Three numbers are chosen at random without replacement from $\{1,2,3, \ldots . .8\}$. The probability that…
Three numbers are chosen at random without replacement from $\{1,2,3, \ldots . .8\}$. The probability that their minimum is $3$ , given that their maximum is $6$ , is
$\frac{3}{8}$
$\frac{1}{5}$
$\frac{1}{4}$
$\frac{2}{5}$
Solution
Let $A$ be the event that maximum is $6$.
$B$ be event that minimum is $3$
$P(A)=\frac{{ }^5 C_2}{{ }^8 C_3}$ (the numbers $ < 6$ are $5$)
$P(B)=\frac{{ }^5 C_2}{{ }^8 C_3}$ (the numbers $>3$ are $5$)
$P(A \cap B)=\frac{{ }^2 C_1}{{ }^8 C_3}$
Required probability is $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{{ }^2 \mathrm{C}_1}{{ }^5 \mathrm{C}_2}=\frac{2}{10}=\frac{1}{5}$.