Chapter 16: Probability

Q41Written answer

The probability of happening of an event\[A\] is \[0.5\] and that of \[B\] is \[0.3\]. If \[A\] and \[B\] are mutually exclusive events, then the probability of neither \[A\] nor \[B\] is ________.

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Worked solution

Ans: Given: Probability of happening of an event\[A\] is \[0.5\] and that of \[B\] is \[0.3\].

Given that,

\[{\text{P}}({\text{A}}) = 0.5,\]

\[{\text{P}}({\text{B}}) = 0.3\]

and \[{\text{P}}({\text{A}} \cap {\text{B}}) = 0\]

\[\because {\text{A}}\] and \[{\text{B}}\] are mutually exclusive events

\[{\text{P}}(\overline{{A}} \cap \overline{{B}} ) = {\text{P}}(\overline{{{A}} \cup {\text{B}}} )\]

\[ = 1 - P(A \cup B)\]

\[ = 1 - [P(A) + P(B)]\]

Substitute the values,

\[{\text{P}}(\overline {\text{A}} \cap \overline {\text{B}} )\]\[ = 1 - (0.5 + 0.3)\]

= 1 - 0.8

= 0.2

Hence, the value of the filler is \[0.2.\]

The probability of happening of an event \[A\] is \[0.5\] and that of \[B\] is \[0.3\]. If A and B are mutually exclusive events, then the probability of neither A nor B is \[0.2.\]

Q42Written answer

Match the proposed probability under Column \[{C_1}\] with the appropriate written description under column \[{C_2}\].

\[{C_1}\] Probability	\[{C_2}\] Written description
\[\left( a \right) 0.95\]	(i) An incorrect assignment
\[\left( b \right) 0.02\]	(ii) No chance of happening
\[\left( c \right) - 0.3\]	(iii) As much chance of happening as not
\[\left( d \right) 0.5\]	(iv) Very likely to happen
\[\left( e \right) 0\]	(v) Very little chance of happening

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Worked solution

Ans: \[0.95 = \] Very likely to happen, so it is close to \[1{\text{ }}.\]

\[0.02 = \] Very little chance of happening as the probability is very low.

\[ - 0.3 = \] an incorrect assignment because probability is never negative.

\[0.5 = \] as much chance of happening as not because sum of chances of happening and not happening is one.

\[0 = \] no chance of happening.

Hence,

(a)$\leftrightarrow (iv),$

(b) $\leftrightarrow (v),$

(c) $\leftrightarrow (i),$

(d) $\leftrightarrow (iii),$

(e) $\leftrightarrow (ii)$

Q43Written answer

Match the following

(a) If \[{E_1}\] and \[{E_2}\] are the two mutually exclusive events	(i) \[{E_1}\cap {E_2} = {E_1}\]
(b) If \[{E_1}\] and \[{E_2}\] are the mutually exclusive and exhaustive events	(ii) \[\left( {{E_1} - {E_2} } \right)\cup \left( {{E_1} \cup {E_2}} \right) = {E_1}\]
(c) If \[{E_1}\] and \[{E_2}\] have common outcomes, then	(iii) \[{E_1} \cap {E_2} = \varphi , {E_1}\cup {E_2} = S\]
(d) If \[{E_1}\] and \[{E_2}\] are two events such that \[{E_1} \subset {E_2}\]	(iv) \[{E_1} \cap {E_2} = \varphi \]

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Worked solution

Ans: (a) If \[{{\text{E}}_1}\] and \[{{\text{E}}_2}\] are mutually exclusive events, then \[{{\text{E}}_1} \cap {{\text{E}}_2} = \phi .\]

(b) If \[{{\text{E}}_1}\] and \[{{\text{E}}_2}\] are mutually exclusive and exhaustive events then \[{{\text{E}}_1} \cap {{\text{E}}_2} = \phi \] and \[{{\text{E}}_1} \cup {{\text{E}}_2} = {\text{S}}\]

(c) If \[{{\text{E}}_1}\] and \[{{\text{E}}_2}\] have common outcomes, then \[\left( {{E_1} - {E_2}} \right) \cup \left( {{E_1} \cap {E_2}} \right) = {E_1}\]

(d) If \[{E_1}\] and \[{E_2}\] are two events such that \[{{\text{E}}_1} \subset {{\text{E}}_2} \Rightarrow {{\text{E}}_1} \cap {{\text{E}}_2} = {{\text{E}}_1}\]

Hence,

\[{\text{}}(a) \leftrightarrow (iv),(b) \leftrightarrow (iii),(c) \leftrightarrow (ii),(d) \leftrightarrow (i)\] .

The Important Topics For Chapter 16 Probability Class 11 Maths

The important topics are described as the following:

Random experiment
Outcome
Sample Space
Event
Types of events
Impossible and Sure Events
Simple or Elementary Event
Compound Event
Mutually exclusive event
Exhaustive events
An axiomatic approach to probability
Probabilities of equally likely outcomes
Addition rule of probability
Addition rule for a mutually exclusive event

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