# A sample space consists of 9 elementary

Question:

A sample space consists of 9 elementary outcomes $e_{1}, e_{2}, \ldots, e_{9}$ whose probabilities are

$P\left(e_{1}\right)=P\left(e_{2}\right)=.08, P\left(e_{3}\right)=P\left(e_{4}\right)=P\left(e_{5}\right)=.1$

$P\left(e_{6}\right)=P\left(e_{7}\right)=.2, P\left(e_{8}\right)=P\left(e_{9}\right)=.07$

Suppose $A=\left\{e_{1}, e_{5}, e_{8}\right\}, B=\left\{e_{2}, e_{5}, e_{8}, e_{9}\right\}$

(a) Calculate $P(A), P(B)$, and $P(A \cap B)$

(b) Using the addition law of probability, calculate P (A U B)

(c) List the composition of the event $A \cup B$, and calculate $P(A \cup B)$ by adding the probabilities of the elementary outcomes.

(d) Calculate $P(\bar{B})$ from $P(B)$, also calculate $P(\bar{B})$ directly from the

elementary outcomes of $\overline{\mathrm{B}}$.

Solution:

Given

S = {e1, e2, e3, e4, e5, e6, e7, e8, e9}

A = {e1, e5, e8} and B = {e2, e5, e8, e9}

P(e1) = P(e2) = .08, P(e3) = P(e4) = P(e5) = .1

P(e6) = P(e7) = .2, P(e8) = P(e9) = .07

(a) To find P (A), P (B) and P (A ⋂ B)

A = {e1, e5, e8}

P (A) = P (e1) + P (e5) + P (e8)

Substituting the values, we get

⇒ P (A) = 0.08 + 0.1 + 0.07

⇒ P (A) = 0.25

B = {e2, e5, e8, e9}

P (B) = P (e2) + P (e5) + P (e8) + P (e9)

Substituting the values, we get

⇒ P (B) = 0.08 + 0.1 + 0.07 + 0.07 [given]

⇒ P (B) = 0.32

Now, we have to find P (A ⋂ B)

A = {e1, e5, e8} and B = {e2, e5, e8, e9}

∴ A ⋂ B = {e5, e8}

⇒ P (A ⋂ B) = P (e5) + P (e8)

= 0.1 + 0.07

= 0.17

(b) To find P (A ⋃ B)

P (A ⋃ B) = P (A) + P (B) – P (A ⋂ B)

from part (a), we have

P (A) = 0.25, P (B) = 0.32 and P (A ⋂ B) = 0.17

Putting the values, we get

P (A ⋃ B) = 0.25 + 0.32 – 0.17

= 0.40

(c) A = {e1, e5, e8} and B = {e2, e5, e8, e9}

∴ A ⋃ B = {e1, e2, e5, e8, e9}

⇒ P (A ⋃ B) = P (e1) + P (e2) + P (e5) + P (e8) + P (e9)

Substituting the values, we get

= 0.08 +0.08 + 0.1 + 0.07 + 0.07

= 0.40

(d) To find $\mathrm{P}(\overline{\mathrm{B}})$

By Complement Rule, we have

$\mathbf{P}(\overline{\mathbf{B}})=\mathbf{1}-\mathbf{P}(\mathbf{B})$

$\Rightarrow \mathrm{P}(\overline{\mathrm{B}})=1-0.32$

$=0.68$

Given: $B=\left\{e_{2}, e_{5}, e_{8}, e_{9}\right\}$

$\therefore \overline{\mathrm{B}}=\left\{\mathrm{e}_{1}, \mathrm{e}_{3}, \mathrm{e}_{4}, \mathrm{e}_{6}, \mathrm{e}_{7}\right\}$

$\therefore \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}\left(\mathrm{e}_{1}\right)+\mathrm{P}\left(\mathrm{e}_{3}\right)+\mathrm{P}\left(\mathrm{e}_{4}\right)+\mathrm{P}\left(\mathrm{e}_{6}\right)+\mathrm{P}\left(\mathrm{e}_{7}\right)$

By substituting the given values, we get

$=0.08+0.1+0.1+0.2+0.2$

$=0.68$