Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
MCQ
Choose the correct answer from the given four options.
In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:
  • A
    $\frac{1}{10}$
  • B
    $\frac{2}{5}$
  • C
    $\frac{9}{20}$
  • D
    $\frac{1}{3}$
✓

Answer

  1. $\frac{2}{5}$

Solution:

Here, $\text{P}_{(\text{Ph})}=\frac{30}{100}=\frac{3}{10}$

$\text{P}_{(\text{M})}=\frac{25}{100}=\frac{1}{4}$

And $\text{P}_{(\text{M}\cap\text{Ph})}=\frac{10}{100}=\frac{1}{10}$

$\therefore\text{P}\Big(\frac{\text{Ph}}{\text{M}}\Big)=\frac{\text{P}(\text{Ph}\cap\text{M})}{\text{P}(\text{M})}$

$=\frac{\frac{1}{10}}{\frac{1}{4}}=\frac{2}{5}$

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 "M.C.Q (1 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

Let $\vec a$ and $\vec b$ be two unit vectors such that $\left| {\vec a\, + \,\vec b} \right| = \sqrt 3 .$ If $\vec c = \vec a\, + \,2\vec b + 3\,(\vec a \times \vec b),$ then $2\left| {\vec c} \right|$ is equal to
→
If the binary operation $\odot$ is defined on the set Q+ of all positive rational numbers by $\text{a}\odot\text{b}=\frac{\text{ab}}4$. Then, $3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$ is equal to:
  1. $\frac{3}{160}$
  2. $\frac{5}{160}$
  3. $\frac{3}{10}$
  4. $\frac{3}{40}$
→
Divide  $20$  into two parts such that the product of one part and the cube of the other is maximum. The two parts are
→
If $y$ is a function of $x$ then  $\frac{{{d^2}\,y}}{{d\,{x^2}}}+ y \frac{{d\,y}}{{d\,x}}=0$ . If $x$ is a function of $y$ then the equation becomes :
→
Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:
  1. -4
  2. 4
  3. -5
  4. -3
→
Let $P(x)=a_0+a_1 x^2+a_2 x^4+\ldots+a_n x^{2 n}$ with all $a_i>0, i=0,1,2, \ldots, n$. Then $P(x)$ has
→
Consider the system of equations

$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$

$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and

$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.

→
The set of points of discontinuity of the function $f(x)$ = $\frac{{\tan \,x.{{\tan }^{ - 1}}\left( {\frac{1}{{x - 1}}} \right)}}{{x\left( {x - 3} \right)\left( {x - 5} \right)}}$ is
→
The objective function Z = 4x + 3y can be maximised subjected to the constraints $ 3\text{x}+4\text{y}\leq24,$ $8\text{x}+6\text{y}\leq48,$ $\text{x}\leq5,\text{y}\leq6;\text{x},\text{y}\leq0.$
  1. At only one point
  2. At two points only.
  3. At an infinite number of points.
  4. None of these
→
If y = x(x - 3)2 decreases for the values of x given by:
→