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Home » A sample space consists of 9 elementary outcomes whose probabilities are Suppose (a) Calculate , and (b) Using the addition law of probability, calculate P (A U B)
A sample space consists of 9 elementary outcomes \mathrm{e}_{1}, \mathrm{e}_{2}, \ldots, \mathrm{e}_{9} whose probabilities are \mathrm{P}\left(\mathrm{e}_{1}\right)=\mathrm{P}\left(\mathrm{e}_{2}\right)=.08, \mathrm{P}\left(\mathrm{e}_{3}\right)=\mathrm{P}\left(\mathrm{e}_{4}\right)=\mathrm{P}\left(\mathrm{e}_{5}\right)=.1 \mathrm{P}\left(\mathrm{e}_{6}\right)=\mathrm{P}\left(\mathrm{e}_{7}\right)=.2, \mathrm{P}\left(\mathrm{e}_{8}\right)=\mathrm{P}\left(\mathrm{e}_{9}\right)=.07 Suppose \mathrm{A}=\left\{\mathrm{e}_{1}, \mathrm{e}_{5}, \mathrm{e}_{8}\right\}, \mathrm{B}=\left\{\mathrm{e}_{2}, \mathrm{e}_{5}, \mathrm{e}_{8}, \mathrm{e}_{9}\right\}
(a) Calculate \mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B}), and \mathrm{P}(\mathrm{A} \cap \mathrm{B})
(b) Using the addition law of probability, calculate P (A U B)
CBSE | Class 11 | Maths | NCERT Exemplar | Probability
A sample space consists of 9 elementary outcomes \mathrm{e}_{1}, \mathrm{e}_{2}, \ldots, \mathrm{e}_{9} whose probabilities are \mathrm{P}\left(\mathrm{e}_{1}\right)=\mathrm{P}\left(\mathrm{e}_{2}\right)=.08, \mathrm{P}\left(\mathrm{e}_{3}\right)=\mathrm{P}\left(\mathrm{e}_{4}\right)=\mathrm{P}\left(\mathrm{e}_{5}\right)=.1 \mathrm{P}\left(\mathrm{e}_{6}\right)=\mathrm{P}\left(\mathrm{e}_{7}\right)=.2, \mathrm{P}\left(\mathrm{e}_{8}\right)=\mathrm{P}\left(\mathrm{e}_{9}\right)=.07 Suppose \mathrm{A}=\left\{\mathrm{e}_{1}, \mathrm{e}_{5}, \mathrm{e}_{8}\right\}, \mathrm{B}=\left\{\mathrm{e}_{2}, \mathrm{e}_{5}, \mathrm{e}_{8}, \mathrm{e}_{9}\right\}
(a) Calculate \mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B}), and \mathrm{P}(\mathrm{A} \cap \mathrm{B})
(b) Using the addition law of probability, calculate P (A U B)

Solution:
It is given that
\begin{array}{l} S=\left\{e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{8}, e_{7}, e_{8}, e_{9}\right\} \\ A=\left\{e_{1}, e_{5}, e_{8}\right\} \text { and } B=\left\{e_{2}, e_{5}, e_{8}, e_{9}\right\} \\ P\left(e_{1}\right)=P\left(e_{2}\right)=.08, P\left(e_{3}\right)=P\left(e_{4}\right)=P\left(e_{5}\right)=.1 \\ P\left(e_{6}\right)=P\left(e_{7}\right)=.2, P\left(e_{8}\right)=P\left(e_{9}\right)=.07 \end{array}

(a) We need to find \mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B}) and \mathrm{P}(\mathrm{A} \cap \mathrm{B})
A=\{{{e_1}, {e_5}, {e_8}}\}
P(A)=P(e_1)+P(e_5)+P(e_8)
Substitute the values, now
\Rightarrow P(A)=0.08+0.1+0.07
\Rightarrow P(A)=0.25
\begin{array}{l} B=\left\{e_{2}, e_{5}, e_{8}, e_{g}\right\} \\ P(B)=P\left(e_{2}\right)+P\left(e_{5}\right)+P\left(e_{8}\right)+P\left(e_{9}\right) \end{array}
Substituting the values,
\begin{array}{l} \Rightarrow P(B)=0.08+0.1+0.07+0.07 \text { [given }] \\ \Rightarrow P(B)=0.32 \end{array}
We now need to find P(A \cap B)
\begin{array}{l} A=\left\{e_{1}, e_{5}, e_{8}\right\} \text { and } B=\left\{e_{2}, e_{5}, e_{8}, e_{9}\right\} \\ \therefore A \cap B=\left\{e_{5}, e_{8}\right\} \\ \Rightarrow P(A \cap B)=P\left(e_{5}\right)+P\left(e_{8}\right) \\ =0.1+0.07 \\ =0.17 \end{array}

(b) We need to find P(A \cup B)
Using the General Addition Rule:
P(A \cup B)=P(A)+P(B)-P(A \cap B)
From the part (a), we have
P(A)=0.25, P(B)=0.32 \text { and } P(A \cap B)=0.17
Putting the values,
\begin{array}{l} P(A \cup B)=0.25+0.32-0.17 \\ =0.40 \end{array}

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