable for a card with the name of a girl = 25
.^. P(card with name of a girl) = P(Girl)
= 25 = 5
40 8
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(!!) a boy?
sol) The no.of outcomes favourable for a card with the name of a boy = 15
Therefore, P(card with name of a boy) = P(Boy)
= 15 = 3
40 8
or
P(Boy) = 1 - P(not Boy)
= 1- P(Girl)
= 1 - 5/8
= 3/8
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Exercise - 13.1
(1) Complete the following statements :
1) Probability of an event E + Probability of the event not E1 = ....................
sol) 1
2) The probability of an event that cannot happen is......... such an event is called..........
sol) zero, impossible event
3) The probability of an event that is certain to happen is........ Such an event is called.........
sol) 1, sure event
4) The sum of the probabilities of all the elementary events of an experiment is............
sol) 1
5) The probability of an event is greater than or equal to..... and less than or equal to....
sol) 0 , 1
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2) Which of the following experiments have equally likely outcomes? Explain.
(1) A driver attempts to start a car. The car starts or does not start.
sol) The car getting started or not started depends on :-
1) How old the car is.
2) Is their fuel in the engine
3) Had met in an accident etc?
So, the factors on which the car will start or not is not the same in both cases.
.^., It is NOT an equally likely event.
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(2) A player attempts to shoot a basketball . She/he shoots or misses the shot.
sol) The amount of shots he/she shoots can go in more times than it misses or vice-versa depending on players ability.
Which is not specified here.
.^. It is NOT an equally likely event
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(3) A trial is made to answer a true-false question. The answer is right or wrong.
sol) The answer can only be right or wrong.
It is equally likely that answer is right or wrong because it is a true false question.
.^., It is an equally likely event.
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(4) A baby is born . It is a boy or girl.
sol) A baby can be a boy or girl.
It is equally likely that the baby is a boy or a girl
.^., It is an equally likely event.
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3) If P(E) = 0.05, what is the probability of "not E1"?
sol) Probability of an event cannot be negative or greater than 1.
i.e, 0 < probability of event E < 1
Given :- P(E) = 0.05
p(E) + p( not E) = 1
p(Not E ) = 1 - p(E)
" = 1 - 0.05
" = 0.95
.^. , the probability of 'not E' is 0.95.
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(4) 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
(1) an orange flavoured candy?
sol) The bag contains lemon flavoured candies only.
It does not contain any orange flavoured candles
So every time. she will take out only lemon flavoured candles.
Probability of orange flavoured candy P (C) = 0
.^., event that Malini will take out an orange flavoured candy is an impossible event.
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(2) a lemon flavoured candy?
sol) Let total number of candles = 10
Number of candies which are lemon flavour = 10
Probability Candy taken out with lemon flavour :-
P (C) = Number of candies which are lemons falvoured
Total number of candies
=> 10
10
=> 1
.^. , event that Malini will take out a lemon flavoured candy is a sure event.
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(5) Rahim takes out at all the hearts from the cards. What is the probability of
1) Picking out an ace from the remaining pack.
sol) There are 3 aces in remaining pack
The number of possible outcomes = 39
Let "E" be the event "the card is an ace"
The no.of outcomes favorable to E = 3
.^. , P(A) = 3 = 0.0769
39
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2) Picking out a diamonds.
sol) There are 13 diamond in pack
The number of possible outcomes = 39
Let E be the event "the card is a diamond"
The number of outcomes favorable to E = 13
.^. , P (D) = 13 = 0.333
39
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3) Picking out a card that is not a heart.
sol) A card that is not a heart = 39
The number of possible outcomes = 39
Let "H" be the event " the card is not an heart".
The number of outcomes favorable to "H" = 39
.^. , P(H) =
39
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4) Picking out the Ace of hearts.
sol) There are no aces of heart in remaining pack
The number of possible outcomes = 39
P(E) = 0 = 0
39
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(6) 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) P(not E) = The probability of 2 students not having the same birthday = 0.992
The probability of 2 student having same birthday = P(E)
We know
P(E) + P(not E) = 1
p(E) = 1 - P(not E)
= 1 - 0.992
= 0.008.
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(7) A die is thrown once. Find the probability of getting
(1) a prime number
sol) Sample space S = { 1,2,3,4,5,6}
No.of outcomes n(S) = 6
Favorable outcomes for prime number P ={2,3,5}
No.of favorable outcomes n(P) = 3
probability P(P) =
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(2) A number lying between 2 and 6
sol) Sample space S = { 1,2,3,4,5,6}
Number of outcomes n(S) = 6
Favorable outcomes for number between 2 and 6 E = {3,4,5}
No.of favorable outcomes n(E) = 3
probability P(E) =
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(3) an odd number.
sol) Sample space S = { 1,2,3,4,5,6}
Number of outcomes n(S) = 6
Favorable outcomes for odd number E = {1,3,5}
No.of favorable outcomes n(E) = 3
probability P(E) =
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(8) What is the probability of drawing out a red king from a deck of cards?
sol) Number of red kings in deck = 2
The number of possible outcomes = 52
let 'E" be the event " the card is red king"
The number of outcomes favorable to E = 2
.^. , P(E) =
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More Applications of probability
Example-8. A box contains 3 blue, 2 white, and 4 red marbles. if a marble is drawn at random from the box, what is the probability that it will be
(!) White?
sol) Saying that a marble is drawn at random means all the marbles are equally likely to be drawn.
.^. The number of possible outcomes
= 3+2+4=9
Let W denote the event' the marble is white',
B denote the event ' the marble is blue' and
R denote the event ' marble is red'
(!) The number of outcomes favourable to the event W = 2
So, P(W) = 2/9
Similarly,
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(!!) Blue?
sol) P(B) = 3/9 = 1 /3
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(!!!) Red?
sol) P(R) = 4/9
Note that
P(W) + P(B) + P(R) = 1
2 + 1 + 4
9 3 9
= 2 + 3 + 4
9 9
= 5 + 4
9 9
= 9/9 = 1
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Example-9: Harpreet tosses two different coins simultaneously (say, one is of rs.1 and other of rs.2). What is the probability that she gets at least one head?
sol) We write
H for ' head'
T for 'Tail'
When two coins are tossed simultaneously, the possible outcomes are
(H,H), (H,T), (T,H), (T,T)
which are all equally likely. Here (H,H) means heads on the first coin (say on rs.1) and also heads on the second coin (Rs.2)
Similarly (H,T) means heads up on the first coin and tail up on the second coin and so on.
The outcomes favourable to the event E,' at least one head' are
(H,H) (H,T) and (T,H)
So, the number of outcomes favourable to E is 3.
.^. P(E) = 3/4
[ Since the total possible outcomes = 4]
i.e., the probability that Harpreet gets at least one head is 3/4
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Check This
Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now.
There are many experiments in which the outcome is number between two given numbers, or in which the outcome is every point within a circle or rectangle, etc.
Can you count the number of all possible outcomes in such cases?As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle.
So, the definition of theoretical probability which you have learnt so far cannot be applied to the present form.
What is the way out? To answer this, let us consider the following example;
Example-10: In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing.
What is the probability that the music will stop within the first half-minute after starting?
sol) Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 and 2.
<-0-----(1/2)|-------1|--------2|--------->
Let E be the event that ' the music is stopped within the first half-minute.'
The outcomes favourable to E are points on the number line from 0 to 1/2
The distance from 0 to 2 is 2, while the distance from 0 to 1/2 is 1/2
Since all the outcomes are equally likely, we can argue that, of the total distance is 2 and the distance favourable to the event E is 1/2
So,
P(E) =
Distance favourable to the event E = (1/2) / 2
Total distance in which outcomes can lie
= 1/4
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We now try to extend this idea of for finding the probability as the ratio of the favourable area to the total area.
Example-11. A missing helicopter is reported to have crashed somewhere in the rectangular region as shown in the figure. What is the probability that it crashed inside the lake shown in the figure?
sol) The helicopter is equally likely to crash anywhere in the region. Area of the entire region where the helicopter can crash
= (4.5 *9 ) km^2 = 40.5 km^2
Area of the lake = (2*3) km^2 = 6 km^2
Therefore,
P(helicopter crashed in the lake)
= 6 = 4
40.5 27
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Example-12. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects.
Jhony, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects.
One shirt is drawn at random from the carton. What is the probability that
(!) It is acceptable to Jhony?
sol) One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.
The number of outcomes favourable (i.e., acceptable) to Jhony = 88
Therefore,
P(Shirt is acceptable to Jhony) = 88 = 0.88
100
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(!!) It is acceptable to Sujatha?
sol) The number of outcomes favourable to Sujatha= 88 + 8 = 96
So
P(Shirt is acceptable to Sujatha) = 96 = 0.96
100
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Example-13. Two dice, one red and one white, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(!) 8
sol) When the red dice shows'1', the white dice could show any one of the numbers 1,2,3,4,5,6.
The same is true when the red dice shows '2','3','4','5', or '6'.
The possible outcomes of the experiment are shown in the figure
the first number in each ordered pair is the number appearing on the red dice and the second number is that on the white dice.
Note:- the pair (1,4) is different from (4,1)
So the number of possible outcomes
n(S) = 6 * 6 = 36
(!) 8
sol) The outcomes favourable to the event' the sum of the two numbers is 8' denoted by E are :
(2,6)(3,5) (4,4) (5,3) ( 6,2)
i.e., the number of outcomes favourable to E is
n(E) = 5
Hence, P(E) = n (E) = 5
n (S) 36
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(!!) 13
sol) As there is no outcome favourable to the event F, ' the sum of two numbers is 13'.
So, P(F) = 0 = 0
36
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(!!!) less than or equal to 12?
sol) As all the outcomes are favourable to the event G, ' sum of two numbers is 12'.
So, P(G) = 36 = 1
36
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Exercise-13.2
(1) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that
sol) Saying that a marble is drawn at random means all the marbles are equally likely to be drawn.
.^. The number of possible outcomes = 3 + 5 = 8
Let "R" denote the event that " the ball is red"
Let "B" denote the event that " the ball is black"
(1) the ball drawn is red ?
sol) The no. of favorable outcomes to the event "R" = 3
So, P(R) = 3
8
(2) not red?
sol) P(Not R) = 1 - P(R)
=> 1 - 3
8
=> 5
8
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(2) A box contains 5 red marbles, 8 whites 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
sol) Saying that a marble is drawn at random means all the marbles are equally likely to be drawn.
.^. The number of possible outcomes = 5 + 8 + 4 = 17
Let "R" denote the event that "the marble is red "
Let "W" denote the event that " the marble is white" and
Let "G" denote the event that " the marble is green"
(1) probability that the marble taken out will be red?
Sol) The no.of outcomes favourable to the event "R" = 5
So, P(R) = 5
17
sol) P (W) = 8
17
(3) probability that the marble taken out will be not green?
sol) P(G) + P(Not G) = 1
P(Not G) = 1 - P(G)
=> 1 - 4
17
=> 13
17
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(3) A kiddy bank contains hundred 50p coins, fifty 1Rs coins, twenty 2rs coins and ten Rs5 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
sol) The number of possible outcomes = 100 + 50 + 20 + 10 = 180
(1) probability that the coin will be a 50 p coin?
sol) Let " E" denote the event "the coin is 50p"
The number of outcomes favourable to the event "E" = 100
SO, P(E) =
The probability that coin is 50p is 5/9.
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(2) probability that the coin Will not be a 5rs coin?
sol) Let "F" denote the event "the coin is not Rs.5"
The number of outcomes favourable to the event "F" = 170
SO, P(F) = 170 = 17
180 18
The probability that coin is not a Rs.5 coin is 17/18
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(4) 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 females fish( see figure). What is the probability that the fish taken out is a male fish?
sol) Saying that a fish is drawn at random means all the fish are equally likely to be drawn.
.^. The number of possible outcomes = 5+8= 13
Let "M" denote that 'the fish is male".
Let "F" denote that " fish is female"
The number of outcomes favourable to the event "M" = 5
So, P(M) = 5
13
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(5) 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 figure) and these are equally likely outcomes. What is the probability that it will point at
(1) probability that it will point at 8?
sol) The probability that it will point at "8" = 1
8
(2) probability that it will point at an odd number?
sol) Let "F" denote the event that "it points at an odd number".
The favourable to the event "F" = 1,3 5,7
The number of outcomes n(F) = 4
So, P(F) = 4 = 1
8 2
(3) probability that it will point at a number greater than 2?
sol) Let "E" denote the event " it point a number greater than 2"
The favourable to the event E = 3,4,5,6,7,8
The number of outcomes n(E) = 6
So, P(E) = 6 = 3
8 4
(4) probability that it will point at a number less than 9?
sol) Let "N" denote the event "it point a number less than 9"
The favourable to the event N = 1,2,3,4,5,6,7,8
The number of outcomes n(N) = 8
So, P(N) = 8/8 = 1
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(6) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(1) a king of red color
sol) There are 2 kings of red colour
The no.of possible outcomes = 52
Let "E" be the event "the card is a red king"
The number of outcomes favourable to E= 2
.^. , P(E) = 2 = 1
52 26
(2) a face card
sol) There are 12 face cards
The number of possible outcomes = 52
Let "E" be the event " the card is a face card"
The number of outcomes favourable to E = 12
.^. , P(E) = 12 = 3
52 13
(3) a red face card
sol) There are 6 red face cards
The number of possible outcomes = 52
Let "F" be the event " the card is a face card".
The number of outcomes favourable to F = 6
.^., P (F) = 6 = 3
52 26
(4) the jack of hearts
sol) There are 1 jack of hearts
The number of possible outcomes = 52
let "E" be the event "the card is the jack of hearts".
The number of outcomes favourable to E = 1
.^. , P(E) = 1
52
(5) a spade
sol) There are 13 spades
The number of possible outcomes = 52
Let 'S' be the event "the card is of spades".
The number of outcomes favourable to S = 13
.^. , P(S) = 13 = 1
52 4
(6) the queen of diamonds
sol) There is 1 queen of diamonds
The number of possible outcomes = 52
Let 'Q" be the event " the card is the queen of diamonds"
The number of outcomes favourable to Q = 1
.^. , P(Q) = 1
52
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(7) Five cards- the ten, jack queen, kind and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(1) What is the probability that the card is the queen?
sol) The number of possible outcomes = 5
The probability that the card is a queen = 1
5
(2) 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) Since the queen is drawn and put aside, now the total no.of outcomes = 5-1 = 4
Let 'A" denote the event "card is ace"
Let "Q" denote the event "card is queen".
P(A) = 1
4
P(Q) = 0 = 0
4
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(8) 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) The number of outcomes = good pens + defective pens = 144
Let "G" denote the event " the pen is good one"
The number of outcomes of event "G" = 132
P(G) = 132 = 11
144 12
The probability that the pen is good one is = 11 / 12
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(9) 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? Suppose the bulb drawn in previous case 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) The number of outcomes = 20
(1) Let "D' denote the event " the bulb is defective"
The number of occurence of "D" = 4
P(D) = 4 = 1
20 5
(2) Bulb drawn previous is not defective
So, the number of outcomes = 19
Let 'G" denote the event " the bulb is good one"
The number of occurence of good bulbs = 15
P(G) = 15
19
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(10) A box contains 90 discs which are numbered from 1 -90 . if one disc is shown at random from the box, find the probability that it bears
(1) a two-digit number
sol) The number of outcomes = 90
favourable outcomes for 2-digit numbers
E = 10,11,12.......90
Number of favourable outcomes n(E) = 81
P(E) =
90 10
(2) a perfect square number
sol) favourable outcomes for perfect square
n(S) = 9
P(S) = 9 = 1
90 10
(3) a number divisible by 5.
sol) Favourable outcomes for a number divisible by "5"
F = 5,10,15,20,,.......90
No.of favourable outcomes n(F) = 18
P(F) = 18 = 1
90 5
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(11) 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?
sol) we know that
Given :- diameter = 1 m
radius = diameter = 0.5 m
2
We have
Area of circle = Pi * r^2
=> 3.14 * (0.5)^2
=> 3.14 * 0.25
=> 0.785 m^2
Total area of the rectangle = 3* 2 = 6 m^2
Now,
Probability that it will land inside the circle :-
P(E) = Area of circle
Total Area of rectangle
=> 0.785
6
=> 0.13
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(12) 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
(1) Probability that She will by it?
sol) The number of outcomes = 144
The number of good pens = 144 - 20 = 124
Let "B" denote that the event "She will buy"
P(B) = 124
144
=> 31 is the probability that she will buy
36
(2) Probability that She will not buy it?
sol) Let "N" denote the event " she will not buy"(If its defective)
P(N) = 20 = 5
144 36
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(13) Two dice are rolled simultaneously and counts are added
(1) complete the table given below:
sol) Two dice are rolled for count 3
=> (first dice) 1 + (second dice) 2 = 3
=> (first dice) 2 + (second dice) 1 = 3
Total 2 times
Two dice are rolled for count 3 :-
=> 2 / 36 = 1 /18 -------(1)
2) Two dice are rolled for count 4
=> 1 + 3 = 4
=> 2+2 = 4
=> 3+ 1 = 4
Total 3 times
=> 3 = 1 ----------(2)
36 12
3) Two dice are rolled for count 5
=> 1 + 4 = 5
=> 2 + 3= 5 1
=> 3 + 2 = 5
=> 4 + 1 = 5
Total 4 times
=> 4 = 1 ------(3)
36 9
Similarly
4) Two dice are rolled for 6 = 5 / 36
5) Two dice are rolled for 7 = 6 / 36 = 1 / 6
6) Two dice are rolled for 9 = 4 / 36 = 1 / 9
7) Two dice are rolled for `10 = 3 / 36 = 1 / 12
8) Two dice are rolled for 11 = 2 / 36 = 1 / 18
(2) 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 answers.
sol) It can be seen that each term does not have a probability of 1 / 11
.^. , it is an incorrect argument.
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(14) A game consists of tossing a one rupee coin 3 times and nothing its outcomes each time. Hanif wins if all the tosses given the same result i.e., three heads or three jails, and loses otherwise. Calculate the probability that Hanif will lose the game.
sol) Possible outcomes :-
{H, H,H}
{H,H,T}
{H,T,H}
{T,H,H}
{T,T,H}
{T,H,T}
{H,T,T}
{T,T,T}
Total possible outcomes = 8
No. of outcomes where Hanif loses the game = 8 - 2 = 6
P(Hanif will lose the game) = No.of outcomes where Hanif loses
Total outcomes
=> 6
8
=> 3
4
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(15) A die is thrown twice. What is the probability that
{Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
(1) 5 will not come up either time?
sol) Total . no of outcomes = 36
Total outcomes where 5 comes up
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (
(5,1) (5,2) (5,3) (5,4) (5,6) = 11 outcomes
Total outcomes where 5 will not come up = 36 - 11 = 25
Probability that 5 will not come up either time
=> Total outcomes where 5 will not come up
Total number of outcomes
=> 25
36
(2) 5 will come up at least once?
sol) Total no.of outcomes = 36
Total outcomes where 5 comes up atleast once = 11
Probability that 5 will come up atleast once
=> Total outcomes where 5 will come up
Total number of outcomes
=> 11
36
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(*) Optional Exercise
1) Two customers Shyam and Ekta are visiting a particular shop in the same week(Tuesday to Saturday).
Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(!) same day
(!!) consecutive days?
(!!!) different days?
sol) Shyam and Ekta can visit the shop on Tuesday, Wednesday, Thursday, Friday & saturday.
Let
Tu= Tuesday
We=Wednesday
Th= Thursday
Fr=Friday
Sa = Saturday.
Then, total possible outcomes are
(Tu, Tu) (Tu, We) (Tu, Th) (Tu, Fr) (Tu,Sa)
(We,Tu) (We,We) (We,Th) (We,Fr) (We,Sa)
(Th,Tu) (Th,We) (Th,Th) (Th,Fr) (Th,Sa)
(Fr,Tu) (Fr,We) (Fr,Th) (Fr,Fr) (Fr,Sa)
(Sa,Tu) (Sa,We) (Sa,Th) (Sa,Fr) (Sa,Sa)
Total number of outcomes = 25
(!) If Shyam and Ekta visit the shop on same day.
sol) Favourable outcomes :-
(Tu,Tu)
(We,We)
(Th,Th)
(Fr,Fr)
(Sa, Sa)
5 outcomes
probability = No.of favourable outcomes
No. of possible outcomes
= 5/25
= 1 /5
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(!!) If Shyam and EKta visit the shop on consecutive days.
sol) Favourable Outcomes :-
(Tu, We)
(We, tu)
(We,Th)
(Th, We)
(Th, Fr)
(Fr,Th)
(Fr, Sa)
(Sa, Fr)
8 outcomes
Probability = No. of favourable outcomes
No. of possible outcomes
= 8/ 25
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(!!!) If Shyam and EKta visit the shop on different days,
Sol) We have
p(E) + P(E) = 1
p(Shyam and Ekta visit on same day) +
P(Shyam and Ekta visit on different days) = 1
5/25 + P (shyam and ekta visit on different days) = 1
P(Shyam and Ekta visit on different days) =
= 1 - 5/25
= 1 - 1/5
= 4/5
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(2) A bag contains 5 red balls and some blue balls. if the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
sol) Given : bag contain red balls= 5
bag contain blue balls = x
Total Number of balls in a bag = x +5
Given :- The probability of drawing a blue ball P(B) is double of a red ball P(R)
P(B) = 2 * P(R)
No. of blue balls = 2 * No. of red balls
Total no. of balls total no. of balls
x = 2 * 5
x+5 x+5
x = 2 * 5
x = 10 are blue balls in the bag.
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(3) A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
sol) Given : Total ball in box = 12
Black balls in box = x
Probability of drawing black ball
P(B) = x
12
Given : If 6 more black balls added in a box
Total Number of balls = 12 + 6 = 18
Number of black balls = x + 6
Given : After adding 6 balls probability of drawing black ball is double
Probability after adding = 2 * probability before
x + 6 = 2 * x
18 12
x + 6 = x
18 6
6 [ x + 6 ] = 18x
6x + 36 = 18x
12x = 36
x =
.^. x = 3
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(4) A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
sol) Given : Total no. of marbles = 24
let green marbles = x
Then, blue marbles = 24 - x
Given, probability of drawing a green marbles = 2/3
No.of green marble = 2
Total no.of marbles 3
x = 2
24 3
x =
3
We got No.of green marbles = 16
Now, blue marbles = 24-x
=> 24 - 16
=> 8
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