7.1 Occurrence of at least one of the independent events

i. If $A$ and $B$ are independent, then probability of occurrence of at least one is $$P(A \cup B) = 1 - P(\overline A )P(\overline B )$$

Proof: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Since $A$ and $B$ are independent,

$$P(A \cap B) = P(A) \times P(B)$$

Therefore,

$$P(A \cup B) = P(A) + P(B) - P(A) \times P(B)$$

Adding and subtracting $1$ on the R.H.S,

$$\begin{equation} \begin{aligned} P(A \cup B) = 1 - 1 + P(A) + P(B) - P(A) \times P(B) \\ P(A \cup B) = 1 - (1 - P(A)) + P(B)(1 - P(A)) \\ P(A \cup B) = 1 - [1(1 - P(A)) - P(B)(1 - P(A))] \\ P(A \cup B) = 1 - (1 - P(A))(1 - P(B)) \\ P(A \cup B) = 1 - P(\bar A) \times P(\bar B) \\\end{aligned} \end{equation} $$

ii. Extending for $n$ independent events,

$$P({A_1} \cup {A_2} \cup {A_3} \cup ..... \cup {A_n}) = 1 - [P({{\bar A}_1})P({{\bar A}_2})P({{\bar A}_3})....P({{\bar A}_n})]$$

Illustration 31. A bag contains $5$ white, $7$ red and $8$ black balls. Four balls are drawn one by one with replacement. What is the probability that at least one is white?

Solution: Let ${A_i}$, where $1 \le i \le 4$, be the event that the ball drawn in the ${i^{th}}$ draw is white.

Since the balls are drawn one by one with replacement, the events, ${A_1}$, ${A_2}$, ${A_3}$, ${A_4}$, are independent.

Therefore,

$$P({A_i}) = {5 \over {20}} = {1 \over 4}$$

where $1 \le i \le 4$

The probability of getting at least one white ball is,

$$\begin{equation} \begin{aligned} P({A_1} \cup {A_2} \cup {A_3} \cup {A_4}) \\ = 1 - [P({{\bar A}_1})P({{\bar A}_2})P({{\bar A}_3})P({{\bar A}_4})] \\ = 1 - [(1 - P({{\bar A}_1}))(1 - P({{\bar A}_2}))(1 - P({{\bar A}_3}))(1 - P({{\bar A}_4}))] \\ = 1 - [(1 - {1 \over 4})(1 - {1 \over 4})(1 - {1 \over 4})(1 - {1 \over 4})] \\ = 1 - \left[ {{3 \over 4} \times {3 \over 4} \times {3 \over 4} \times {3 \over 4}} \right] \\ = 1 - {\left[ {{3 \over 4}} \right]^4} \\\end{aligned} \end{equation} $$

Illustration 32. The probability that a teacher will give an un-announced test during any class meeting is ${1 \over 5}$. If a student is absent twice, what is the probability that he will miss at least one test?

Solution: Let ${A_i}$ be the event that the student misses ${i^{th}}$ test, where $(i = 1,2)$. Then ${A_1}$ and ${A_2}$ are independent events. The probability that he misses the test the first time is the probability that the teacher has kept the test and he did not turn up. The probability that she could have kept on the first time is ${1 \over 5}$. The same is for the second time.

Thus,

$$\begin{equation} \begin{aligned} P({A_1}) = {1 \over 5} \\ P({A_2}) = {1 \over 5} \\\end{aligned} \end{equation} $$

The probability that he misses at least one = Probability of missing the first or the second

$$\begin{equation} \begin{aligned} P({A_1} \cup {A_2}) \\ = 1 - P({{\bar A}_1}) \times P({{\bar A}_2}) \\ = 1 - \left( {1 - {1 \over 5}} \right)\left( {1 - {1 \over 5}} \right) \\ = 1 - \left[ {{4 \over 5} \times {4 \over 5}} \right] \\ = {9 \over {25}} \\\end{aligned} \end{equation} $$