Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
Question
Given two independent events A and B such that $\text{P}(\text{A})=0.3,\ \text{P}(\text{B})=0.6.\ \text{Find}$ $\text{P}(\text{A}\ \text{and not}\ \text{B})$
✓

Answer

It is given that $\text{P} (\text{A})=0.3\ \text{and}\ \text{P}(\text{B})=0.6$
Also, A and B are independent events.
P(A and not B) $=\text{P}(\text{A}\cap\text{B}')$
$ =\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})$
= 0.3 - 0.18
= 0.12

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 "1 Marks Question" 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

Find the area under the given curves and given lines: $y = x^4, x = 1, x = 5$ and x-axis
→
A matrix A, of order $3 × 3$, has determinant $4$. Find the value of $|3A|$.
→
If $\begin{bmatrix} 3 \text{x}& 7 \$0.3em] -2 & 4\$0.3em] \end{bmatrix} = \begin{bmatrix} 8 & 7 \$0.3em] 6 & 4\$0.3em] \end{bmatrix}, $find the value of x.
→
Write the position vector of the point which divides the join of points with position vectors $3\overrightarrow{\text{a}} - 2\overrightarrow{\text{b}} \text{and } 2\overrightarrow{\text{a}} + 3\overrightarrow{\text{b}}$ in the ratio $2:1.$
→
Show that the relation R in the set {1, 2 ,3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
→
Show that the function defined by $f\left( x \right) = \left| {\cos x} \right|$ is a continuous function.
→
If $\vec{\text{p}}$ is a unit vector and $(\vec{\text{x}}-\vec{\text{p}})\cdot(\vec{\text{x}}+\vec{\text{p}})=80,$ then find $|\vec{\text{x}}|.$
→
Find: $\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
→
Examine the following functions for continuity.
f(x) = x - 5
→
Compute the following:
$\begin{bmatrix}\text{a} & \text{b} \\-\text{b} & \text{a} \end{bmatrix}+\begin{bmatrix}\text{a} & \text{b} \\\text{b} & \text{a} \end{bmatrix}$
→