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

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Solution

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

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

Event X : From bag A , white ball is drawn

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

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Solution

$\mathrm{a}, \mathrm{b} \in\{1,2,3, \ldots, 99,100\}$
Total cases $=100 \times 99$
Favourable cases $a-b \geq 10$
No of Pairs of (a,b)

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

$b=3, a \geq 13$	$a \rightarrow 13,14,15, \ldots, 99,100$	88
$\vdots$
$b=88, a \geq 98$	$a \rightarrow 98,99,100$	3
$b=89, a \geq 99$	$a \rightarrow 99,100$	2
$b=90, a \geq 100$	$a \rightarrow 100$	1

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

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

Required probability $=\frac{45 \times 91}{100 \times 99}=\frac{91}{220}=\frac{\mathrm{m}}{\mathrm{n}}$

\[ \mathrm{m}+\mathrm{n}=91+220=311 \text { Answer }(311) \]

[shahkb4]

Solution

$\mathrm{n} \in\{1,2,3,4,5,6\}$
$\Delta=\left|\begin{array}{ccc}1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1\end{array}\right|$
$\Delta=-(\mathrm{n}-1)(\mathrm{n}-2)$
For unique solution $\Delta \neq 0 \Rightarrow n \neq 1, n \neq 2$
favourable cases $n=\{3,4,5,6\}$
Required Probability $=\frac{4}{6}=\frac{\mathrm{k}}{6}$
$\mathrm{k}=4, \mathrm{n}=3,4,5,6$
required sum $=4+3+4+5+6=22$ Answer(4)

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Solution

$\sum \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=0+2 \mathrm{k}+\mathrm{k}+3 \mathrm{k}+2 \mathrm{k}^{2}+2 \mathrm{k}+\mathrm{k}^{2}+\mathrm{k}+7 \mathrm{k}^{2}=1$
$\Rightarrow 10 \mathrm{k}^{2}+9 \mathrm{k}-1=0 \Rightarrow(10 \mathrm{k}-1)(\mathrm{k}+1)=0 \Rightarrow \mathrm{k}=\frac{1}{10}, \mathrm{k} \neq-1$
$\mathrm{P}(3<\mathrm{x} \leq 6)=\mathrm{P}(\mathrm{x}=4)+\mathrm{P}(\mathrm{x}=5)+\mathrm{P}(\mathrm{x}=6)$ $=2 \mathrm{k}^{2}+2 \mathrm{k}+\mathrm{k}^{2}+\mathrm{k}=3 \mathrm{k}^{2}+3 \mathrm{k}=\frac{3}{100}+\frac{3}{10}=0.33$ Answer(4)

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Solution

First Method:

$\mathrm{P}(\mathrm{ab}$ 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$)$
$=\frac{{ }^{16} \mathrm{C}_{2}}{{ }^{50} \mathrm{C}_{2}}+\frac{{ }^{16} \mathrm{C}_{1} \times{ }^{34} \mathrm{C}_{1}}{{ }^{50} \mathrm{C}_{2}}=\frac{8.15+16.34}{25(49)}=\frac{664}{1225}$
$3 \mathrm{~K}=\{3,6,9, \ldots, 48\}$

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

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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)

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Solution

Exponent of $7=\left[\frac{101}{7}\right]+\left[\frac{101}{49}\right]+\left[\frac{101}{49 \times 7}\right]+\cdots$

\[ =14+2+0=16 \]

$101!=7^{16}(\lambda)$ Answer(1)

[shahkb4]

Solution

available floors $\rightarrow 4^{\text {th }}, 5^{\text {th }}, \ldots 9^{\text {th }}, 10^{\text {th }}$
Required ways $=\left({ }^{7} \mathrm{C}_{3}\right) 3!=\frac{765}{6} \times 6=7(30)=210$ Answer(210)

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Solution

Using 5,5,1,2,7,8,6,4 digits

\[ \mathrm{x}=\underset{\substack{\text { Repeating } \\ \text { digit }}}{{ }^{9} \mathrm{C}_{1}} \times \underset{\substack{\text { Non repeating } \\ \text { digits }}}{{ }^{8} \mathrm{C}_{7}} \times \underset{\text { arrangement }}{\left(\frac{9!}{2!}\right)} \]

$x=9.8 \frac{9!}{2!}=36(9!)$

Using 1,1,7,7,4,2,5,6,9 digits
$\mathrm{y}=\underset{\substack{\text { repeating } \\ \text { digits }}}{{ }^{9} \mathrm{C}_{2}} \times \underset{\substack{\text { Non-repeating } \\ \text { digits }}}{{ }^{7} \mathrm{C}_{5}} \times \frac{9!}{2!2!}$
$y=\left(\frac{9.8}{2}\right)\left(\frac{7.6}{2}\right)\left(\frac{9!}{2!2!}\right)=9(7) 3.9!$
$\frac{x}{y}=\frac{36(9!)}{9(7)(3) 9!} \Rightarrow \frac{x}{y}=\frac{4}{21} \Rightarrow 21 x=4 y$ Answer(3)

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Solution

A, D, P, U,U,Y
A □ such words $=6!/ 2!=360$
D $\_\_\_\_$ such words $=6!/ 2!=360$
$\mathrm{P}—-$ such words $=6!/ 2!=360$
R $\_\_\_\_$ such words $=6!/ 2!=360$
Till $\overline{\mathrm{n}} \overline{\mathrm{ow}} \overline{1} 4 \overline{4} 0$ words

U

U A □ such words $=5!=120$
left → D, P, R, U, Y
Till now 1560 words
U D A P $\_\_\_\_$ such words $=3!=6$
left R,V,Y
U D A R $\_\_\_\_$ such words $=3!=6$
left P,U,Y
U D A U $\_\_\_\_$ such words $=3!=6$
left R,P,Y
Till now 1578 words
UDAYPRU $\rightarrow 1579{ }^{\text {th }}$ word
UDA Y P U R $\rightarrow 1580{ }^{\text {th }}$ word Answer(1)

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Solution

5	x	y	z

$\quad \mathrm{x}, \mathrm{y}, \mathrm{z} \in\{0,1,2,5,9\}$

\[ \frac{5+x+y+z}{3}=1+\frac{x+y+z+2}{3} \in I \]

$\mathrm{x}+\mathrm{y}+\mathrm{z}$	$(\mathrm{x}, \mathrm{y}, \mathrm{z})$	ways
1	$0,0,1$	$3!/ 2!=3$
4	$0,2,2$	$3!/ 2!=3$
4	$1,1,2$	$3!/ 2!=3$
7	$0,2,5$	$3!=6$
7	$1,1,5$	$3!/ 2!=3$
10	$0,1,9$	$3!=6$
10	$0,5,5$	$3!/ 2!=3$
13	$9,2,2$	$3!/ 2!=3$
16	$2,5,9$	$3!=6$
19	$9,9,1$	$3!/ 2!=3$
19	$5,5,9$	$3!/ 2!=3$

Total possible ways of $\mathrm{x}, \mathrm{y}, \mathrm{z}=42$ Answer(42)

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Solution

0,1,2,3,4,5,6,7,8,9
$\mathrm{n}(\mathrm{S})={ }^{10} \mathrm{C}_{4} \times 1=\frac{10 \cdot 9 \cdot 8 \cdot 7}{24}=210$
xyzwt $=20$
Case I: 5, 4, 1, 1, 1
Such numbers formed $=\frac{5!}{3!}=\frac{120}{6}=20$
Case II: 5, 2, 2, 1, 1
Such numbers formed $=\frac{5!}{2!2!}=\frac{120}{4}=30$
$\mathrm{n}(\mathrm{P})=20+30=50$
$\mathrm{n}(\mathrm{s})+\mathrm{n}(\mathrm{p})=210+50=260$ answer(260)

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Solution

assuming all oranges are identical
Let children $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}$ get number of oranges $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}$
$\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=16$ where $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4} \geq 1$
let $\mathrm{x}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}^{\prime}+1 ; \mathrm{x}_{1}^{\prime}+\mathrm{x}_{2}^{\prime}+\mathrm{x}_{3}^{\prime}+\mathrm{x}_{4}^{\prime}=12$
Number of non-negative integral solutions ${ }^{n+p-1} C_{p-1}={ }^{12+4-1} C_{4-1}$
$={ }^{15} \mathrm{C}_{3}=455$ Answer(4)

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Solution

$(\mathrm{P}, \mathrm{P}, \mathrm{P})(\mathrm{QQ})(\mathrm{RR}) \mathrm{S}, \mathrm{T}, \mathrm{U}, \mathrm{V}$
(i) 3 same, 1 diff. $(\mathrm{P}, \mathrm{P}, \mathrm{P}, \mathrm{R}) \rightarrow 1 \times{ }^{6} \mathrm{C}_{1} \times\left(\frac{4!}{3!}\right)=6 \times 4=24$
(ii) 2 same, 2 diff. ( $\mathrm{Q}, \mathrm{Q}, \mathrm{P}, \mathrm{T}) \rightarrow{ }^{3} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2} \times\left(\frac{4!}{2!}\right)=3 \times 15 \times 12=450+90=540$
(iii) 2 same, 2 same $(\mathrm{P}, \mathrm{P}, \mathrm{R}, \mathrm{R}) \rightarrow{ }^{3} \mathrm{C}_{2} \times\left(\frac{4!}{2!2!}\right)=\frac{3 \times 24}{4}=18$
(iv) 4 different $(\mathrm{P}, \mathrm{R}, \mathrm{U}, \mathrm{V}) \rightarrow{ }^{7} \mathrm{C}_{4}(4!)=\frac{7.6 .5}{6} \times 24=42(20)=840$

Total such words $=24+540+18+840=1422$ Answer(1422)

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