Question

A, B, C are aiming to shoot a ballon, A will succeed $4$ times out of $5$ attempts. The chance of B to shoot the ballon is $3$ out of $4$ and that of C is $2$ out of $3$. If the three aim the ballon simultaneously, then find the probability that atleast two of them hit the balloon.

Answer 1

Let the event of A hitting the ballon be $A$ and the event of A missing the balloon be $\overline{A}$ .

Similarly, we define $B,\overline{B}$ and $C,\overline{C}$ for B and C, respectively.

The event of atleast 2 hitting the balloon can be denoted as

$(A,B,\overline{C})\cup (A,\overline{B},C)\cup (\overline{A},B,C)\cup (A,B,C)$

Let the events below be denoted as

$(A,B,\overline{C})\equiv X$

$(A,\overline{B},C)\equiv Y$

$(\overline{A},B,C)\equiv Z$

$(A,B,C)\equiv T$.

The events $X,Y,Z,T$ are mutually exclusive. Therefore, the probability

$P(X\cup Y\cup Z\cup T)=P\left(X\right)+P\left(Y\right)+P\left(Z\right)+P\left(T\right)$

$=P(A,B,\overline{C})+P(A,\overline{B},C)+P(\overline{A},B,C)+P(A,B,C)$

Thus, we have:

P(atleast 2 hitting the balloon) = $P(A,B,\overline{C})+P(A,\overline{B},C)+P(\overline{A},B,C)+P(A,B,C)$

Given that:

$P\left(A\right)=\frac{4}{5}$

$P\left(B\right)=\frac{3}{4}$

$P\left(C\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{A}\right)=1-\frac{4}{5}=\frac{1}{5}$

$P\left(\overline{B}\right)=1-\frac{3}{4}=\frac{1}{4}$

$P\left(\overline{C}\right)=1-\frac{2}{3}=\frac{1}{3}$

The events A, B, C are independent. Hence,

$P(A,B,C)=P\left(A\right)\times P\left(B\right)\times P\left(C\right)$

$=\frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}=\frac{2}{5}$

Similarly,

$P(A,B,\overline{C})=P\left(A\right)\times P\left(B\right)\times P\left(\overline{C}\right)$

$=\frac{4}{5}\times \frac{3}{4}\times \frac{1}{3}=\frac{1}{5}$

$P(A,\overline{B},C)=P\left(A\right)\times P\left(\overline{B}\right)\times P\left(C\right)$

$=\frac{4}{5}\times \frac{1}{4}\times \frac{2}{3}=\frac{2}{15}$

$P(\overline{A},B,C)=P\left(\overline{A}\right)\times P\left(B\right)\times P\left(C\right)$

$=\frac{1}{5}\times \frac{3}{4}\times \frac{2}{3}=\frac{1}{10}$

The probability of atleast 2 of them hitting the balloon:

$P(A,B,\overline{C})+P(A,\overline{B},C)+P(\overline{A},B,C)+P(A,B,C)$

$=\frac{2}{5}+\frac{1}{5}+\frac{2}{15}+\frac{1}{10}$

$=\frac{5}{6}$