In this question paper, it covers the important questions from the chapter Probability.

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Questions:

Generic 2

15 x 2 = 30

One natural number is selected from the first 20 natural numbers. What is the probability that it is either a prime number or an odd number?

In a single throw of two die, what is the probability that the sum of the numbers on the two faces is either 9 or 11.

In a single throw of two die, find the probability of getting the sum neither 9 nor 11.

A coin is tossed twice, what is the probability that atleast one tail occurs?

Two die are thrown. The events A, B, C are as follows:
getting an odd number on the first die.

A die is thrown twice. Each time the number appearing on it is recorded. Describe the events.
A: both numbers are odd
B: both numbers are even
C: sum of the numbers is less than 6
Also find A u B, A n B, A u C and A n C. Which pairs of events are mutually exclusive?

Abox contains 5 slips marked 1, 2, 3,4 and 5.Write the sample space of the experiment when
(i) one slip is drawn
(ii) one slip is drawn and not replaced then a second slip is drawn.

A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment.

In a lottery, a person choose 4 different natural numbers at random from 1 to 30 and there four numbers match with the six numbers already fixed by the lottery committee, he wins the prize.What is the probability of winning the prize in the game?

10.

Find the probability that in a random arrangement of the letters of the word 'SOCIAL' vowels come together.

11.

If A and B are two events associated with a random experiment such that P(A)=0.3.P(B)=0.2 and \(P(A\cap B)=0.1\) ,then find the value of \(P(A\cap B)\)

12.

How many are two-digital positive integers multiple of 3?

13.

Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.

14.

A and B are two events that P(A) = 0.54, P(B) = 0.69 and \(P(A\cap B)\) = 0.35. Find \(P(B\cap A')\)

15.

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children.
List the eight elements in the sample space whose outcomes are all possible genders of the three children.

Generic 3

5 x 3 = 15

16.

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e. from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

17.

From the employees of a company, 5persons are selected to represents them in the managing committee of the company's particulars of five persons are as follows

S.no	Name	Sex	Age in year
1 2 3 4 5	Harish Rohan Sheetal Alice Salim	M M F F M	30 33 46 28 41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

18.

Fill in the blanks in following table:
\(P(A)\quad P(B)\quad P(A\cap B)\quad P(A\cup B)\)
\( i)\quad \frac { 1 }{ 3 } \quad \frac { 1 }{ 5 } \quad \frac { 1 }{ 15 } \quad ...\)
\(ii)\quad 0.35\quad ...\quad ...\quad 0.25\quad 0.6\)
\(iii)\quad 0.5\quad 0.35\quad ...\quad 0.7\)

19.

A bag contains 6 discs of which 4 red,3 are blue and 2 are yellow.The discs are similar in shape and size.A disc is drawn at random from the bag.Calculate the probability that it will be
either red or blue

20.

A bag contains 6 discs of which 4 red,3 are blue and 2 are yellow.The discs are similar in shape and size.A disc is drawn at random from the bag.Calculate the probability that it will be
blue

Generic 5

1 x 5 = 5

21.

A die is thrown, find the probability of following events:
(i) A prime number will appear.
(ii) A number greater than or equal to 3 will appear.
(iii) A number less than or equal to one will appear.
(iv) A number more than 6 will appear.
(v) A number less than 6 will appear.

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Answers Key:

Generic 2

15 x 2 = 30

\(\frac { 11 }{ 20 } \)

\(\frac { 1 }{ 6 } \)

\(\frac { 5 }{ 6 } \)

Here S = {HH, HT, TH, TT}
\(\therefore \) number of possible out comes n(S) = 4
Let E be the event of getting at least one tail
\(\therefore \) n(E) =3
\(\therefore \) Probability of getting at least one tail
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 3 }{ 4 } \)

A = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2),(5, 3), (5, 4), (5, 5), (5, 6)}

A = {(I. 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
B = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
C = {(I, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, I)}
\(A\cup B\)= {(I, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2) (4, 4), (4, 6), (5, 1),
(5,,3), (5, 5), (6, 2), (6, 4), (6, 6)}
\(A\cap B=\phi \)
\(A\cup C\) = {(I, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 5), (4, 1),
(5, 1), (5, 3), (5, 5)}
\(A\cap C\) = {(I, 1), (1, 3), (3, I)}
A and B, B and C are mutually exclusive events.

(i) {1,2,3,4,5}
(ii) {(1,2),(1,3),1,4),(1,5),(2,1),(2,3),(2,4),(2,5),(3,1),(3,2),(3,4),(3,5),(4,1),(4,2),(4,3),(4,5),(5,1),(5,2),(5,3),(5,4)}

The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R₁, R₂ and the 3 black balls as B₁, B₂, and B₃.
The sample space of this experiment is given by
S = {TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}

\(\frac { 1 }{ ^{ 30 }{ C }_{ 4 } } \)

10.

Total outcomes = 6!
Favourable outcomes = 4! x 3!
\(\frac { 1 }{ 5 } \)

11.

0.2

12.

13.

\(We\ have,\ P(\overline { E } \cup \overline { F } )=0.25\)
\(\Rightarrow P(\overline { E\cap F) } =0.25\)
\(\Rightarrow 1-P(E\cap F)=0.25\)
\( \Rightarrow P(E\cap F)=1-0.25=0.75\neq 0\)
\(\therefore \) E and F are not mutually exclusive events.

14.

\(P(B\cap A')=P(B)-P(A\cap B)=0.69-0.35=0.34\)

15.

Let B denotes a boy and G denotes a girl. Then, all possible genders are expressed as
S={BBB,BBG,BGB,GBB,BGG,GBG,GGB,GGG}

Generic 3

5 x 3 = 15

16.

There are total 10 digits from 0 to 9. Since the digits cannot be repeated.
So the first place may filled in 10 ways, second place in 9 ways, third place in 8 ways and fourth place in 7 ways.
∴ Number of possible outcomes =10×9×8×7=5040
The lock of suitcase can be opened in 1 way only
∴ Number of favourable cases = 1
Thus required probability \(=\frac { 1 }{ 5040 } \)

17.

Here total number of persons = 5 one spokesperson is selected out of 5 persons in \(^{ 5 }C_{ 1 }\)=ways
Let A be the event that person is male and Bbe the event that person is over 35 years. There are 3male and one person can be selected in \(^{ 3 }C_{ 1 }\)ways
\(\therefore \quad P(A)=\frac { ^{ 3 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 3 }{ 5 } \)
There are 2 person who are over 35 years. So one person can be selected in \(^{ 2 }C_{ 1 }\) ways
\(\therefore \ P(B)=\frac { ^{ 2 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 2 }{ 5 } \)
There is 1 person who is male and over 35 years.
\(\therefore P(A\cap B)=\frac { ^{ 1 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 1 }{ 5 } \)
\(\text{Now P(either male or over 35 years)}=P(A\cup B)\)
\(=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 3 }{ 5 } +\frac { 2 }{ 5 } -\frac { 1 }{ 5 } =\frac { 3+2-1 }{ 5 } =\frac { 4 }{ 5 } \)

18.

\(\text{i) here} P(A)=\frac { 1 }{ 3 } ,P(B)=\frac { 1 }{ 5 }\)
\(\text{and }P(A\cap B)=\frac { 1 }{ 15 } \)
we know that
\( P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 1 }{ 3 } +\frac { 1 }{ 5 } -\frac { 1 }{ 15 } \)
\(=\frac { 5+3-1 }{ 15 } =\frac { 7 }{ 15 } \)
\(\text{Here } P(A)=0.35,(P(A\cap B)=0.25\)
\(\text{and } P(A\cup B)=0.6\)
\(we\quad know\quad that\)
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\( \therefore \ 0.6=0.35+P(B)-0.25\)
\(\Rightarrow \ 0.6=0.1+P(B)\)
\(\Rightarrow \ P(B)=0.6-0.1=0.5\)
\(iii)\quad Here\quad P(A)=0.5,P(B)=0.35\)
\(and\quad P(A\cup B)=0.7\)
We know that
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\therefore \quad 0.7=0.5+0.35-P(A\cap B)\)
\(\Rightarrow \ 0.7=0.85-P(A\cap B)\)
\(\Rightarrow P(A\cap B)=0.85-0.7=0.15\)

19.

\(\frac { 7 }{ 9 } \)

20.

\(\frac { 1 }{ 3 } \)

Generic 5

1 x 5 = 5

21.

Here the sample space S = {I, 2, 3,4,5,6}
\(\therefore \) n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5} \(\Rightarrow \) n(A) = 3
\(Thus\ P(A)=\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event of getting a number greater than or equal to 3
B = {3, 4, 5, 6} \(\Rightarrow \) n(B) = 4
\(Thus\ P(B)=\frac { n(B) }{ n(S) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Let C be the event of getting a number less than or equal to 1
C = {I} \(\Rightarrow \) n(C) = 1
\(Thus\ P(C)=\frac { n(C) }{ n(S) } =\frac { 1 }{ 6 } \)
(iv) Let D be the event of getting a number more than 6
\(D=\phi \Rightarrow n(D)=0\)
\(Thus\ P(D)=\frac { n(D) }{ n(S) } \frac { 0 }{ 6 } =0\)
Let E be the event of getting a number less than 6
E = {I, 2, 3, 4, 5} \(\Rightarrow \) n(E) = 5
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)