Chapter 14: Probability

Ans: 1,2,3,4,5 and 6 are the possible outcomes when the die is thrown.

Therefore, S = {1,2,3,4,5,6} 

According to the question,

E is event die shows 4

\[E = \left( 4 \right)\]

F is die shows even number

\[F = \left( {2,4,6} \right)\]

\[E \cap F = \left( 4 \right) \cap \left( {2,4,6} \right)\]

\[E \cap F = 4\]

Therefore, we can conclude that E and F are not mutually exclusive event.

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

A: a number less than 7

All the numbers in the die are less than 7,

\[A = \left( {1,2,3,4,5,6} \right)\]

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

B: a number greater than 7

There is no number on dice greater than 7.

\[B = \phi \]

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

C : a number multiple of 3

Multiple of 3 between 1 to 6 is 3 and 6

Therefore,

\[C = \left( {3,6} \right)\]

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

D: a number less than 4

\[D = \left( {1,2,3} \right)\]

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

E: an even number greater than 4

There are only one number which is even greater than 4.

Therefore,

\[E = \left( 6 \right)\]

Ans: Let us 1,2,3,4,5 and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

F: a number not less than 3

\[F = \left( {3,4,5,6} \right)\]

Also find \[A \cup B,{\text{ }}A\; \cap B,{\text{ }}B \cup C,{\text{ }}E\; \cap {\text{ }}F,{\text{ }}D\; \cap {\text{ }}E,{\text{ }}D--E,{\text{ }}A--C,{\text{ }}E\; \cap {\text{ }}F',{\text{ }}F'\]

Ans: As we have to find, \[{\text{ }}A \cup B,{\text{ }}A\; \cap B,{\text{ }}B \cup C,{\text{ }}E\; \cap {\text{ }}F,{\text{ }}D\; \cap {\text{ }}E,{\text{ }}D--E,{\text{ }}A--C,{\text{ }}E\; \cap {\text{ }}F',{\text{ }}F'\]

Therefore,

\[A\; \cap B{\text{ }} = \;\left( {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right)\; \cap \;\left( \phi  \right)\]

\[A\; \cap B = {\text{ }}\left( \phi  \right)\]

\[B \cup C{\text{ }} = \;\left( \phi  \right) \cup \;\left( {3,{\text{ }}6} \right)\]

\[B \cup C = \left( {3,6} \right)\]

\[E\; \cap F = \;\left( 6 \right)\; \cap \;\left( {3,4,5,6} \right)\]

\[E\; \cap F = \left( 6 \right)\]

\[D\; \cap {\text{ }}E{\text{ }} = \;\left( {1,2,3} \right)\; \cap \;\left( 6 \right)\]

\[D\; \cap {\text{ }}E = \left( \phi  \right)\]

\[D--E{\text{ }} = \;\left( {1,2,3} \right)--\left( 6 \right)\]

\[D--E = \;\left( {1,2,3} \right)\]

\[A--C = \left( {1,2,3,4,5,6} \right)--\;\;\left( {3,6} \right)\]

\[A--C = \left( {1,2,4,5} \right)\]

\[F' = S--F\]

\[F' = \left( {1,2,3,4,5,6} \right)--\left( {3,4,5,6} \right)\]

\[F' = \left( {1,2} \right)\]

\[E\; \cap F' = \;\left( 6 \right)\; \cap \;\left( {1,2} \right)\]

\[E\; \cap F' = \;\left( \phi  \right)\]

Ans: Let \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

As in this question pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: The sum is greater than 8

$A=   (3,6) ,( 4,5),(4,6),( 5,4),(5,5) \\$ 

       $(5,6) ,( 6,3),(6,4),(6,5),(6,6) \\$ 

B: 2 occur on the die

$B=   (2,1) ,( 2,2),(2,3),( 2,4),(2,5) \\$ 

       $(1,2) ,( 3,2),(4,2),(5,2),(6,2),(2,6) \\$ 

C: the sum is at least 7 and multiple of 3

\[C =  \left\{ {\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right),\left( {6,6,} \right)}   \right\}\]

Now, we can find pairs of the events which are mutually exclusive or not.

(i) \[A \cap \;B = \phi \]

As, there is no common element in A and B

Therefore, A and B are mutually exclusive

(ii) \[B\; \cap \;C = \;\phi \]

Since there is no common element between

Therefore, B and C are mutually exclusive.

(iii) \[A\; \cap \;C =  \left\{ {\left( {3,6} \right),{\text{ }}\left( {4,5} \right),{\text{ }}\left( {5,4} \right),{\text{ }}\left( {6,3} \right),{\text{ }}\left( {6,6} \right)}   \right\}\]

\[A\; \cap \;C\; \ne \phi \]

As, A and C have common elements.

Therefore A and C are not mutually exclusive.

Ans: As three coins are tossed once so the possible sample space will be,

\[S{\text{ }} =  \left\{ HHH,HHT,HTH,THH, HTT,THT,TTH,TTH   \right\}\]

According to question,

A: ‘three heads’

\[A = \left( {HHH} \right)\]

B: two heads and one tail

\[B = \left( {HHT,THH,HTH} \right)\]

C :three tails

\[C = \left( {TTT} \right)\]

D: a head shows on the first coin

\[D = \left( {HHH,HHT,HTH,HTT} \right)\]

(i) Mutually Exclusive

\[A\; \cap \;B = \left( {HHH} \right)\; \cap \;\left( {HHT,THH,HTH} \right)\]

\[A\; \cap \;B = \phi \]

Therefore, A and B are mutually exclusive.

\[A \cap \;C = \left( {HHH} \right)\; \cap \;\left( {TTT} \right)\]

\[A \cap \;C = \phi \]

Therefore, A and C are mutually exclusive.

\[A \cap D{\text{ }} = \left( {HHH} \right)\; \cap \;\left( {HHH,HHT,HTH,HTT} \right)\]

\[A \cap D = \left( {HHH} \right)\]

\[A \cap D{\text{ }} \ne \;\phi \]

Therefore, A and D are not mutually exclusive

\[B \cap C = \left( {HHT,HTH,THH} \right) \cap \left( {TTT} \right)\]

\[B \cap C = \phi \]

Therefore, B and C are mutually exclusive.

\[B \cap D = \left( {HHT,THH,HTH} \right) \cap \left( {HHH,HHT,HTH,\;HTT} \right)\]

\[B \cap D = \left( {HHT,HTH} \right)\]

\[B \cap D{\text{ }} \ne \;\phi \]

Therefore, B and D are not mutually exclusive.

\[C \cap D = \left( {TTT} \right)\; \cap \;\left( {HHH,HHT,HTH,\;HTT} \right)\]

\[C \cap D = \;\phi \]

Therefore, C and D are mutually exclusive.

(ii) Simple Event

If an event has only one outcome then it is called a simple event.

\[A = \left( {HHH} \right)\]

\[C{\text{ }} = {\text{ }}\left( {TTT} \right)\]

Both A and C have only one element,

Therefore, they are simple events.

(iii) Compound Events

If an event has more than one outcome then it is called a Compound event

\[B = \left( {HHT,HTH,THH} \right)\]

\[D = \left( {HHH,HHT,HTH,\;HTT} \right)\]

Both B and D have more than one element,

Therefore, they are compound events.

Solution coming soon.

Ans: Let 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question is given that pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: Getting an even number on first die

$A =   (2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

       $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

B: Getting odd number on first die

$B =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$

C: Getting the sum of the number on dice \[ \leqslant 5\]

$C =   (1,1) ,( 1,2),(1,3),( 1,4),(2,1)\\$ 

      $ (2,2),(2,3),(3,1),(3,2),(4,1) \\$

Ans: $A = (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

Ans: $B=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

       $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

Ans: \[A{\text{ or }}B\left( {A \cup B} \right)\] =$ (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) \\$

Ans: \[A{\text{ and }}B\left( {A \cap B} \right) = \phi \]

Ans: \[A{\text{ but not }}C\left( {A - C} \right)\] =$    (2,4),(2,5),(2,6),(4,2),(4,3),(4,2),(4,5) \\$

 $ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

Ans: \[B{\text{ or }}C\left( {B \cup C} \right)\] =$ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\$

 $(2,1),(2,2),(2,3),(3,1),(3,2),(3,3) \\$

 $(3,4),(3,5),(3,6),(4,1),(5,1),(5,2) \\$

 $(5,3),(5,4),(5,5),(5,6)\\$

Ans: \[B{\text{ and }}C\left( {B \cap C} \right) =  \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {3,1} \right),\left( {3,2} \right)}   \right\}\]

Ans: \[A \cap B' \cap C' = A \cap A \cap C'\]

\[A \cap A \cap C' = A \cap C'\]

\[A \cap C' =  \left\{  \left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right)\\   \left( {4,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)     \right\}\]

Ans: Let 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question is given that pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: Getting an even number on first die

$A=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

B: Getting odd number on first die

$B=(1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

C: Getting the sum of the number on dice \[ \leqslant 5\]

$C =   (1,1) ,( 1,2),(1,3),( 1,4),(2,1)\\$ 

      $ (2,2),(2,3),(3,1),(3,2),(4,1) \\$

Ans: \[\left( {A \cap B} \right) = \phi \]

Therefore, A & B are mutually exclusive.

Ans: \[\left( {A \cap B} \right) = \phi \]

Therefore, A & B are mutually exclusive.

\[A \cup B \]= $(1,1)(1,2)(1,3),(1,4),(1,5),(1,6)\\$

    $ (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) \\$

\[A \cup B = S\]

Therefore, A and B are exhaustive also.

Ans: $A = (2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

$B'=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

Therefore, \[A = B'\]

Ans: As we have,

\[A \cap C =  \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {4,1} \right)}   \right\}\]

\[A \cap C \ne \phi \]

Therefore, A and C are not mutually exclusive

Ans: As we have,

\[A \cap B'\; = A \cap A = A\]

\[\therefore A \cap B' \ne \phi \]

Therefore, A and \[B'\] are not mutually exclusive.

Ans: $A'=(1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

$B'=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

$C=(1,1),(1,2),(1,3),(1,4),(2,1)\\$

$(2,2),(2,3),(3,1),(3,2),(4,1)\\$

\[A' \cap B' = \phi \]

But, \[B' \cap C =  \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right)}   \right\}\]

\[B' \cap C \ne \phi \]

\[A'{\text{ and }}B'\] are mutually exclusive.

\[B'{\text{ and }}C\] are not mutually exclusive.

Therefore, \[A',B',C\] are not mutually exclusive.

\[A' \cup B' \cup C\] =  $(1,1)(1,2)(1,3),(1,4),(1,5),(1,6)\\$

    $ (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) \\$

\[A' \cup B' \cup C = S\]

Therefore, \[A',B',C\] are exhaustive.