Download App

Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
Coloured balls are distributed in four boxes as shown in the following table:
Box
Colour
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→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ prove that $2\text{x}\frac{\text{dy}}{\text{dx}}=\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}$
→
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
→
Evaluate the following integrals:
$\int\text{cosec}^43\text{x}\text{ dx}$
→
Differentiate the following functions with respect to x:
$\sqrt{\frac{1-\text{x}^2}{1+\text{x}^2}}$
→
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
→
Evaluate the following integrals:
$\int\frac{\text{x}}{3\text{x}^4-18\text{x}^2+11}\text{dx}$
→
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
→
Find the shortest distance between the lines
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and, $\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\mu\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
→
If $\int\limits_{0}^{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16},$ find the value of k.
→
Solve the following differential equations: $\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
→