2 Marks Questions

Question 12 Marks

Two dice are thrown. If it is known that the sum of numbers on the die was less than 6, then find the probability of getting a sum 3.

Answer

Let,
$E_1=$ Event that sum of numbers on the dice was less than 6
and $E_2=$ Event that sum of numbers on the dice is 3.
$\begin{aligned} \therefore E_1= & \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3), \\ & (3,1),(3,2),(4,1)\}\end{aligned}$
$\Rightarrow \quad n\left(E_1\right)=10$
and $\quad E_2=\{(1,2),(2,1)\}$
$\Rightarrow \quad n\left(E_2\right)=2$
$\therefore$ Required Probability $=\frac{n\left(E_2\right)}{n\left(E_1\right)}=\frac{2}{10}$

View full question & answer→

Question 22 Marks

If $P(A)=0.4, P(B)=0.8$ and $P\left(\frac{B}{A}\right)=0.6$, then find the $P(A \cup B)$.

Answer

Given, $P(A)=0.4, P(B)=0.8$ and $P\left(\frac{B}{A}\right)=0.6$
$\because \quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$
$\Rightarrow \quad 0.6=\frac{P(A \cap B)}{0.4}$
$\Rightarrow \quad P(A \cap B)=0.6 \times 0.4=0.24$
Now, $\quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=0.4+0.8-0.24$
$=1.2-0.24=0.96$

View full question & answer→

Question 32 Marks

If $A$ and $B$ are any two events such that $P(A)+P(B)$ $-P(A$ and $B)=P(A)$, then show that $P\left(\frac{A}{B}\right)=1$.

Answer

Given $\quad P(A)+P(B)-P(A$ and $B)=P(A)$
$\Rightarrow \quad P(A)+P(B)-P(A \cap B)=P(A)$
$\Rightarrow \quad P(B)-P(A \cap B)=0$
$\Rightarrow \quad P(A \cap B)=P(B)$
$\Rightarrow \quad \frac{P(A \cap B)}{P(B)}=1$
$\Rightarrow \quad P\left(\frac{A}{B}\right)=1$.
Hence Proved.

View full question & answer→

Applied Maths 2 Marks Questions

Test yourself on this topic