probabilities of hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$. Find the probability that the target
(a)is hit exactly by one of them.
(b) is not hit by any one of them.
(c) is hit.
(d) is exactly hit by two of them.
probabilities of hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$. Find the probability that the target
(a)is hit exactly by one of them.
(b) is not hit by any one of them.
(c) is hit.
(d) is exactly hit by two of them.
event B: B can hit the target,
event C: C can hit the target.
$\therefore \quad P ( A )=\frac{3}{4}, P ( B )=\frac{1}{2}, P ( C )=\frac{5}{8}$
$\therefore \quad P \left( A ^{\prime}\right)=1- P ( A )=1-\frac{3}{4}=\frac{1}{4}$
$P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{2}=\frac{1}{2}$
$P\left(C^{\prime}\right)=1-P(C)=1-\frac{5}{8}=\frac{3}{8}$
Since A, B, C are independent events,
A’, B’, C’ are also independent events.
(a) Let event W: Target is hit exactly by one of them.
$\begin{array}{r} P ( W )= P \left( A \cap B ^{\prime} \cap C ^{\prime}\right) \cup P \left( A ^{\prime} \cap B \cap C ^{\prime}\right) \\ \cup P \left( A ^{\prime} \cap B ^{\prime} \cap C \right) \\ = P ( A ) \cdot P ^{\prime}\left( B ^{\prime}\right) \cdot P \left( C ^{\prime}\right)+ P \left( A ^{\prime}\right) \cdot P ( B ) \cdot P \left( C ^{\prime}\right)\end{array}$
$+ P \left( A ^{\prime}\right) \cdot P \left( B ^{\prime}\right) \cdot P ( C )$
$\begin{aligned} & =\left(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{5}{8}\right) \\ & =\frac{9}{64}+\frac{3}{64}+\frac{5}{64}=\frac{17}{64}\end{aligned}$
Let event X: Target is not hit by any one of them.
∴ P(X) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
$\begin{aligned} & =\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8} \\ & =\frac{3}{64}\end{aligned}$
Let event Y: Target is hit.
∴ P(Y) = 1 – P(target is not hit by any one of them)
$\begin{aligned} & =1-\frac{3}{64} \\ & =\frac{61}{64}\end{aligned}$
Let event Z: Target is hit by exactly two of them.
$\begin{aligned} P(Z)=P\left(A \cap B \cap C^{\prime}\right) \cup P\left(A \cap B^{\prime} \cap C\right) \\ \cup P\left(A^{\prime} \cap B \cap C\right) \\ =P(A) \cdot P(B) \cdot P\left(C^{\prime}\right)+P(A) \cdot P\left(B^{\prime}\right) \cdot P(C)\end{aligned}$
$+ P \left( A ^{\prime}\right) \cdot P ( B ) \cdot P ( C )$
$\begin{aligned} & =\left(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{3}{4} \times \frac{1}{2} \times \frac{5}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{5}{8}\right) \\ & =\frac{9}{64}+\frac{15}{64}+\frac{5}{64}=\frac{29}{64}\end{aligned}$
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$\frac{x+i y}{2+3 i}+\frac{2+i}{2-3 i}=\frac{9}{13}(1+i)$