[ Given three indentical bags each containing 10 balls, whose colours are as follows : ] […

Question

$$ \text { Given three indentical bags each containing } 10 \text { balls, whose colours are as follows : } $$

$$ \[\begin{array}{lccc} \text { Red } & \text { Blue } & \text { Green } \ \text { Bag I } & 3 & 2 & 5 \ \text { Bag II } & 4 & 3 & 3 \ \text { Bag III } & 5 & 1 & 4 \end{array}$$ \]
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 $\left(\frac{1}{p}+\frac{1}{q}\right)$ 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 $\left(\frac{1}{p} + \frac{1}{q}ight)$:

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

Therefore,

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

Solutions · 1