Question
[Maximum mark: 5]
Let $A$ and $B$ be events such that $\mathrm{P}(A)=0.35, \mathrm{P}(B)=0.5$ and $\mathrm{P}(A \cup B)=0.65$.
Find $\mathrm{P}(A \mid B)$.
Answer/Explanation
Using the combined events formula, we have
$
\begin{aligned}
\mathrm{P}(A \cup B) & =\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cap B) \\
0.65 & =0.35+0.5-\mathrm{P}(A \cap B) \\
\mathrm{P}(A \cap B) & =0.85-0.65 \\
\mathrm{P}(A \cap B) & =0.2
\end{aligned}
$
Hence, using the conditional probability formula, we get
$
\begin{aligned}
\mathrm{P}(A \mid B) & =\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \\
& =\frac{0.2}{0.5} \\
& =\frac{2}{5}
\end{aligned}
$
Question
[Maximum mark: 5]
Let $f(x)=x^4+5 x^3+a x^2+b x+2$, for $x \in \mathbb{R}$, where $a, b$ are constants. The remainder when $f(x)$ is divided by $(x-1)$ is 6 , and the remainder when $f(x)$ is divided by $(x+2)$ is $-6$. Find the value of $a$ and the value of $b$.
Answer/Explanation
Using the remainder theorem, we have
$
\begin{aligned}
f(1) & =6 \\
1^4+5(1)^3+a(1)^2+b(1)+2 & =6 \\
1+5+a+b+2 & =6 \\
a+b & =-2 \quad \quad\quad\quad\quad \text { (1) }
\end{aligned}
$
and
$
\begin{aligned}
f(-2) & =-6 \\
(-2)^4+5(-2)^3+a(-2)^2+b(-2)+2 & =-6 \\
16-40+4 a-2 b+2 & =-6 \\
4 a-2 b & =16 \quad \quad\quad\quad\quad \text { (2) }
\end{aligned}
$
Hence, solving the system of linear equations (1)-(2) for $a$ and $b$, we get $a=2$ and $b=-4$