Maths MCQ

$\begin{aligned} & \text {(b) : Since, } n=5, p=\frac{1}{3} \\
& \therefore q=1-p=1-\frac{1}{3}=\frac{2}{3} \\
& \Rightarrow P(2<X<4)=P(X=3) \\
& ={ }^5 C_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2=\frac{10 \times 4}{3^5}=\frac{40}{243}\end{aligned}$

(a) : Probability of at least one failure $=1-P($ no failure $)=1-p^5$
Now $1-p^5 \geq \frac{31}{32}$
$\Rightarrow p^5 \leq \frac{1}{32}$ thus $p \leq \frac{1}{2} \therefore p \in(0,1 / 2]$

(a) : Let $X=$ number of defective bulbs $p=$ probability that bulb is defective
$
\begin{aligned}
& \therefore p=\frac{10}{100}=\frac{1}{10} ; \quad q=1-p=\frac{9}{10} \\
& \text { Given, } n=5 \quad \therefore \quad X \sim B\left(5, \frac{1}{10}\right) \\
& P[X \leq 1]=P(X=0)+P(X=1) \\
& ={ }^5 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5+{ }^5 C_1\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^4 \\
& \quad=\left(\frac{9}{10}\right)^4\left[\frac{9}{10}+\frac{5}{10}\right] \\
& =\left(\frac{9}{10}\right)^4\left[\frac{14}{10}\right]=\frac{7}{5}\left(\frac{9}{10}\right)^4
\end{aligned}
$

(c) : $P($ solving a problem $)=\frac{4}{5}$
$
P(\text { not solving a problem })=1-\frac{4}{5}=\frac{1}{5}
$
Let $X$ be the random variable representing number of problem that candidate is unable to solve.
Now, $P(X<2)=P(X=0)+P(X=1)$
$
\begin{aligned}
& ={ }^{50} C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^{50}+{ }^{50} C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^{49} \\
& =\left(\frac{4}{5}\right)^{50}+\frac{50}{5}\left(\frac{4}{5}\right)^{49}=\left(\frac{4}{5}\right)^{49}\left[\frac{4+50}{5}\right]=\frac{54}{5}\left(\frac{4}{5}\right)^{49}
\end{aligned}
$

(c) : Since the function is p.m.f so,
$
\begin{aligned}
& \Rightarrow \sum_{x=0}^3 \frac{K \cdot 2^x}{x !}=1 \\
& \Rightarrow K+2 K+2 K+\frac{4}{3} K=1 \Rightarrow \frac{19 K}{3}=1 \Rightarrow K=\frac{3}{19}
\end{aligned}
$

(a) : Given $n=4$ and
$
\begin{aligned}
& 2 P(X=3)=3 P(X=2) \\
& \Rightarrow 2\left({ }^4 C_3 p^3 q\right)=3\left({ }^4 C_2 p^2 q^2\right) \\
& \Rightarrow 8 p^3 q=18 p^2 q^2 \Rightarrow 4 p=9 q
\end{aligned}
$
We know that $p=1-q$
$
\Rightarrow 4(1-q)=9 q \Rightarrow q=\frac{4}{13}
$

(d) : Let $p$ denotes the probability of getting an odd number i.e., 1,3 or 5 .
$\therefore p=\frac{3}{6}=\frac{1}{2}$ and $q$ be the probability of failure, i.e., getting an even number.
$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}$, Here, $n=5$
$\therefore \quad P($ at least 4 successes $)=P(4)+P(5)$
$={ }^5 C_4\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^4+{ }^5 C_5\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^5$
$=\left(\frac{1}{2}\right)^5[5+1]=\frac{6}{32}=\frac{3}{16}$

(a) : We have given $E(X)+\operatorname{Var}( X )=\frac{270}{49}$
$
\begin{aligned}
& \Rightarrow n p+n p q=\frac{270}{49} \Rightarrow 6 p(1+q)=\frac{270}{49} \\
& \Rightarrow 6 p(1+1-p)=\frac{270}{49} \quad(\because p+q=1) \\
& \Rightarrow 6 p(2-p)=\frac{270}{49} \Rightarrow 12 p-6 p^2=\frac{270}{49} \\
& \Rightarrow p=\frac{9}{7} \text { or } \frac{5}{7} \Rightarrow p=\frac{5}{7} \quad(\because 0<p<1)
\end{aligned}
$

(b) : Given, mean, $E(X)=18$
And $\operatorname{Var}(X)=12$
But $E(X)=n p$ and $\operatorname{Var}(X)=n p q$
$\therefore n p=18$ and $n p q=12$
Now, $\frac{n p q}{n p}=\frac{12}{18} \Rightarrow q=\frac{2}{3} \quad \therefore p=1-q=1-\frac{2}{3}=\frac{1}{3}$

(b) : Here $n=4$,
$p=$ probability of getting perfect square $=\frac{2}{6}=\frac{1}{3}$
$q=$ probability of failure $=\frac{4}{6}=\frac{2}{3}$
Let $X=$ Number on die is perfect square, then
$
\begin{aligned}
& P(X=0)={ }^4 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^4=\frac{16}{81} \\
& \therefore \quad P(X \geq 1)=1-P(X=0)=1-\frac{16}{81}=\frac{65}{81}
\end{aligned}
$

(d) In binomial distribution, $V(X)=n p q$ and
$E(X)=n p$, Here, $n=10, p=0.4 \therefore q=1-p=0.6$
$
\begin{array}{ll}
\therefore & V(X)=10 \times 0.4 \times 0.6=2.4 \text { and } E(X)=10 \times 0.4=4 \\
\therefore & V(X)=E\left(X^2\right)-[E(X)]^2 \Rightarrow 2.4=E\left(X^2\right)-(4)^2 \\
\Rightarrow & E\left(X^2\right)=2.4+16=18.4
\end{array}
$

(b) : Mean $=n p=18$ ...(i)
Variance $=n p q=12$ ...(ii)
Dividing (ii) by (i), we get
$
\frac{n p q}{n p}=\frac{12}{18} \Rightarrow q=\frac{2}{3}
$
Now, $p=1-q \Rightarrow p=\frac{1}{3}$
From (i), $n p=18 \Rightarrow n\left(\frac{1}{3}\right)=18 \Rightarrow n=54$
$\therefore \quad$ Values of $X$ are $0,1,2, \ldots . . . .54$
$\therefore \quad$ Total possible values of $X$ are 55

$\begin{aligned} & \text {(a) : Here, } n=10, p(\text { success })=\frac{1}{2}, q(\text { failure })=\frac{1}{2} \\
& \therefore \quad \text { Required probability }={ }^{10} C_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^3 \\
& +{ }^{10} C_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^2+{ }^{10} C_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^0 \\
& =\left(\frac{1}{2}\right)^{10}[120+45+10+1]=\frac{11}{64}\end{aligned}$

(b): $\because E(X)=n p=5$ ...(i)
and $\operatorname{Var}(X)=n p q=2.5$ ...(ii)
From (i) and (ii), we get
$
\frac{n p q}{n p}=\frac{2.5}{5} \Rightarrow q=\frac{1}{2}
$
Since, $p+q=1 \Rightarrow q=1-\frac{1}{2}=\frac{1}{2}$
$\therefore n=\frac{5}{1} \times 2=10$ (From (i))
$\Rightarrow P(X<1)=P(X=0)={ }^{10} C _0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{10}=\left(\frac{1}{2}\right)^{10}$