Permutation and Combination based Probability Questions in English

(D) Total number of bulbs = $50$.
Number of defective bulbs = $5$.
Number of non-defective bulbs = $50 - 5 = 45$.
We need to draw $3$ bulbs without replacement.
The probability that the first bulb is non-defective is $P(E_1) = \frac{45}{50}$.
After drawing one non-defective bulb,there are $44$ non-defective bulbs left out of $49$ total bulbs.
The probability that the second bulb is non-defective is $P(E_2|E_1) = \frac{44}{49}$.
After drawing two non-defective bulbs,there are $43$ non-defective bulbs left out of $48$ total bulbs.
The probability that the third bulb is non-defective is $P(E_3|E_1 \cap E_2) = \frac{43}{48}$.
The probability that all $3$ bulbs are non-defective is $P = \frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}$.
$P = \frac{9}{10} \times \frac{44}{49} \times \frac{43}{48} = \frac{9 \times 11 \times 43}{10 \times 49 \times 12} = \frac{3 \times 11 \times 43}{10 \times 49 \times 4} = \frac{1419}{1960}$.

(B) Total number of ways to select $2$ students out of $100$ is ${ }^{100} C_2$.
Number of ways in which both you and your friend are placed in the first section (strength $40$) is ${ }^{40} C_2$.
Number of ways in which both you and your friend are placed in the second section (strength $60$) is ${ }^{60} C_2$.
Since these are mutually exclusive events,the total number of favorable ways is ${ }^{40} C_2 + { }^{60} C_2$.
Therefore,the required probability is $\frac{{ }^{40} C_2 + { }^{60} C_2}{{ }^{100} C_2}$.

(A) The total number of people is $9 + 5 = 14$.
We need to form a committee of $4$ members.
The total number of ways to select $4$ members from $14$ is given by $^{14}C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
We want the probability that the committee contains at least one woman.
It is easier to calculate the complement: the probability that the committee contains no women (i.e.,all $4$ members are men).
The number of ways to select $4$ men from $9$ men is $^{9}C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
The probability of selecting no women is $P(\text{No women}) = \frac{126}{1001} = \frac{18}{143}$.
The probability of selecting at least one woman is $1 - P(\text{No women}) = 1 - \frac{18}{143} = \frac{125}{143}$.

(A) For the quadratic expression $x^2 + bx + c > 0$ to hold for all $x \in R$,the discriminant $D$ must be less than $0$.
$D = b^2 - 4c < 0 \Rightarrow b^2 < 4c$.
Since $b$ and $c$ are chosen from $\{1, 2, \ldots, 9\}$ without replacement $(b \neq c)$,we count the pairs $(b, c)$ satisfying $b^2 < 4c$:

$b$	Possible $c$ values $(c \neq b)$	Count
$1$	$2, 3, 4, 5, 6, 7, 8, 9$	$8$
$2$	$3, 4, 5, 6, 7, 8, 9$	$7$
$3$	$4, 5, 6, 7, 8, 9$	$6$
$4$	$5, 6, 7, 8, 9$	$5$
$5$	$7, 8, 9$	$3$
$6$	None $(6^2=36, 4c \leq 32)$	$0$

Total favorable outcomes $= 8 + 7 + 6 + 5 + 3 = 29$.
Total possible outcomes $= 9 \times 8 = 72$.
Probability $= \frac{29}{72}$.

(A) Total number of floors available $= 7$.
Each of the $5$ persons can choose any of the $7$ floors to exit.
Total number of ways for $5$ persons to exit $= 7^5$.
If all $5$ persons leave at different floors,the number of ways is the number of permutations of $7$ floors taken $5$ at a time,which is ${}^7P_5$.
${}^7P_5 = \frac{7!}{(7-5)!} = \frac{7 \times 6 \times 5 \times 4 \times 3}{1} = 2520$.
Required probability $= \frac{{}^7P_5}{7^5} = \frac{2520}{16807} = \frac{360}{2401}$.

(A) Total number of balls $= 9 + 4 = 13$.
Total ways to draw $3$ balls from $13$ is $^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.
The probability of getting at least one white ball is $1 - P(\text{no white ball})$.
If no white ball is drawn,all $3$ balls must be black.
Number of ways to draw $3$ black balls $= ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
$P(\text{no white ball}) = \frac{84}{286} = \frac{42}{143}$.
$P(\text{at least one white ball}) = 1 - \frac{42}{143} = \frac{101}{143}$.
Therefore,option $A$ is correct.

(A) Total number of balls $= 8 + 3 + 9 = 20$.
Number of ways to select $3$ balls out of $20$ is given by $n(S) = {}^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 20 \times 19 \times 3 = 1140$.
Number of ways to select $2$ red balls from $8$ and $1$ white ball from $3$ is $n(E) = {}^{8}C_2 \times {}^{3}C_1 = \frac{8 \times 7}{2 \times 1} \times 3 = 28 \times 3 = 84$.
Probability $P(E) = \frac{n(E)}{n(S)} = \frac{84}{1140} = \frac{84 \div 12}{1140 \div 12} = \frac{7}{95}$.

(A) Total number of balls $= 6 + 2 + 8 = 16$.
Total ways to draw $3$ balls $= ^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.
$A$. Probability that none of the balls is white (i.e.,all $3$ are from $6$ red and $8$ blue balls):
$P(A) = \frac{^{14}C_3}{^{16}C_3} = \frac{14 \times 13 \times 12}{16 \times 15 \times 14} = \frac{13}{20} = III$.
$B$. Probability of getting $2$ white and $1$ blue ball:
$P(B) = \frac{^{2}C_2 \times ^{8}C_1}{^{16}C_3} = \frac{1 \times 8}{560} = \frac{1}{70} = I$.
$C$. Probability of getting $2$ blue and $1$ white ball:
$P(C) = \frac{^{8}C_2 \times ^{2}C_1}{^{16}C_3} = \frac{28 \times 2}{560} = \frac{56}{560} = \frac{1}{10} = IV$.
$D$. Probability of getting $1$ red,$1$ white and $1$ blue ball:
$P(D) = \frac{^{6}C_1 \times ^{2}C_1 \times ^{8}C_1}{^{16}C_3} = \frac{6 \times 2 \times 8}{560} = \frac{96}{560} = \frac{6}{35} = II$.
Thus,the correct match is $A-III, B-I, C-IV, D-II$.

(A) Total number of balls = $4 + 3 + 5 = 12$.
Total ways to draw $3$ balls = $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
$(A)$ Probability of getting $1$ red,$1$ white and $1$ blue ball:
Ways = $^4C_1 \times ^3C_1 \times ^5C_1 = 4 \times 3 \times 5 = 60$.
Probability = $\frac{60}{220} = \frac{3}{11}$ (Matches $(iv)$).
$(B)$ Probability of getting $2$ white and $1$ blue ball:
Ways = $^3C_2 \times ^5C_1 = 3 \times 5 = 15$.
Probability = $\frac{15}{220} = \frac{3}{44}$ (Matches $(i)$).
$(C)$ Probability of getting $2$ red and $1$ white ball:
Ways = $^4C_2 \times ^3C_1 = 6 \times 3 = 18$.
Probability = $\frac{18}{220} = \frac{9}{110}$ (Matches $(v)$).
$(D)$ Probability that none of the balls is white (i.e.,all $3$ are from red and blue balls,total $4+5=9$):
Ways = $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Probability = $\frac{84}{220} = \frac{21}{55}$ (Matches $(ii)$).
Therefore,the correct matching is: $A \rightarrow (iv), B \rightarrow (i), C \rightarrow (v), D \rightarrow (ii)$.

(A) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.
Using the digits ${1, 2, 3, 4, 5}$,the possible two-digit combinations divisible by $4$ are $12, 24, 32, 52$.
There are $4$ such favorable combinations for the last two digits.
The remaining $3$ positions can be filled by the remaining $3$ digits in $3!$ ways.
Total number of favorable outcomes $= 4 \times 3!$.
Total number of possible five-digit numbers $= 5!$.
Probability $= \frac{4 \times 3!}{5!} = \frac{4 \times 3!}{5 \times 4 \times 3!} = \frac{1}{5}$.

(C) The total number of ways to choose $2$ numbers from $8$ is given by ${}^8C_2 = \frac{8 \times 7}{2} = 28$.
Let the two chosen numbers be $x$ and $y$ such that $x < y$. We want the probability that $x < 4$.
Case $I$: If $x = 1$,then $y$ can be any of the remaining $7$ numbers $(\{2, 3, 4, 5, 6, 7, 8\})$. Number of ways $= 7$.
Case $II$: If $x = 2$,then $y$ can be any of the remaining $6$ numbers $(\{3, 4, 5, 6, 7, 8\})$. Number of ways $= 6$.
Case $III$: If $x = 3$,then $y$ can be any of the remaining $5$ numbers $(\{4, 5, 6, 7, 8\})$. Number of ways $= 5$.
Total favourable cases $= 7 + 6 + 5 = 18$.
Therefore,the required probability $= \frac{18}{28} = \frac{9}{14}$.

(A) Total number of students = $15 + 5 = 20$.
Number of ways to select $3$ students from $20$ is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Number of ways to select $2$ boys from $15$ and $1$ girl from $5$ is $^{15}C_2 \times ^5C_1$.
$^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
$^{5}C_1 = 5$.
Number of favorable outcomes = $105 \times 5 = 525$.
Required probability = $\frac{525}{1140} = \frac{35}{76}$.

(A) The word '$PROBABILITY$' contains $11$ letters,where $B$ appears $2$ times and $I$ appears $2$ times.
Total number of arrangements $= \frac{11!}{2!2!}$.
To find the number of arrangements where both $B$'s are together,we treat the two $B$'s as a single unit $(BB)$.
Now,we have $10$ units to arrange: $(BB), P, R, O, A, I, L, I, T, Y$.
Since $I$ appears $2$ times,the number of arrangements where $B$'s are together $= \frac{10!}{2!}$.
Required probability $= \frac{\frac{10!}{2!}}{\frac{11!}{2!2!}} = \frac{10! \times 2! \times 2!}{2! \times 11!} = \frac{2}{11}$.

(C) Total number of people = $4 + 2 = 6$.
The total number of ways to arrange $6$ people in a row is $n(S) = 6! = 720$.
To find the number of ways where the $2$ girls sit side by side,we treat the $2$ girls as a single unit.
Now,we have $4$ boys and $1$ unit (the $2$ girls),which makes $5$ units in total.
These $5$ units can be arranged in $5!$ ways.
The $2$ girls within their unit can be arranged in $2!$ ways.
So,the number of favorable outcomes is $n(E) = 5! \times 2! = 120 \times 2 = 240$.
The probability $P$ is given by $\frac{n(E)}{n(S)} = \frac{5! \times 2!}{6!} = \frac{120 \times 2}{720} = \frac{240}{720} = \frac{1}{3}$.

(B) The total number of ways to select two distinct numbers from $50$ is $^{50}C_2 = \frac{50 \times 49}{2} = 1225$.
For the product $ab$ to be divisible by $3$,at least one of the numbers must be a multiple of $3$.
It is easier to calculate the complement: the probability that the product $ab$ is $NOT$ divisible by $3$.
This happens if neither $a$ nor $b$ is a multiple of $3$.
In the set ${1, 2, \dots, 50}$,the multiples of $3$ are ${3, 6, \dots, 48}$,which are $16$ numbers.
The numbers that are $NOT$ multiples of $3$ are $50 - 16 = 34$ numbers.
The number of ways to choose two distinct numbers that are not multiples of $3$ is $^{34}C_2 = \frac{34 \times 33}{2} = 17 \times 33 = 561$.
The probability that the product is not divisible by $3$ is $\frac{561}{1225}$.
Therefore,the probability that the product $ab$ is divisible by $3$ is $1 - \frac{561}{1225} = \frac{1225 - 561}{1225} = \frac{664}{1225}$.