Question
The probability that A hits a target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$, What is the probability that the target will be hit, if each one of A and B shoots at the target?

Answer

Given,
Probability that A hits a target $=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}=\frac{1}{3}$
Probability that B hits the targer $=\frac{2}{5}$
$\Rightarrow\ \text{P(B)}=\frac{2}{5}$
P (Target will be hit)
= 1 - P (target will not be hit)
= 1 - P (Niether A non B hits the target)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})$
$=1-[1-\text{P(A)}][1-\text{P}(\overline{\text{B}})]$
$=1-\Big[1-\frac{1}{3}\Big]\Big[1-\frac{2}{5}\Big]$
$=1-\frac{2}{3},\frac{3}{5}$
$=1-\frac{2}{5}$
$=\frac{2}{5}$
Required probability $=\frac{2}{5}$

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

Similar questions

A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin?
Evaluate: $\int\limits^\frac{3}{2}_{0}|x \cos\pi x| dx$
Draw a rough sketch of the region {(x, y) : y2 < 3x, 3x2 + < 16} and find the area by the region using mwthod of integration.
Evaluate the following intregals:
$\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
Show that the lines $\vec{\text{r}}=(2\hat{\text{i}}-3\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$ and $\vec{\text{r}}=(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}})+\mu(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ are coplanar. Also, find the equation of the plane containing them.
If $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}},$ prove that $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}$
Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abcissa and ordinate of the point.
If $\text{y}=\text{e}^{\tan^{-1}\text{x}},$ prove that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
Find the equation of the plane through (3, 4, -1) which is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+2=0$
If $A =\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then, prove that $A^3-6 A^2+7 A+2 I=O$ and hence find $A^{-1}$.