Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
Question
Choose the correct answer from the given four options.
The probability that exactly two of the three balls were red, the first ball being red, is:
  1. $\frac{1}{3}$
  2. $\frac{4}{7}$
  3. $\frac{15}{28}$
  4. $\frac{5}{28}$
✓

Answer

  1. $\frac{4}{7}$

Solution:

Let E1 = Event that first ball being red

And E2 = Event that exactly two of three balls being red

$\therefore\text{P}(\text{E}_1)=\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}$

$=\frac{60+60+60+30}{336}=\frac{210}{336}$

$\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\text{R}}+\text{P}{_\text{R}}\cdot\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{120}{336}$

$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$

$=\frac{\frac{120}{336}}{\frac{210}{336}}=\frac{4}{7}$

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

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is:
→
If $f(x) = A\sin \left( {\frac{{\pi x}}{2}} \right) + B,$ $f'\left( {\frac{1}{2}} \right) = \sqrt 2 $ and $\int_0^1 {f(x)\,dx = \frac{{2A}}{\pi },} $ then the constants $A$ and $B$ are respectively
→
If $A$ and $B$ are two independent events such that $P\,(A) = \frac{1}{2},\,\,P(B) = \frac{1}{5},$ then
→
The distance of the plane through the intersection of the planes ax + by + cz +d = 0 and lx + my + nz + P = 0 and parallel to the line y = 0, z = 0
→
If $A = \left[ {\begin{array}{*{20}{c}}1&2&{ - 1}\\3&0&{\,\,2}\\4&5&{\,\,0}\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\0&1&3\end{array}} \right],$then $AB$ is
→
The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is
→
If $y=\cos ^{-1}\left(e^x\right)$, then $\frac{d y}{d x}$ is :
→
$\int {\frac{{{x^{\frac{1}{3}}}}}{{{{(1 + {x^{\frac{2}{3}}})}^3}}}}dx$ is equal to   (where $C$ is constant of integration)
→
The radius of a cylinder is increasing at the rate of  $5\ cm/min$, so that its volume is constant. When its radius is $5\ cm$ and height is $3\ cm$,  then the rate of decreases of its height is .......... $cm/min$.
→
The value of $\int {\frac{{dx}}{{3 - 2x - {x^2}}}} $ will be
→