Probability 4 Question 7
7. One of the two boxes, box I and box II was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II, is $\frac{1}{3}$, then the correct option(s) with the possible values of $n _1, n _2, n _3$ and $n _4$ is/are
(a) $n _1=3, n _2=3, n _3=5, n _4=15$
(b) $n _1=3, n _2=6, n _3=10, n _4=50$
(c) $n _1=8, n _2=6, n _3=5, n _4=20$
(d) $n _1=6, n _2=12, n _3=5, n _4=20$
Show Answer
Answer:
Correct Answer: 7. (b)
Solution:
$$ \begin{array}{|ll|} n _1 & \text { Red } \\ n _2 & \text { Black } \end{array} \mid \underbrace{\left|\begin{array}{ll} n _3 & \text { Red } \\ n _4 & \text { Black } \end{array}\right|} _{\text {Box I }} $$
Let $\quad A=$ Drawing red ball
$$ \therefore \quad P(A)=P\left(B _1\right) \cdot P\left(A / B _1\right)+P\left(B _2\right) \cdot P\left(A / B _2\right) $$
$$ =\frac{1}{2} \frac{n _1}{n _1+n _2}+\frac{1}{2} \frac{n _3}{n _3+n _4} $$
Given, $\quad P\left(B _2 / A\right)=\frac{1}{3}$
$$ \begin{aligned} & \Rightarrow \quad \frac{P\left(B _2\right) \cdot P\left(B _2 \cap A\right)}{P(A)}=\frac{1}{3} \\ & \Rightarrow \quad \frac{\frac{1}{2} \frac{n _3}{n _3+n _4}}{\frac{1}{2} \frac{n _1}{n _1+n _2}+\frac{1}{2} \frac{n _3}{n _3+n _4}}=\frac{1}{3} \\ & \Rightarrow \quad \frac{n _3\left(n _1+n _2\right)}{n _1\left(n _3+n _4\right)+n _3\left(n _1+n _2\right)}=\frac{1}{3} \end{aligned} $$
Now, check options, then clearly options (a) and (b) satisfy.
