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Home » A sample space consists of 9 elementary outcomes whose probabilities are Suppose (a) List the composition of the event A U B, and calculate P (A U B) by adding the probabilities of the elementary outcomes. (b) Calculate P (B) from P (B), also calculate P (B directly from the elementary outcomes of .
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) List the composition of the event A U B, and calculate P (A U B) by adding the probabilities of the elementary outcomes.
(b) Calculate P (B) from P (B), also calculate P (B ) directly from the elementary outcomes of \overline{\mathrm{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) List the composition of the event A U B, and calculate P (A U B) by adding the probabilities of the elementary outcomes.
(b) Calculate P (B) from P (B), also calculate P (B ) directly from the elementary outcomes of \overline{\mathrm{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) A=\left\{e_{1}, e_{5}, e_{8}\right\} and B=\left\{e_{2}, e_{5}, e_{8}, e_{g}\right\}
\therefore A \cup B=\left\{e_{1}, e_{2}, e_{5}, e_{8}, e_{9}\right\}
\Rightarrow P(A \cup B)=P\left(e_{1}\right)+P\left(e_{2}\right)+P\left(e_{5}\right)+P\left(e_{8}\right)+P\left(e_{9}\right)
Substituting the values,
\begin{array}{l} =0.08+0.08+0.1+0.07+0.07 \\ =0.40 \end{array}

(b) We need to find \mathrm{P}(\overline{\mathrm{B}})
Using the Complement Rule, we have
\mathbf{P}(\overline{\mathbf{B}})=\mathbf{1}-\mathbf{P}(\mathbf{B})
\Rightarrow \mathrm{P}(\overline{\mathrm{B}})=1-0.32
=0.68
On substituting the given values, we obtain
=0.08+0.1+0.1+0.2+0.2
=0.68

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