If P(A)=1 2 , P(B)=0, then P(A | B) is - Vidyadip

MCQ

If $P(A)=\frac{1}{2}, P(B)=0,$ then $P(A | B)$ is

A
$0$
✓
Not defined
C
$\frac {1}{2}$
D
$1$

✓

Answer

Correct option: B.

Not defined

b
It is given that $P(A)=\frac{1}{2}$ and $P(B)=0$

$P(A | B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{0}$

Therefore, $\mathrm{P}(\mathrm{A} | \mathrm{B})$ is not defined.

Thus, the correct answer is $B$.

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 13. probability→13. 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 R be the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$ Then, R is:

Symmetric.
Reflexive.
Transitive.
An equivalence relation.

Let $A$ be the area of the region $\left\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2 x(1-x)\right\}$ Then $540\,A$ is equal to

Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is

Area of circle $x^2+y^2=4$ :

If $f(x)=\frac{4 x+3}{6 x-4}, x \neq \frac{2}{3}$ and $(f \circ f)(x)=g(x)$, where $\mathrm{g}: \mathbb{R}-\left\{\frac{2}{3}\right\} \rightarrow \mathbb{R}-\left\{\frac{2}{3}\right\}$, then $(gogog) (4)$ is equal to

The value of $\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$ is:

5²
0
5¹³
5⁹

If $a, b, c$ be positive real numbers and the value of $\theta = {\tan ^{ - 1}}\sqrt {\frac{{a(a + b + c)}}{{bc}}} + {\tan ^{ - 1}}\sqrt {\frac{{b(a + b + c)}}{{ca}}} + {\tan ^{ - 1}}\sqrt {\frac{{c(a + b + c)}}{{ab}}} $,then $\tan \theta $ is equal to

The probability that a leap year will have 53 fridays or 53 Saturdays is.

$\frac{2}{7}$
$\frac{3}{7}$
$\frac{4}{7}$
$\frac{1}{7}$

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

Linear programming model which involves funds allocation of limited investment is classified as:

Ordination budgeting model
Capital budgeting models
Funds investment models
Funds origin models.