Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
MCQ
If $\text{P}(\text{A}\cup\text{B})=0.8$ and $\text{P}(\text{A}\cap\text{B})=0.3$ then $\text{P}(\overline{\text{A}})=\text{P}(\overline{\text{B}})=$
  • A
    0.3
  • B
    0.5
  • C
    0.7
  • ✓
    0.9
✓

Answer

Correct option: D.
0.9
If $\text{P}(\text{A}\cup\text{B})=0.8\text{ P}(\text{A}\cap\text{B})=0.3,$

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\ \text{P(A)}+\text{P(B)}=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\ \text{P(A)}+\text{P(B)}=1.1$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\big[\text{P(A)}+\text{P(B)}\big]$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-1.1$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=0.9$

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

Consider the function $f:(0, \infty) \rightarrow R$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $\mathrm{m}+\mathrm{n}$ is
→
$\int {\frac{{\log x -log^2\ x+ x^2}}{{{x^3}}}} dx\,\,is\, $         (where $C$ is constant of integration)
→
If ${I_n} = \int_0^{\pi /4} {{{\tan }^n}\theta \,d\theta ,} $ then ${I_8} + {I_6}$ equals
→
Let $A =\{1,2,3,4, \ldots .10\}$ and $B =\{0,1,2,3,4\}$ The number of elements in the relation $R =\{( a , b )$ $\left.\in A \times A : 2( a - b )^2+3( a - b ) \in B \right\}$ is $.........$.
→
If $A$ is a skew-symmetric matrix of order 3 , then the value of $|A|$ is
→
Let$\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:
→
If $A=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right]$, then $(A-I)(A+I)=O$ for
→
If $f(x)$ satisfies the relation $f\left( {\frac{{5x - 3y}}{2}} \right)\, = \,\frac{{5f(x) - 3f(y)}}{2}\,\forall x,y\in R$ $f(0) = 1, f '(0) = 2$ then period of $sin \ (f(x))$ is
→
If $\alpha=3 \sin ^{-1}\left(\frac{6}{11}\right)$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$, where the inverse trigonometric functions take only the principal values, then the correct option$(s)$ is(are)

$(A)$ $\cos \beta > 0$ $(B)$ $\sin \beta < 0$ $(C)$ $\cos (\alpha+\beta) > 0$ $(D)$ $\cos \alpha < 0$

→
A homogeneous dofferential equation of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by making the substitution:
→