Question

A bag is Randomly selected, If drawn ball is red, then probability that ball is selected from bag-I is p. If ball drawn is green then probability that ball is selected from bag-III is q. Then \(\frac{1}{p}+\frac{1}{q}\) equals to

	Red	Blue	Green
Bag-I	3	3	4
Bag-II	4	3	3
Bag-III	5	2	3

(1) \(\frac{22}{3}\)

(2) \(\frac{22}{5}\)

(3) \(\frac{11}{3}\)

(4) \(\frac{11}{5}\)

Answer 1

Correct option is: (1) \(\frac{22}{3}\)  

\(p\left(B_{1} / R\right)=\frac{p\left(B_{1}\right) \cdot p\left(R / B_{1}\right)}{p(R)}\)

\(=\frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}}=\frac{1}{4}=p\) 

\(p\left(B_{3} / G\right)=\frac{p\left(B_{3}\right) \cdot p\left(G / B_{3}\right)}{p(G)}\)

\(=\frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}}=\frac{3}{10}=q\)

\(\frac{1}{p}+\frac{1}{q}=4+\frac{10}{3}=\frac{22}{3}\)