Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 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, E1, E2, E3 and E4 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 "4 Marks" 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 possible, find BA and AB, where $\text{A}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}.$
→
Evaluate the following definite integral as limit of sum:

$\int_\limits{0}^{4}\text{(x}+\text{e}^{2\text{x}})\ \text{dx}$

→
Find the area of the region bounded by the ellipse $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1.$
→
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
→
Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}\big)$
→
Evaluate the following integrals:
$\int(2\text{x}+3)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$
→
Find the integrals of the functions in Exercises:
$\sin\text{x } \sin2\text{x }\sin3\text{x}$
→
Using matrices, solve the following system of equations:

$\text{x + y + z} = 3, \text{x} - 2{\text{y}} + 3{z}=2\ \text{and} \ 2 \text{x - y + z} = 2 $

 

→
Solve the following differential equation:

$4\frac{\text{dy}}{\text{dx}}+8\text{y}=5\text{e}^{-3\text{x}}$

→
If $\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix},$ find A16.
→