Arun and Ved appeared for an interview for two vacancies. The probability of Arun's selection is $1 / 4$, and that of Ved's rejection is $2 / 3 .$ Find the probability that at least one of them will be selected.

Home » Arun and Ved appeared for an interview for two vacancies. The probability of Arun’s selection is , and that of Ved’s rejection is Find the probability that at least one of them will be selected.

Arun and Ved appeared for an interview for two vacancies. The probability of Arun’s selection is $1 / 4$ , and that of Ved’s rejection is $2 / 3 .$ Find the probability that at least one of them will be selected.

Arun and Ved appeared for an interview for two vacancies. The probability of Arun’s selection is $1 / 4$ , and that of Ved’s rejection is $2 / 3 .$ Find the probability that at least one of them will be selected.

Given : let A denote the event ‘Arun is selected’ and let B denote the event ‘ved is selected’.
Therefore, $\mathrm{P}(\mathrm{A})=\frac{1}{4}$ and $\mathrm{P}(\bar{B})=\frac{2}{3} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{1}{3}$ and $\mathrm{P}(\bar{A})=\frac{3}{4}$
Also, $A$ and $B$ are independent.A and not $B$ are independent, not $A$ and $B$ are independent.
Now,
$\mathrm{P}$ (atleast one of them getting selected) $=\mathrm{P}$ (selecting only Arun ) $+\mathrm{P}$ (selecting only ved) $+\mathrm{P}$ (selecting both)
$\begin{array}{l} =P(A \text { and } \operatorname{not} B)+P(B \text { and } \operatorname{not} A)+P(A \text { and } B) \\ =P(A \cap \bar{B})+P(B \cap \bar{A})+P(A \cap B) \\ =P(A) \times P(\bar{B})+P(B) \times P(\bar{A})+P(A) \times P(B) \\ =\left(\frac{1}{4} \times \frac{2}{3}\right)+\left(\frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{1}{3}\right) \\ =\frac{2}{12}+\frac{3}{12}+\frac{1}{12} \\ =\frac{1}{2} \end{array}$
Therefore, The probability that atleast one of them will get selected is $\frac{1}{2}$