Maths MCQ

(d) : We have, $E(X)=\Sigma x P(X=x)$
$
\begin{aligned}
& =1 \times P(X=1)+2 \times P(X=2)+\ldots \ldots+n \times P(X=n) \\
& =\frac{2}{n(n+1)}+2 \times \frac{2 \times 2}{n(n+1)}+\ldots \ldots .+n \frac{2 \times n}{n(n+1)} \\
& =\frac{2}{n(n+1)}\left[1^2+2^2+\ldots . .+n^2\right] \\
& =\frac{2}{n(n+1)}\left[\frac{n(n+1)(2 n+1)}{6}\right]=\frac{2 n+1}{3}
\end{aligned}
$

(a) : Let $X$ denotes the number of factors of the number obtained.
Number of factors of $4=3$
Number of factors of $6=4$
Number of factors of 2, 3, $5=2$
$
\begin{aligned}
& \therefore \quad P(X=1)=\frac{1}{6}, P(X=2)=\frac{3}{6}=\frac{1}{2}, \\
& P(X=3)=\frac{1}{6}
\end{aligned}
$

,$P(X=4)=\frac{1}{6}$

(a) : Total number of cards $=52$
Total number of queens $=4$
$
\begin{aligned}
& P(\text { getting a queen })=\frac{4}{52}=\frac{1}{13} \\
& P(\text { not getting a queen })=1-\frac{1}{13}=\frac{12}{13}
\end{aligned}
$
Let $X$ be the random variable denoting the number of queens drawn in 2 trails.
$P(X=0)=P($ non queen $) \times P($ non queen $)$
$
=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}
$
$P(X=1)=P($ queen $) \times P($ non queen $)+P($ non queen $)$
$P$ (queen)
$
=\frac{1}{13} \times \frac{12}{13}+\frac{12}{13} \times \frac{1}{13}=\frac{24}{169}
$
$P(X=2)=P($ queen $) \times P($ queen $)$
$
=\frac{1}{13} \times \frac{1}{13}=\frac{1}{169}
$

(c) : $P(a<x<b)=\int_a^b f(x) d x$
$
\therefore P\left(\frac{1}{4}<x<\frac{1}{3}\right)=\int_{1 / 4}^{1 / 3} 3\left(1-2 x^2\right) d x=\left(3 x-\frac{6 x^3}{3}\right)_{1 / 4}^{1 / 3}
$
$\begin{aligned} & \Rightarrow\left(3 x-2 x^3\right)_{1 / 4}^{1 / 3}=\frac{3 \times 1}{3}-\frac{2 \times 1}{27}-\frac{3}{4}+\frac{2}{64} \\ & =1-\frac{2}{27}-\frac{3}{4}+\frac{1}{32}=\frac{864-64-648+27}{864}=\frac{179}{864}\end{aligned}$

(b) :

(b) : $\because P(X<a)=P(X>a) \Rightarrow \int_0^a f(x) d x=\int_a^1 f(x) d x$
$
\begin{aligned}
& \Rightarrow 6 \int_0^a x(1-x) d x=6 \int_a^1 x(1-x) d x \\
& \Rightarrow \int_0^a\left(x-x^2\right) d x=\int_a^1\left(x-x^2\right) d x \\
& \Rightarrow\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^a=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_a^1 \\
& \Rightarrow \frac{a^2}{2}-\frac{a^3}{3}=\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{a^2}{2}-\frac{a^3}{3}\right) \Rightarrow 4 a^3-6 a^2+1=0
\end{aligned}
$
$\therefore \quad$ From the given options $a=\frac{1}{2}$ satisfies the above equation.

$\begin{array}{ll}& \text { (a) : } \because \sum_{x=1}^6 P(X=x)=1 \\ \therefore & P(X=1)+P(X=2)+\ldots+P(X=6)=1 \\ \Rightarrow & K+3 K+5 K+7 K+8 K+K=1 \\ \Rightarrow & 25 K=1 \Rightarrow K=\frac{1}{25} \\ \therefore & P(2 \leq X<5)=P(X=2)+P(X=3)+P(X=4) \\ =& 3 K+5 K+7 K=15 K=\frac{15}{25}=\frac{3}{5}\end{array}$

(c) : For $X$ heads, he gets amount $Y=2 X$

Expected gain, $E(Y)=\Sigma Y P(Y)$
$
=0\left(\frac{1}{8}\right)+2\left(\frac{3}{8}\right)+4\left(\frac{3}{8}\right)+6\left(\frac{1}{8}\right)=\frac{6+12+6}{8}=3
$

$\begin{aligned} & \text { (b) }: P(X \leq 1)=F(1)=0.2 \\
& P(X \leq 2)=P(X=1)+P(X=2)=F(2) \\
\Rightarrow \quad & P(X=2)=0.37-0.2=0.17 \\
& P(X \leq 3)=P(X=1)+P(X=2)+P(X=3)=F(3) \\
\Rightarrow \quad & P(X=3)=0.48-0.2-0.17=0.11\end{aligned}$
Similarly, $P(X=4)=0.14, P(X=5)=0.23$ and
$
P(X=6)=0.15
$

$\begin{aligned} & P(3<X \leq 5)=P(X=4)+P(X=5) \\ & =0.14+0.23=0.37\end{aligned}$

(d) : $X=$ no. of defective pens.
$\therefore \quad X$ can take values $0,1,2$
$
\begin{aligned}
& P(X=0)=\frac{{ }^4 C_2}{{ }^6 C_2}=\frac{4 \times 3}{6 \times 5}=\frac{2}{5}=\frac{6}{15} \\
& P(X=1)=\frac{{ }^2 C_1 \times{ }^4 C_1}{{ }^6 C_2}=\frac{2 \times 4 \times 2}{6 \times 5}=\frac{8}{15} \\
& P(X=2)=\frac{{ }^2 C_2}{{ }^6 C_2}=\frac{1 \times 2}{6 \times 5}=\frac{1}{15}
\end{aligned}
$

$
\begin{aligned}
& E(X)=\sum X_i P_i=\frac{10}{15}=\frac{2}{3} \\
& E\left[X^2\right]=\sum X_i^2 P_i=\frac{12}{15}=\frac{4}{5}
\end{aligned}
$
Standard deviation $=\sqrt{E\left(X^2\right)-[E(X)]^2}$
$
=\sqrt{\left(\frac{4}{5}\right)-\left(\frac{2}{3}\right)^2}=\sqrt{\frac{4}{5}-\frac{4}{9}}=\sqrt{\frac{36-20}{45}}=\sqrt{\frac{16}{45}}=\frac{4}{3 \sqrt{5}}
$

(b) : Let the random variable $X$ be defined as the number of aces in a draw of 2 cards simultaneous, then $X$ can take the values $0,1,2$. The corresponding probabilities are:
$
\begin{aligned}
P(X=0) & =P(\text { drawing no ace }) \\
& =P(\text { drawing both non-ace cards }) \\
& =\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \cdot 47}{1 \cdot 2} \times \frac{1 \cdot 2}{52 \cdot 51}=\frac{188}{221}
\end{aligned}
$
$P(X=1)=P($ drawing one ace and one non-ace card $)$
$
=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4.48}{\frac{52.51}{1.2}}=\frac{32}{221}
$
$P(X=2)=P$ (drawing both aces)
$
=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \cdot 3}{1 \cdot 2} \times \frac{1 \cdot 2}{52.51}=\frac{1}{221}
$
The probability distribution of number of aces is