10 gathered 3 2 5 4 3 3 5 1 4 gathered A person chooses a bag at random and takes out a ba…

Question

$10$

$\begin{gathered} \quad  \quad  \ 3 \quad  2 \quad  5 \ 4 \quad  3 \quad  3 \ 5 \quad  1 \quad  4 \end{gathered}$

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is $q$, then the value of $(\frac{1}{p}+\frac{1}{q})$ is:

Accepted Answer

Probability that a Red ball comes from Bag I ($p$):

$p({B}_{1}/R)=\frac{p({B}_{1})\cdot p(R/{B}_{1})}{p(R)}$

Substituting the values,

$p({B}_{1}/R)=\frac{\frac{1}{3}	imes \frac{3}{10}}{\frac{1}{3}	imes \frac{3}{10}+\frac{1}{3}	imes \frac{4}{10}+\frac{1}{3}	imes \frac{5}{10}}$

Simplify to find $p$:

$=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)}=\frac{1}{4}$

So, $p=\frac{1}{4}$.

Probability that a Green ball comes from Bag III ($q$):

$p({B}_{3}/G)=\frac{p({B}_{3})\cdot p(G/{B}_{3})}{p(G)}$

Substitute the values,

$p({B}_{3}/G)=\frac{\frac{1}{3}	imes \frac{4}{10}}{\frac{1}{3}	imes \frac{5}{10}+\frac{1}{3}	imes \frac{3}{10}+\frac{1}{3}	imes \frac{4}{10}}$

Simplify to find $q$:

$=\frac{\frac{2}{15}}{\frac{1}{3}	imes 1.2}=\frac{1}{3}$

So, $q=\frac{1}{3}$.

Calculation of $(\frac{1}{p}+\frac{1}{q})$:

$\frac{1}{p}=4,\, \frac{1}{q}=3$

Therefore,

$(\frac{1}{p}+\frac{1}{q})=4+3=7$

Solutions · 1