[shahkb4]

Solution

[t]{|l|l|l|l|l|} x	0	1	2	3
P(x)	$\frac{2 \mathrm{a}+1}{30}$	$\frac{8 \mathrm{a}-1}{30}$	$\frac{4 \mathrm{a}+1}{30}$	b

\[ \begin{aligned} & \sum P(x)=\frac{2 a+1}{30}+\frac{8 a-1}{30}+\frac{4 a+1}{30}+b=1 \\ & \Rightarrow 14 a+1+30 b=30 \Rightarrow 14 a+30 b=29 \ldots \\ & \text { Variance }=\sigma^{2}=\frac{\sum x_{i}^{2} P\left(x_{i}\right)}{\sum P\left(x_{i}\right)}-\mu^{2} \\ & \Rightarrow \sigma^{2}+\mu^{2}=\frac{\sum x_{i}^{2}\left(P\left(x_{i}\right)\right)}{1} \\ & \Rightarrow 2=0+1\left(\frac{8 a-1}{30}\right)+4\left(\frac{4 a+1}{30}\right)+9 b \\ & \Rightarrow 2=\frac{8 a-1+16 a+4+270 b}{30} \\ & \Rightarrow 60=24 a+270 b+3 \Rightarrow 24 a+270 b=57 \\ & \Rightarrow 8 a+90 b=19 \ldots(2) \\ & (\text { Eq. }(1)) 3-\text { Eq. }(2) \\ & 42 a+90 b-8 a-90 b=29 \times 3-19 \\ & \Rightarrow 34 a=87-19=68 \Rightarrow a=2, b=\frac{1}{30} \\ & \frac{a}{b}=\frac{2}{1 / 30}=60 \text { Answer }(3) \end{aligned} \]

[shahkb4]

Solution

$\mathrm{S}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\}$
Set form by all subset $\mathrm{P}(\mathrm{S})=\{\{ \},\{\mathrm{a}\}, \ldots,\{\mathrm{a}, \mathrm{b}\}, \ldots\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\}\}$

A, $\mathrm{B} \in \mathrm{P}(\mathrm{S})$
Every element has 4 choices
Total ways $=4^{5}$
$\mathrm{A} \cap \mathrm{B}=\phi$
favourable choices $=3 \times 3 \times 3 \times 3 \times 3=3^{5}$

\[ \begin{aligned} & \mathrm{P}=\frac{3^{5}}{4^{5}}=\frac{3^{5}}{2^{10}}=\frac{3^{\mathrm{p}}}{2^{\mathrm{q}}} \\ & \mathrm{P}=5, \mathrm{q}=10 \\ & \mathrm{P}+\mathrm{q}=15 \text { Answer(15) } \end{aligned} \]

[t]{|l|l|} A	B
✓	✓
✓	x
x	✓
x	x

[shahkb4]

Solution

\[ \begin{aligned} & \mathrm{k}, 2 \mathrm{k}, 3 \mathrm{k}, \ldots \ldots, 500 \mathrm{k}, 501 \mathrm{k}, \ldots, 1000 \mathrm{k} \\ & \text { Median }=\frac{500 \mathrm{k}+501 \mathrm{k}}{2}=1001 \frac{\mathrm{k}}{2}=\overline{\mathrm{x}} \\ & \text { Mean deviation about mean }=\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}=500 \\ & \Rightarrow \frac{\left[\frac{999|\mathrm{k}|}{2}+\frac{997|\mathrm{k}|}{2}+\cdots+\frac{|\mathrm{k}|}{2}+\frac{|\mathrm{k}|}{2}+\frac{3|\mathrm{k}|}{2}+\cdots+\frac{999|\mathrm{k}|}{2}\right]}{1000}=500 \\ & \Rightarrow|\mathrm{k}|+3|\mathrm{k}|+5|\mathrm{k}|+\cdots+997|\mathrm{k}|+999|\mathrm{k}|=500 \times 1000 \\ & \Rightarrow \frac{500}{2}[|\mathrm{k}|+999|\mathrm{k}|]=500 \times 1000 \\ & \Rightarrow 1000|\mathrm{k}|=2 \times 1000 \Rightarrow|\mathrm{k}|=2 \\ & \mathrm{k}^{2}=|\mathrm{k}|^{2}=4 \text { Answer }(2) \end{aligned} \]

[shahkb4]

Solution

\[ \begin{aligned} & \text { Mean }=\frac{\sum x_{i}}{8}=\frac{-10-7-1+x+y+9+2+16}{8}=\frac{7}{2} \\ & \Rightarrow x+y=19 \quad \ldots(1) \\ & \text { Variance }=\frac{1}{n} \sum x_{i}^{2}-\mu^{2}=\frac{293}{4} \\ & =\frac{1}{8}\left(100+49+1+x^{2}+y^{2}+81+4+256\right)-\left(\frac{7}{2}\right)^{2}=\frac{293}{4} \end{aligned} \]

\[ \begin{aligned} & \Rightarrow x^{2}+y^{2}+406+85=8\left[\frac{49+293}{4}\right]=684 \\ & \Rightarrow x^{2}+y^{2}=684-491=193 \text { from }(1) \\ & \Rightarrow x^{2}+(19-x)^{2}=193 \\ & \Rightarrow(x-12)(x-7)=0 \\ & x=12 \mid x=7 \\ & y=7 \mid y=12 \end{aligned} \]

required numbers $\mathrm{x}, \mathrm{y}, \mathrm{x}+\mathrm{y}+1,|\mathrm{x}-\mathrm{y}|$ are 12, 7, 20, 5

\[ \text { Mean }=\frac{12+7+20+5}{4}=11 \text { Answer }(1) \]

[shahkb4]

Solution

[t]{|l|l|l|l|l|} class	$x_{i}$	$f_{i}$	$x_{i} f_{i}$	$f_{i} x_{i}^{2}$
4-8	6	3	18	$18 \times 6$
8-12	10	$\lambda$	$10 \lambda$	$100 \lambda$
12-16	14	4	56	14 × 56
16-20	18	$\frac{7}{14+\lambda}$	126	18 × 126

\[ \begin{aligned} & \mu=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{18+10 \lambda+56+126}{14+\lambda} \\ & \Rightarrow \mu=\frac{200+10 \lambda}{14+\lambda}=\frac{10(\lambda+14)+60}{14+\lambda} \\ & \Rightarrow \mu=10+\frac{60}{14+\lambda} ; \lambda \in \mathrm{N} \\ & \lambda=1, \mu=10+4=14 ; \end{aligned} \]

\[ \begin{aligned} & \lambda=6, \mu=10+3=13 \\ & \lambda=16, \mu=10+2=12 \\ & \lambda=46, \mu=10+1=11 \end{aligned} \]

\[ \begin{aligned} & \text { Variance }=\frac{\sum x_{i}^{2} f_{i}}{\sum f_{i}}(\text { Mean })^{2}=19 \\ & \Rightarrow \frac{108+100 \lambda+14 \times 56+18 \times 126}{14+\lambda}-\mu^{2}=19 \\ & \frac{108+100 \lambda+560+4 \times 56+1260+8 \times 126}{14+\lambda}-\left(10+\frac{60}{14+\lambda}\right)^{2}=19 \\ & \Rightarrow \frac{3160+100 \lambda}{14+\lambda}-\left(10+\frac{60}{14+\lambda}\right)^{2}=19 \end{aligned} \]

we get $\lambda=6, \mu=10+3=13$

\[ \therefore \lambda+\mu=19 \text { Answer (4) } \]

[shahkb4]

Solution

\[ \text { old Mean }=\frac{\sum \mathrm{x}_{1}}{10}=10 \Rightarrow \sum \mathrm{x}_{1}=100 \]

\[ \begin{aligned} & \text { New Mean }=\frac{\sum x_{i}-\alpha+\beta}{10}=\frac{100-\alpha+\beta}{10}=10.1 \\ & \Rightarrow 100-\alpha+\beta=101 \Rightarrow \beta=1+\alpha \\ & \text { Old Variance }=\frac{1}{n} \sum x_{i}^{2}-(\text { Mean })^{2}=2 \\ & \Rightarrow \frac{1}{10} \sum x_{i}^{2}-100=2 \Rightarrow \sum x_{i}^{2}=1020 \\ & \text { New Variance }=\frac{1}{10}\left(\sum x_{i}^{2}-\alpha^{2}+\beta^{2}\right)-\binom{\text { New }}{\text { Mean }}^{2}=1.99 \\ & \Rightarrow \frac{1020-\alpha^{2}+\beta^{2}}{10}-(10.1)^{2}=1.99 \\ & \Rightarrow \frac{1020-\alpha^{2}+\beta^{2}}{10}=\frac{101 \times 101}{100}+\frac{199}{100} \\ & \Rightarrow 1020-\alpha^{2}+\beta^{2}=\frac{1}{10}\left(100^{2}+1+200+199\right) \\ & \Rightarrow 1020-\alpha^{2}+(1+\alpha)^{2}=1040 \\ & \Rightarrow \alpha=\frac{19}{2}, \beta=1+\alpha=\frac{21}{2} \\ & \alpha+\beta=\frac{19+21}{2}=20 \text { Answer(4) } \end{aligned} \]

[shahkb4]

Solution

Event $\mathrm{E}_{1}$ : white ball is transferred
Event $\mathrm{E}_{2}$ : Black ball is transferred

\[ P\left(E_{1}\right)=\frac{6}{6+4}=\frac{3}{5}, P\left(E_{2}\right)=\frac{4}{6+4}=\frac{2}{5} \]

Event X : From bag A , white ball is drawn

\[ \begin{aligned} & P(X)=P\left(E_{1}\right) P\left(\frac{X}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{X}{E_{2}}\right) \\ & =\frac{3}{5}\left(\frac{10}{18}\right)+\frac{2}{5}\left(\frac{9}{18}\right)=\frac{30+18}{5 \times 18}=\frac{8}{15}=\frac{p}{q} \\ & \therefore p+q=23 \text { Answer(2) } \end{aligned} \]

[shahkb4]

Solution

(10 Defective, 90 Non-defective)
Required Probability of getting atleast 7 defective bulbs out 8 bulbs $=\mathrm{P}($ Exactly 7 defective bulbs $)+\mathrm{P}($ all 8 defective bulbs $)$

\[ ={ }^{8} \mathrm{C}_{7}\left(\frac{10}{100}\right)^{7}\left(\frac{90}{100}\right)+\left(\frac{10}{100}\right)^{8}=8\left(\frac{1}{10}\right)^{7}\left(\frac{9}{16}\right)+\left(\frac{1}{100}\right)^{8}=\frac{72+1}{108}=\frac{73}{10^{8}} \text { Answer(4) } \]

[shahkb4]

Solution

A bag contains 10 balls out of which k are red and (10-k) are black
There are 8 possibilities as

\[ \begin{aligned} & \frac{\mathrm{K}=0\left(\mathrm{E}_{0}\right) 0 \text { Red } 10 \text { Black }}{\mathrm{K}=1\left(\mathrm{E}_{1}\right) 1 \text { Red } 9 \text { Black }} \\ & \hline \mathrm{K}=2\left(\mathrm{E}_{2}\right) 2 \text { Red } 8 \text { Black } \\ & \mathrm{K}=3\left(\mathrm{E}_{3}\right) 3 \text { Red } 7 \text { Black } \\ & \mathrm{K}=7\left(\mathrm{E}_{7}\right) 7 \text { Red } 3 \text { Black } \\ & \mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{E}_{3}\right)=\ldots=\mathrm{P}\left(\mathrm{E}_{7}\right)=\mathrm{p} \\ & \mathrm{P}\binom{\text { all } 3}{\text { are black }}=\mathrm{p} \frac{{ }^{10} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\cdots+\mathrm{p} \frac{{ }^{3} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}} \\ & \mathrm{P}\left(\mathrm{E}_{1} / \text { All } 3 \text { Blacks }\right)=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{8} \mathrm{C}_{3}+\ldots .+{ }^{3} \mathrm{C}_{3}} \\ & \left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}\right) \\ & { }^{3} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots+{ }^{8} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{4} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{5} \mathrm{C}_{4}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{6} \mathrm{C}_{4}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{3}={ }^{11} \mathrm{C}_{4} \\ & \operatorname{Pr} \text { ob. }=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{4}}=\frac{9.8 .7}{6\left(\frac{11.10 .9 .8}{24}\right)}=\frac{14}{55} \text { Answer(4) } \end{aligned} \]

[shahkb4]

Solution

\[ \begin{aligned} & \text { Expectation } \mathrm{E}(\mathrm{x})=\sum \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=\frac{263}{15} \\ & \Rightarrow \frac{28 \times 2 \mathrm{k}+30 \mathrm{k}+64 \mathrm{k}+102 \mathrm{k}+36 \mathrm{k}+76 \mathrm{k}+120 \mathrm{k}+42 \mathrm{k}}{7 \times 15}=\frac{263}{15} \\ & \Rightarrow 526 \mathrm{k}=263 \times 7 \Rightarrow \mathrm{k}=\frac{7}{2} \end{aligned} \]

[t]{|l|l|l|l|l|l|l|l|l|} X	14	15	16	17	18	19	20	21
P(x)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$

\[ P(x<20)=\frac{2+1+2+3+1+2}{15}=\frac{11}{15} \text { Answer }(3) \]

[shahkb4]

Solution

\[ a, b \in\{1,2,3, \ldots, 99,100\} \]

Total cases $=100 \times 99$
Favourable cases $\mathrm{a}-\mathrm{b} \geq 10$
No of Pairs of $(\mathrm{a}, \mathrm{b})$

\[ \begin{array}{ll} b=1, a \geq 11 & a \rightarrow 11,12,13, \ldots, 99,100 \\ b=2, a \geq 12 & a \rightarrow 12,13,14, \ldots, 99,100 \end{array} \]

90
89

\[ \begin{array}{ll} b=3, a \geq 13 & a \rightarrow 13,14,15, \ldots, 99,100 \\ \vdots & \\ b=88, a \geq 98 & a \rightarrow 98,99,100 \\ b=89, a \geq 99 & a \rightarrow 99,100 \\ b=90, a \geq 100 & a \rightarrow 100 \end{array} \]

No. of favourable ordered pair $=1+2+3+\ldots .+89+90$

\[ =\frac{90(90+1)}{2}=45 \times 91 \]

Required probability $=\frac{45 \times 91}{100 \times 99}=\frac{91}{220}=\frac{\mathrm{m}}{\mathrm{n}}$
$\mathrm{m}+\mathrm{n}=91+220=311$ Answer(311)

[shahkb4]

Solution

\[ \begin{aligned} & \text { Exponent of } 7=\left[\frac{101}{7}\right]+\left[\frac{101}{49}\right]+\left[\frac{101}{49 \times 7}\right]+\cdots \\ & \quad=14+2+0=16 \\ & 101!=7^{16}(\lambda) \text { Answer }(1) \end{aligned} \]

[shahkb4]

Solution

\[ 40=8 \times 5=2^{3} \times 5 \]

Exponent of 2 in $60!=\left[\frac{60}{2}\right]+\left[\frac{60}{4}\right]+\left[\frac{60}{8}\right]+\left[\frac{60}{16}\right]+\left[\frac{60}{32}\right]+\left[\frac{60}{64}\right]+\cdots$

\[ =30+15+7+3+1+0+\cdots=56 \]

Exponent of 5 in $60!=\left[\frac{60}{5}\right]+\left[\frac{60}{25}\right]+\left[\frac{60}{125}\right]+\cdots=12+2+0=14$

\[ 60!=2^{56} 5^{14}(\mathrm{~K})=2^{14}\left(2^{42}\right) 5^{14} \mathrm{~K}=2^{14}\left(2^{3} \times 5\right)^{14} \mathrm{~K}=2^{14}(40)^{14} \mathrm{~K} \]

∴ Maximum value of n is 14
Answer (4)

[shahkb4]

Solution

First Method:
P(ab is divisible by 3)
$=\mathrm{P}(\mathrm{a} \& \mathrm{~b}$ both are multiple of 3$)+\mathrm{P}($ Exactly are of $\mathrm{a} \& \mathrm{~b}$ is multiple of 3$)$

\[ \begin{aligned} & =\frac{{ }^{16} \mathrm{C}_{2}}{{ }^{50} \mathrm{C}_{2}}+\frac{{ }^{16} \mathrm{C}_{1} \times{ }^{34} \mathrm{C}_{1}}{{ }^{50} \mathrm{C}_{2}}=\frac{8.15+16.34}{25(49)}=\frac{664}{1225} \\ & 3 \mathrm{~K}=\{3,6,9, \ldots, 48\} \end{aligned} \]

Second Method:
Number which are not multiple of $3=50-16=34$
$\mathrm{P}(\mathrm{ab}$ is divisible by 3$)=1-\mathrm{P}(\mathrm{ab}$ is not divisible by 3$)$

\[ =1-\frac{{ }^{34} \mathrm{C}_{2}}{{ }^{50} \mathrm{C}_{2}}=1-\frac{17 \times 93}{25 \times 49}=\frac{1225-561}{1225}=\frac{664}{1225} \text { Answer(2) } \]

[shahkb4]

Solution

\[ \begin{aligned} & \sum P\left(x_{i}\right)=0+2 k+k+3 k+2 k^{2}+2 k+k^{2}+k+7 k^{2}=1 \\ & \Rightarrow 10 k^{2}+9 k-1=0 \Rightarrow(10 k-1)(k+1)=0 \Rightarrow k=\frac{1}{10}, k \neq-1 \\ & P(3

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