Coloured balls are distributed in four boxes as shown in the foll… - Vidyadip

Question

Coloured balls are distributed in four boxes as shown in the following table:

Box	Colour
Box	Black	White	Red	Blue
I II III IV	3 2 1 4	4 2 2 3	5 2 3 1	6 2 1 5

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.

✓

Answer

Let $A, E_1, E_2, E_3$ and $E_4$ denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{18}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_4}\Big)=\frac{4}{13}$
Using Bayes' theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{4}\times\frac{1}{7}}{\frac{1}{4}\times\frac{3}{18}+\frac{1}{4}\times\frac{2}{8}+\frac{1}{4}\times\frac{1}{7}+\frac{1}{4}\times\frac{4}{13}}$
$=\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{1}{13}}=\frac{156}{947}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Probability→Probability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix},\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix},$ verify that $(AB)^T = B^TA^T$

Two probability distributions of the discrete random variable X and Y are given below.

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

$\text{Y}$	$0$	$1$	$2$	$3$
$\text{P}(\text{Y})$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$	$\frac{1}{10}$

Prove that $E(Y^2) = 2E(X)$.

Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{2\text{x}+\text{x}^2}{\text{x}}, & \text{x} \neq0\\0,&\text{ x} = 0\end{cases}\text{at x}=0$

Evaluate $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$.

If $\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$ show that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$

Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$

If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that, $\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$

Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$

Evaluate the following intregals:
$\int\frac{1}{(\text{x}^2+1)(\text{x}^2+2)}\ \text{dx}$

Find the value of x for which the matrix product
$\begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-\text{x}&14\text{x}&7\text{x}\\0&1&0\\\text{x}&-4\text{x}&-2\text{x}\end{bmatrix}$ equal an identity matrix.