Probability
4.0 Intersection and Union of Sets of Events
4.0 Intersection and Union of Sets of Events
Let ${E_1},{E_2},....,{E_n}$ be $n$ events, then
- $$P({E_1} \cup {E_2} \cup .... \cup {E_n})$$ represents the probability of occurrence of at least one of the events from ${E_1},{E_2},....,{E_n}$
- $$P({E_1} \cap {E_2} \cap .... \cap {E_n})$$ represents the occurrence of all the events together.
$$P({E_1} \cup {E_2} \cup .... \cup {E_n}) = \sum\limits_{i = 1}^n {P({E_i})} - \sum\limits_{1 \le i < j \le n}^n {P({E_i} \cap {E_j})} + \sum\limits_{1 \le i < j < k \le n}^n {P({E_i} \cap {E_j} \cap {E_k})} - .... + {( - 1)^n}P({E_1} \cap {E_2} \cap .... \cap {E_n})$$
Proof by induction using Sets:
Let there be two events, say $A$ and $B$,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
First $A$ occurs, then $A \cap B$ also is included. Then $B$ occurs, this will also include $A \cap B$. Thus $A \cap B$ is included twice. To remove duplication one is removed.
Let us now consider three events, $A$, $B$ and $C$.
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
First $A$ occurs, then $A \cap B$ and $A \cap C$ also is included. Then $B$ occurs, this will also include $A \cap B$ and $B \cap C$. Then $C$ occurs this will include $A \cap C$ and $B \cap C$. In each case $A \cap B \cap C$ is included twice. On removing duplicates, $A \cap B \cap C$ gets excluded completely. Hence it is added separately.
Hence for $n$ events.
Illustration 12. The probability that a person will get an award for music is ${2 \over 5}$ and the probability that he will not get an award for lyrics is ${4 \over 7}$. If the probability of getting at least one award is ${2 \over 3}$, what is the probability that he will get both the awards.
Solution: Let event $A$ be that the person gets award for music and event $B$ be getting award for lyrics.
$P(A) = {2 \over 5}\;$
$P(B) = 1 - {4 \over 7} = {3 \over 7}$
$$\begin{equation} \begin{aligned} P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ {2 \over 3} = {2 \over 5} + {3 \over 7} - P(A \cap B) \\ P(A \cap B) = {2 \over 5} + {3 \over 7} - {2 \over 3} = {{17} \over {105}} \\\end{aligned} \end{equation} $$
Illustration 13. In a hospital unit there are $8$ nurses and $5$ physicians. Out of these $7$ nurses and $3$ physicians are females. If a staff is selected, find the probability that the staff is a nurse or a male.
Solution:
| Staff | Females | Males | Total |
| Nurses | $7$ | $1$ | $8$ |
| Physicians | $3$ | $2$ | $5$ |
| Total | $10$ | $3$ | $13$ |
$$\begin{equation} \begin{aligned} P(nurse) = {8 \over {13}} \\ P(male) = {3 \over {13}} \\ P(male\;nurse) = {1 \over {13}} \\ P(nurse\;or\;male) = {8 \over {13}} + {3 \over {13}} - {1 \over {13}} \\ P(nurse\;or\;male) = {{10} \over {13}} \\\end{aligned} \end{equation} $$
Illustration 14. In a game, a player has to roll three dice, and at least one of the dice has to show $2$ in order to proceed to the next level. What is the probability of the player to proceed to the next level?
Solution:
On rolling three dice, the total possible outcomes, i.e. the size of the sample space is $216$.
Case $1$: Let us consider the player getting the number $2$.
- If he gets in the first die. There are three vacant places. The first place is always occupied by $2$, the next two positions, each can be filled in $6$ ways. Thus there are $6 \times 6 = 36$ total possible ways.
Thus, $$P(getting\;2\;in\;{1^{st}}) = {{36} \over {216}}$$
- There are three vacant places. The second place is always occupied by $2$, the next two positions, each can be filled in $6$ ways. Thus there are $6 \times 6 = 36$ total possible ways.
Thus, $$P(getting\;2\;in\;{2^{nd}}) = {{36} \over {216}}$$
- There are three vacant places. The third place is always occupied by $2$, the next two positions, each can be filled in $6$ ways. Thus there are $6 \times 6 = 36$ total possible ways.
Thus,$$P(getting\;2\;in\;{3^{rd}}) = {{36} \over {216}}$$
- When he gets in the first two places, the last place can be filled in $6$ ways.
Thus, $$P(getting\;2\;in\;{1^{st}}\;and\;{2^{nd}}) = {6 \over {216}}$$
- Similarly, when he gets in last two, or in the first and the last places.
Thus, $$P(getting\;2\;in\;{2^{nd}}\;and\;{3^{rd}}) = {6 \over {216}}$$
$$P(getting\;2\;in\;{1^{st}}\;and\;{3^{rd}}) = {6 \over {216}}$$
- When he gets $2$ in all three places, the number of outcomes possible is $1$.
$$P(getting\;2\;in\;all\;the\;three) = {1 \over {216}}$$
Thus the probability that he gets $2$ in at least one of the three dice is
$$P(getting\;2\;in\;atleast\;one) = P(getting\;2\;in\;{1^{st}}) + P(getting\;2\;in\;{2^{nd}}) + P(getting\;2\;in\;{3^{rd}}) - P(getting\;2\;in\;{1^{st}}\;and\;{2^{nd}})$$$$- P(getting\;2\;in\;{2^{nd}}\;and\;{3^{rd}}) - P(getting\;2\;in\;{1^{st}}\;and\;{3^{rd}}) + P(getting\;2\;in\;all\;the\;three)$$
$$\begin{equation} \begin{aligned} = \frac{{36}}{{216}} + \frac{{36}}{{216}} + \frac{{36}}{{216}} - \frac{6}{{216}} - \frac{6}{{216}} - \frac{6}{{216}} + \frac{1}{{216}} \\ = \frac{{108}}{{216}} - \frac{{18}}{{216}} + \frac{1}{{216}} \\ = \frac{{91}}{{216}} \\\end{aligned} \end{equation} $$
