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Let $\mathrm{U}_{1}$ and $\mathrm{U}_{2}$ be two urns such that $\mathrm{U}_{1}$ contains 3 white and 2 red balls, and $\mathrm{U}_{2}$ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from $\mathrm{U}_{1}$ and put into $\mathrm{U}_{2} .$ However, if tail appears then 2 balls are drawn at random from $\mathrm{U}_{1}$ and put into $\mathrm{U}_{2} .$ Now 1 ball is drawn at random from $\mathrm{U}_{2} .$
A box $\mathrm{B}_{1}$ contains 1 white ball, 3 red balls and 2 black balls. Another box $\mathrm{B}_{2}$ contains white balls, 3 red balls and 4 black balls. A third box $B_{3}$ contains 3 white balls, 4 red balls and 5 black balls.
Box 1 contains three cards bearing numbers, 1,2,3 ; box 2 contains five cards
bearing numbers 1,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1,2,3.
Let $n_{1}$ and $n_{2}$ be the number of red and black balls respectively, in box I. Let $n_{3}$ and $n_{4}$ be the number of red and black balls, respectively, in box II.
Q. A fair die is tossed repeatedly until a six is obtained. Let X denote the
number of tosses required.
(A) The probability that X = 3 equals –
(A) $\frac{25}{216}$
(B) $\frac{25}{36}$
(C) $\frac{5}{36}$
(D) $\frac{125}{216}$
(B) The probability that $\mathrm{X} \geq 3$ equals –
(A) $\frac{125}{216}$
(B) $\frac{25}{36}$
(C) $\frac{5}{36}$
(D) $\frac{25}{216}$
(C) The conditional probability that X 6 given X > 3 equals –
(A) $\frac{125}{216}$
(B) $\frac{25}{216}$
(C) $\frac{5}{36}$
(D) $\frac{25}{36}$
[JEE 2009, 4+4+4]
Ans. ( (a) A (b) B (c) D ) (A) $\mathrm{P}(\mathrm{X}=3)=\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{3}\right)=\frac{25}{216}$
(B) $P(X \leq 2)=\frac{1}{6}+\frac{5}{6} \times \frac{1}{6}=\frac{11}{36}$
(C) For $\mathrm{X} \geq 6$ the probability is \[\begin{array}{l}
{\frac{5^{5}}{6^{6}}+\frac{5^{5}}{6^{7}}+\ldots \ldots+\infty=\frac{5^{5}}{6^{6}}\left(\frac{1}{1-5 / 6}\right)=\left(\frac{5}{6}\right)^{5}} \\ {\text { For } \mathrm{X}>3} \\
{\frac{5^{3}}{6^{4}}+\frac{5^{4}}{6^{5}}+\frac{5^{5}}{6^{6}}+\ldots \ldots \infty=\left(\frac{5}{6}\right)^{3}} \end{array}\]
Hence the conditional probability is $\frac{(5 / 6)^{6}}{(5 / 6)^{3}}=\frac{25}{36}$
Q. (A) Let $\omega$ be a complex cube root of unity with $\omega \neq 1 .$ A fair die is thrown three times. If $\mathrm{r}_{1}, \mathrm{r}_{2}$ and $\mathrm{r}_{3}$ are the numbers obtained on the die, then the probability that $\omega^{\mathrm{r}_{1}}+\omega^{\mathrm{r}_{2}}+\omega^{\mathrm{rs}}=0$ is –
(A) $\frac{1}{18}$
(B) $\frac{1}{9}$
(C) $\frac{2}{9}$
(D) $\frac{1}{36}$
(B) A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $A$ and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac{3}{4} .$ If the signal received at station $\mathrm{B}$ is green, then the probability that the original signal was green is –
(A) $\frac{3}{5}$
(B) $\frac{6}{7}$
(C) $\frac{20}{23}$
(D) $\frac{9}{20}$
[JEE 2010, 3+5]
Ans. ( (a) C (b) C )
(A) $\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}$ can be from the set $(3,6),(1,4)$ or $(2,5)$ which can be done in $2 \times 2 \times 2=$
8 ways and these can be arranged in $3 !$ ways
\[ \text { Probability }=\frac{3 ! \times 8}{216}=\frac{2}{9} \]
(B) $\mathrm{C}:$ Correct signal is transmitted
$\overline{\mathrm{C}}:$ false signal is transmitted
$\mathrm{G}:$ Original signal is green
$\mathrm{R}:$ Original signal is red
$\mathrm{K}:$ Signal received at station $\mathrm{B}$ is green. $\mathrm{P}(\mathrm{G} / \mathrm{K})=\frac{\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{K} / \mathrm{G})}{\mathrm{P}(\mathrm{K})}$
$\mathrm{P}(\mathrm{G} / \mathrm{K})=\frac{\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{K} / \mathrm{G})}{\mathrm{P}(\mathrm{K})}$
$=\frac{\mathrm{P}(\mathrm{GCC})+\mathrm{P}(\mathrm{GC} \overline{\mathrm{C}})}{\mathrm{P}(\mathrm{GCC})+\mathrm{P}(\mathrm{G} \overline{\mathrm{C}} \overline{\mathrm{C}})+\mathrm{P}(\mathrm{RC} \overline{\mathrm{C}})+\mathrm{P}(\mathrm{R} \overline{\mathrm{C}} \mathrm{C})}$
$=\frac{\frac{4}{5} \times \frac{3}{4} \times \frac{3}{4}+\frac{4}{5} \times \frac{1}{4} \times \frac{1}{4}}{\frac{4}{5} \times \frac{3}{4} \times \frac{3}{4}+\frac{4}{5} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{5} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{5}+\frac{1}{4} \times \frac{3}{4}}$
$=\frac{40}{46}=\frac{20}{23}$
Paragraph for Question 6 and 7
Q. The probability of the drawn ball from $\mathrm{U}_{2}$ being white is –
(A) $\frac{13}{30}$
(B) $\frac{23}{30}$
(C) $\frac{19}{30}$
(D) $\frac{11}{30}$
[JEE 2011, 3+3]
Ans. (B)
Required probability
$=\frac{1}{2}\left(\frac{3}{5} \cdot 1+\frac{2}{5} \cdot \frac{1}{2}\right)+\frac{1}{2}\left(\frac{^{3} \mathrm{C}_{2}}{^{5} \mathrm{C}_{2}} \cdot 1+\frac{^{2} \mathrm{C}_{2}}{^{5} \mathrm{C}_{2}} \cdot \frac{1}{3}+\frac{^{3} \mathrm{C}_{1}^{2} \mathrm{C}_{1}}{^{5} \mathrm{C}_{2}} \cdot \frac{2}{3}\right)$
$=\frac{1}{2}\left(\frac{4}{5}\right)+\frac{1}{2}\left(\frac{3}{10}+\frac{1}{30}+\frac{2}{5}\right)=\frac{2}{5}+\frac{11}{30}=\frac{23}{30}$
Q. Given that the drawn ball from $\mathrm{U}_{2}$ is white, the probability that head appeared on the coin is –
(A) $\frac{17}{23}$
(B) $\frac{11}{23}$
(C) $\frac{15}{23}$
(D) $\frac{12}{23}$
[JEE 2011, 3+3]
Ans. (D)
Required probability
$=\frac{2 / 5}{2 / 5+11 / 30} \quad$ (using Baye's theorem)
$=\frac{12}{23}$
Q. Let $\mathrm{E}$ and $\mathrm{F}$ be two independent events. The probability that exactly one of them occurs $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25} .$ If $\mathrm{P}(\mathrm{T})$ denotes the probability of occurrence of the event $\mathrm{T}$, then
[JEE 2011, 4M]
[JEE 2011, 4M]
Ans. (A,D)
Q. A ship is fitted with three engines $\mathrm{E}_{1}, \mathrm{E}_{2}$ and $\mathrm{E}_{3} .$ The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{4} .$ For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and $\mathrm{X}_{1}, \mathrm{X}_{2}, \mathrm{X}_{3}$ denotes respectively the events that the engines $\mathrm{E}_{1}, \mathrm{E}_{2}$ and $\mathrm{E}_{3}$ are functioning. Which of the following is (are) true?
(A) $\mathrm{P}\left[\mathrm{X}_{1}^{\mathrm{c}} | \mathrm{X}\right]=\frac{3}{16}$
(B) $\mathrm{P}[\text { Exactly two engines of ship are functioning } | \mathrm{X}]=\frac{7}{8}$
(C) $\mathrm{P}\left[\mathrm{X} | \mathrm{X}_{2}\right]=\frac{5}{16}$
(D) $\mathrm{P}\left[\mathrm{X} | \mathrm{X}_{1}\right]=\frac{7}{16}$
[JEE 2012, 4M]
Ans. (B,D)
Q. Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4},$ each having six faces numbered $1,2,3,4,5$ and 6 are rolled simultaneously. The probability that $\mathrm{D}_{4}$ shows a number appearing on one of $\mathrm{D}_{1}$ $\mathrm{D}_{2}$ and $\mathrm{D}_{3}$ is –
(A) $\frac{91}{216}$
(B) $\frac{108}{216}$
(C) $\frac{125}{216}$
(D) $\frac{127}{216}$
[JEE 2012, 4M]
Ans. (A)
$1-\frac{^{6} \mathrm{C}_{1} \cdot 5^{3}}{6^{4}}=\frac{91}{216}$
Q. Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X} | \mathrm{Y})=\frac{1}{2}, \mathrm{P}(\mathrm{Y} | \mathrm{X})=\frac{1}{3}$ and $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{1}{6} \cdot$ Which of the following is (are) correct?
(A) $P(X \cup Y)=\frac{2}{3}$
(B) X and Y are independent
(C) $\mathrm{P}(\mathrm{E})=\frac{2}{5}, \mathrm{P}(\mathrm{F})=\frac{1}{5}$
(D) $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$
[JEE 2012, 4M]
Ans. (A,B)
Q. Four persons independently solve a certain problem correctly with probabilities $\frac{1}{2}, \frac{3}{4}, \frac{1}{4}, \frac{1}{8}$ Then the probability that the problem is solved correctly by at least one of them is
(A) $\frac{235}{256}$
(B) $\frac{21}{256}$
(B) $\frac{21}{256}$
(D) $\frac{253}{256}$
[JEE(Advanced) 2013, 2M]
Ans. (A)
$\mathrm{P}($ Problem is solved by at least one of them)
$=1-\mathrm{P}(\text { solved by none })$
$=1-\left(\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8}\right)=1-\frac{21}{256}=\frac{235}{256}$
Q. Of the three independent events $\mathrm{E}_{1}, \mathrm{E}_{2}$ and $\mathrm{E}_{3}$, the probability that only $\mathrm{E}_{1}$ occurs is $\alpha$ only $\mathrm{E}_{2}$ occurs is $\beta$ and only $\mathrm{E}_{3}$ occurs is $\gamma .$ Let the probability p that none of events $\mathrm{E}_{1} \cdot \mathrm{E}_{2}$ or $\mathrm{E}_{3}$ occurs satisfy the equations $(\alpha-2 \beta) \mathrm{p}=\alpha \beta$ and $(\beta-3 \gamma) \mathrm{p}=2 \beta \gamma .$ All the given probabilities are assumed of lie in the interval $(0,1) .$ Then
$\frac{\text { Pr obability of occurrence of } \mathrm{E}_{1}}{\text { Probability of occurrence of } \mathrm{E}_{3}}=$
[JEE-Advanced 2013, 4, (–1)]
Ans. 6
Q. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $\mathrm{B}_{2}$ is
(A) $\frac{116}{181}$
(B) $\frac{126}{181}$
(C) $\frac{65}{181}$
(D) $\frac{55}{181}$
[JEE(Advanced) 2013, 3, (–1)]
Ans. (D)
Q. If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3},$ the probability that all 3 drawn balls are of the same colour is
(A) $\frac{82}{648}$
(B) $\frac{90}{648}$
(C) $\frac{558}{648}$
(D) $\frac{566}{648}$
[JEE(Advanced) 2013, 3, (–1)]
Ans. (A)
Q. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is –
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{3}$
(D) $\frac{3}{4}$
[JEE(Advanced)-2014, 3(–1)]
Ans. (A)
Q. The probability that $\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}$ is odd, is –
(A) $\frac{29}{105}$
(B) $\frac{53}{105}$
(C) $\frac{57}{105}$
(D) $\frac{1}{2}$
[JEE(Advanced)-2014, 3(–1)]
Ans. (B)
Q. The probability that $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}$ are in an arithmetic progression, is –
(A) $\frac{9}{105}$
(B) $\frac{10}{105}$
(C) $\frac{11}{105}$
(D) $\frac{7}{105}$
[JEE(Advanced)-2014, 3(–1)]
Ans. (C)
Q. The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is
[JEE 2015, 4M, –0M]
Ans. 8
Q. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is $\frac{1}{3},$ then the correct option(s) with the possible values of $\mathrm{n}_{1}, \mathrm{n}_{2}, \mathrm{n}_{3}$ and $\mathrm{n}_{4}$ is (are)
(A) $\mathrm{n}_{1}=3, \mathrm{n}_{2}=3, \mathrm{n}_{3}=5, \mathrm{n}_{4}=15$
(B) $\mathrm{n}_{1}=3, \mathrm{n}_{2}=6, \mathrm{n}_{3}=10, \mathrm{n}_{4}=50$
(C) $\mathrm{n}_{1}=8, \mathrm{n}_{2}=6, \mathrm{n}_{3}=5, \mathrm{n}_{4}=20$
(D) $\mathrm{n}_{1}=6, \mathrm{n}_{2}=12, \mathrm{n}_{3}=5, \mathrm{n}_{4}=20$
[JEE 2015, 4M, –0M]
Ans. (A,B)
Required probability $=\frac{\left(\frac{\mathrm{n}_{3}}{\mathrm{n}_{3}+\mathrm{n}_{4}}\right)}{\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{3}}{\mathrm{n}_{3}+\mathrm{n}_{4}}}=\frac{1}{3}$
now check options.
Q. All ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\frac{1}{3},$ then the correct option(s) with the possible values of $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ is (are)
(A) $\mathrm{n}_{1}=4$ and $\mathrm{n}_{2}=6$
(B) $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$
(C) $\mathrm{n}_{1}=10$ and $\mathrm{n}_{2}=20$
(D) $\mathrm{n}_{1}=3$ and $\mathrm{n}_{2}=6$
[JEE 2015, 4M, –0M]
Ans. (C,D)
Required probability
$=\frac{\mathrm{n}_{1}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)} \frac{\left(\mathrm{n}_{1}-1\right)}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}-1\right)}+\frac{\mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)} \frac{\mathrm{n}_{1}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}-1\right)}=\frac{1}{3}$
$\Rightarrow \frac{\mathrm{n}_{1}^{2}+\mathrm{n}_{1} \mathrm{n}_{2}-\mathrm{n}_{1}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\left(\mathrm{n}_{1}+\mathrm{n}_{2}-1\right)}=\frac{1}{3}$
now check options.
Q. A computer producing factory has only two plants $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ Plant $\mathrm{T}_{1}$ produces $20 \%$ and plant $\mathrm{T}_{2}$ produces $80 \%$ of the total computers Jproduced. $7 \%$ of computers produced in the factory turn out to be defective. It is known what $\mathrm{P}\left(\text { computer turns out to be defective given that is produced in plant } \mathrm{T}_{1} \text { ) }=\right.$ $10 \mathrm{P}\left(\text { computer turns out to be defective given that it is produced in plant } \mathrm{T}_{2} \text { ) where } \mathrm{P}(\mathrm{E})\right.$ denotes the probability of an event E. A computer produces in the factory is randomly eselected and it does not turn out to be defective. Then the probabality that it is produced plant $\mathrm{T}_{2}$ is
(A) $\frac{36}{73}$
(B) $\frac{47}{79}$
(C) $\frac{78}{93}$
(D) $\frac{75}{83}$
[JEE(Advanced)-2016]
Ans. (C)
$\mathrm{P}\left(\mathrm{T}_{1}\right)=\frac{20}{100}$
$\mathrm{P}\left(\mathrm{T}_{2}\right)=\frac{80}{100}$
Let $\mathrm{P}\left(\frac{\mathrm{D}}{\mathrm{T}_{2}}\right)=\mathrm{x}$
$\mathrm{P}\left(\frac{\mathrm{D}}{\mathrm{T}_{1}}\right)=10 \mathrm{x}$
$P(D)=\frac{7}{100} \quad(\text { given })$
$\mathrm{P}\left(\mathrm{T}_{1}\right) \mathrm{P}\left(\frac{\mathrm{D}}{\mathrm{T}_{1}}\right)+\mathrm{P}\left(\mathrm{T}_{2}\right) \mathrm{P}\left(\frac{\mathrm{D}}{\mathrm{T}_{2}}\right)=\frac{7}{100}$
$\frac{20}{100} \times 10 x+\frac{80}{100} \times x=\frac{7}{100}$
$\mathrm{x}=\frac{1}{40}$
$P\left(\frac{D}{T_{2}}\right)=\frac{1}{40} \Rightarrow P\left(\frac{\bar{D}}{T_{2}}\right)=\frac{39}{40}$
$P\left(\frac{D}{T_{1}}\right)=\frac{10}{40} \Rightarrow P\left(\frac{\bar{D}}{T_{1}}\right)=\frac{30}{40}$
$P\left(\frac{T_{2}}{\bar{D}}\right)=\frac{\frac{80}{100} \times \frac{39}{40}}{\frac{20}{100} \times \frac{30}{40}+\frac{80}{100} \times \frac{39}{40}}=\frac{78}{93}$
Q. P(X > Y) is-
(A) $\frac{1}{4}$
(B) $\frac{5}{12}$
(C) $\frac{1}{2}$
$(D) \frac{7}{12}$
[JEE(Advanced)-2016]
Ans. (B)
$\mathrm{P}(\mathrm{X}>\mathrm{Y})=\mathrm{P}\left(\mathrm{T}_{1} \text { win }\right) \mathrm{P}\left(\mathrm{T}_{1} \text { win }\right)+\mathrm{P}\left(\mathrm{T}_{1} \text { win) } \mathrm{P}(\text { match draw })+\mathrm{P}(\text { match draw }) . \mathrm{P}\left(\mathrm{T}_{1} \text { win }\right)\right.$
Q. P(X = Y) is-
(A) $\frac{11}{36}$
(B) $\frac{1}{3}$
(C) $\frac{13}{36}$
(D) $\frac{1}{2}$
[JEE(Advanced)-2016]
Ans. (C)
$\mathrm{P}(\mathrm{X}=\mathrm{Y})=\mathrm{P}(\text { match draw }) \mathrm{P}(\text { match Draw })+\mathrm{P}\left(\mathrm{T}_{1} \text { win) } \mathrm{P}\left(\mathrm{T}_{2} \text { win }\right)+\mathrm{P}\left(\mathrm{T}_{2} \text { win) } \mathrm{P}\left(\mathrm{T}_{1} \text { win }\right)\right.\right.$
$=\frac{1}{6} \times \frac{1}{6}+\frac{1}{2} \times \frac{1}{3}+\frac{1}{3} \times \frac{1}{2}=\frac{13}{36}$
Q. Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X})=\frac{1}{3}, \mathrm{P}(\mathrm{X} | \mathrm{Y})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{Y} | \mathrm{X})=\frac{2}{5} .$ Then
[JEE(Advanced)-2017]
[JEE(Advanced)-2017]
Ans. (A)
$P(x)=\frac{1}{3} ; \frac{P(X \cap Y)}{P(Y)}=\frac{1}{2} ; \frac{P(Y \cap X)}{P(X)}=\frac{2}{5}$
from this information, we get
$\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{2}{15} ; \mathrm{P}(\mathrm{Y})=\frac{4}{15}$
$\therefore \mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}$
$\mathrm{P}(\overline{\mathrm{X}} / \mathrm{Y})=\frac{\mathrm{P}(\overline{\mathrm{X}} \cap \mathrm{Y})}{\mathrm{P}(\mathrm{Y})}=\frac{\mathrm{P}(\mathrm{Y})-\mathrm{P}(\mathrm{X} \cap \mathrm{Y})}{\mathrm{P}(\mathrm{Y})}$
$\Rightarrow \mathrm{P}(\overline{\mathrm{X}} / \mathrm{Y})=1-\frac{2 / 15}{4 / 15}=\frac{1}{2}$
Q. Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is
(A) $\frac{36}{55}$
(B) $\frac{6}{11}$
(C) $\frac{5}{11}$
(D) $\frac{1}{2}$
[JEE(Advanced)-2017]
Ans. (A)
Q. The probability that, on the examination day, the student $\mathrm{S}_{1}$ gets the previously allotted seat $\mathrm{R}_{1}$ and $\mathrm{NONE}$ of the remaining students gets the seat previously allotted to him/her is –
(A) $\frac{3}{40}$
(B) $\frac{1}{8}$
(C) $\frac{7}{40}$
$(D) \frac{1}{5}$
Ans. (C)
Q. For $\mathrm{i}=1,2,3,4,$ let $\mathrm{T}_{\mathrm{i}}$ denote the event that the students $\mathrm{S}_{\mathrm{i}}$ and $\mathrm{S}_{\mathrm{i}+1}$ do $\mathrm{NOT}$ sit adjacent to each other on the day of the examination. Then the probability of the event $\mathrm{T}_{1} \cap \mathrm{T}_{2} \cap \mathrm{T}_{3} \cap$ $\mathrm{T}_{4}$ is-
(A) $\frac{1}{15}$
(B) $\frac{1}{10}$
(C) $\frac{7}{60}$
(D) $\frac{1}{5}$
Ans. (C)
$\begin{aligned} \mathrm{n}\left(\mathrm{T}_{1} \cap \mathrm{T}_{2} \cap \mathrm{T}_{3} \cap \mathrm{T}_{4}\right) &=\mathrm{Total}-\mathrm{n}\left(\overline{\mathrm{T}}_{1} \cup \overline{\mathrm{T}}_{2} \cup \overline{\mathrm{T}}_{3} \cup \overline{\mathrm{T}}_{4}\right) \\ &=5 !-\left(^{4} \mathrm{C}_{1} 4 ! 2 !-\left(^{3} \mathrm{C}_{1} \cdot 3 ! 2 !+^{3} \mathrm{C}_{1} 3 ! 2 ! 2 !\right)+\left(^{2} \mathrm{C}_{1} 2 ! 2 !+^{4} \mathrm{C}_{1} \cdot 2 \cdot 2 !\right)-2\right) \\ &=14 \\ \text { Probability }=\frac{14}{5 !}=\frac{7}{60} \end{aligned}$
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