2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of ${{125} \over 4}{p_2}$ is ___________.
The value of ${{125} \over 4}{p_2}$ is ___________.
Correct Answer: 24.5
Explanation:
p2 = 1 $-$ p (all three numbers are > 40)
$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$
So, ${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$
$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$
So, ${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is ............
Correct Answer: 6
Explanation:
It is given that the probability, a missile hits a target successfully $p = {3 \over 4}$, so the probability to not hits the target is ${1 \over 4}$. And it is also given that to destroy the target completely, at least three successful hits are required.
Now, according to the question, let the minimum number of missiles required to fired is n, so
$^n{C_3}{p^3}{q^{n - 3}}{ + ^n}{C_4}{p^4}{q^{n - 4}} + ...{ + ^n}{C_n}{p^n}\, \ge \,0.95$
$ \Rightarrow 1 - \left\{ {^n{C_0}{{\left( {{1 \over 4}} \right)}^n}{ + ^n}{C_1}\left( {{3 \over 4}} \right){{\left( {{1 \over 4}} \right)}^{n - 1}}{ + ^n}{C_2}{{\left( {{3 \over 4}} \right)}^2}{{\left( {{1 \over 4}} \right)}^{n - 2}}} \right\}\, \ge \,0.95$
$ \Rightarrow 1 - {{95} \over {100}}\, \ge \,{1 \over {{4^n}}} + {{3n} \over {{4^n}}} + {{n(n - 1)} \over 2}{9 \over {{4^n}}}$
$ \Rightarrow {{{4^n}} \over {20}}\, \ge \,{{2 + 6n + 9{n^2} - 9n} \over 2}$
$ \Rightarrow 10(9{n^2} - 3n + 2)\, \le \,{4^n}$
Now, at n = 3, LHS = 720, RHS = 64
at n = 4, LHS = 1340, RHS = 256
at n = 5, LHS = 2120, RHS = 1024
at n = 6, LHS = 3080, RHS = 4096
Hence, n = 6 missiles should be fired.
Now, according to the question, let the minimum number of missiles required to fired is n, so
$^n{C_3}{p^3}{q^{n - 3}}{ + ^n}{C_4}{p^4}{q^{n - 4}} + ...{ + ^n}{C_n}{p^n}\, \ge \,0.95$
$ \Rightarrow 1 - \left\{ {^n{C_0}{{\left( {{1 \over 4}} \right)}^n}{ + ^n}{C_1}\left( {{3 \over 4}} \right){{\left( {{1 \over 4}} \right)}^{n - 1}}{ + ^n}{C_2}{{\left( {{3 \over 4}} \right)}^2}{{\left( {{1 \over 4}} \right)}^{n - 2}}} \right\}\, \ge \,0.95$
$ \Rightarrow 1 - {{95} \over {100}}\, \ge \,{1 \over {{4^n}}} + {{3n} \over {{4^n}}} + {{n(n - 1)} \over 2}{9 \over {{4^n}}}$
$ \Rightarrow {{{4^n}} \over {20}}\, \ge \,{{2 + 6n + 9{n^2} - 9n} \over 2}$
$ \Rightarrow 10(9{n^2} - 3n + 2)\, \le \,{4^n}$
Now, at n = 3, LHS = 720, RHS = 64
at n = 4, LHS = 1340, RHS = 256
at n = 5, LHS = 2120, RHS = 1024
at n = 6, LHS = 3080, RHS = 4096
Hence, n = 6 missiles should be fired.
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ..........
Correct Answer: 8
Explanation:
Let an event E of sum of outputs are perfect square (i.e., 4 or 9), so
E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
and an event F of sum of outputs are prime numbers
(i.e., 2, 3, 5, 7, 11) so
F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}
and event T of sum of outputs are odd numbers
(i.e., 3, 5, 7, 9, 11)
T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}
Now, required probability $p = P(T/E)$
$ = {{P(T \cap E)} \over {P(E)}}$
where, ${P(T \cap E)}$ = probability of occurring perfect square odd number before prime
$ = \left( {{4 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{4 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{4 \over {36}}} \right) + ....\infty $
$ = {{{4 \over {36}}} \over {1 - {{14} \over {36}}}} = {4 \over {22}} = {2 \over {11}}$
and P(E) = probability of occurring perfect square before prime
$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $
$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$
$ \therefore $ $P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$
E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
and an event F of sum of outputs are prime numbers
(i.e., 2, 3, 5, 7, 11) so
F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}
and event T of sum of outputs are odd numbers
(i.e., 3, 5, 7, 9, 11)
T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}
Now, required probability $p = P(T/E)$
$ = {{P(T \cap E)} \over {P(E)}}$
where, ${P(T \cap E)}$ = probability of occurring perfect square odd number before prime
$ = \left( {{4 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{4 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{4 \over {36}}} \right) + ....\infty $
$ = {{{4 \over {36}}} \over {1 - {{14} \over {36}}}} = {4 \over {22}} = {2 \over {11}}$
and P(E) = probability of occurring perfect square before prime
$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $
$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$
$ \therefore $ $P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 1 Offline
Let S be the sample space of all 3 $ \times $ 3 matrices with entries from the set {0, 1}. Let the events E1 and E2 be given by
E1 = {A$ \in $S : det A = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
If a matrix is chosen at random from S, then the conditional probability P(E1 | E2) equals ...............
E1 = {A$ \in $S : det A = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
If a matrix is chosen at random from S, then the conditional probability P(E1 | E2) equals ...............
Correct Answer: 0.5
Explanation:
Given sample space (S) of al 3 $ \times $ 3 matrices with entries from the set {0, 1} and events
E1 = {A$ \in $S : det(A) = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
For event E2, means sum of entries of matrix A is 7, then we need seven 1s and two 0s.
$ \therefore $ Number of different possible matrices = ${{9!} \over {7!2!}}$
$ \Rightarrow $ n(E2) = 36
For event E1, |A| = 0, both the zeroes must be in same row/column.
$ \therefore $ Number of matrices such that their determinant is zero = $6 \times {{3!} \over {2!}}$ = 18 = $n({E_1} \cap {E_2})$
$ \therefore $ Required probability,
$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {{n({E_1} \cap {E_2})} \over {n({E_2})}}$
$ = {{18} \over {36}}$
$ = {1 \over 2} = 0.50$
E1 = {A$ \in $S : det(A) = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
For event E2, means sum of entries of matrix A is 7, then we need seven 1s and two 0s.
$ \therefore $ Number of different possible matrices = ${{9!} \over {7!2!}}$
$ \Rightarrow $ n(E2) = 36
For event E1, |A| = 0, both the zeroes must be in same row/column.
$ \therefore $ Number of matrices such that their determinant is zero = $6 \times {{3!} \over {2!}}$ = 18 = $n({E_1} \cap {E_2})$
$ \therefore $ Required probability,
$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {{n({E_1} \cap {E_2})} \over {n({E_2})}}$
$ = {{18} \over {36}}$
$ = {1 \over 2} = 0.50$
2015
JEE Advanced
Numerical
JEE Advanced 2015 Paper 1 Offline
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
Correct Answer: 8
Explanation:
Let the coin is tossed $n$ times.
$\because p$ (at least 2 heads) $=1-[p$ (one heads) $+p$ (No heads)$\}$
As we know, by binominal probability theorem the probability of getting $r$ success in $n$ trials with $p$ being the probability of success and $q$ be the probability of failure, is given by ${ }^n \mathrm{C}_r(p)^r(q)^{n-r}$.
Let head be considered as the success and tail be the failure probability of getting head in a toss $=p=\frac{1}{2}$ and probability of getting tail in a toss $=q=\frac{1}{2}$
$\begin{aligned} & \therefore \mathrm{P}(\text { one head })={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{n-1} \\ & ={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^n \\ & \Rightarrow \mathrm{P} \text { (one head) }=n\left(\frac{1}{2}\right)^n \\ & \text { Similarly, } \mathrm{P}(\text {No heads})={ }^n \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^n \\ & \Rightarrow \mathrm{P}(\text { No heads })=\left(\frac{1}{2}\right)^n \quad\left\{\because n_{c_o}=1\right\} \\ & \therefore \mathrm{P} \text { (at least } 2 \text { heads) }=1-\left\{n\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^n\right\} \end{aligned}$
$\begin{aligned} & \qquad=1-\frac{(n+1)}{2^n} \\ & \because P(\text { at least } 2 \text { heads }) \geq 0.96 \\ & \Rightarrow 1-\frac{(n+1)}{2^n} \geq 0.96 \\ & \Rightarrow 0.04 \geq \frac{n+1}{2^n} \\ & \Rightarrow \frac{1}{25} \geq \frac{n+1}{2^n} \\ & \Rightarrow \frac{2^n}{n+1} \geq 25 \\ & \Rightarrow n \geq 8 \end{aligned}$
So, the minimum number of times a fair coin to be tossed is 8.
Hint:
(i) P (at least 2 heads) $=1-\{\mathrm{P}$ (one head} + $\mathrm{P}$(No heads)}
(ii) Use binomial probability theorem and simplify it.
2013
JEE Advanced
Numerical
JEE Advanced 2013 Paper 1 Offline
Of the three independent events ${E_1},{E_2}$ and ${E_3},$ the probability that only ${E_1}$ occurs is $\alpha ,$ only ${E_2}$ occurs is $\beta $ and only ${E_3}$ occurs is $\gamma .$ Let the probability $p$ that none of events ${E_1},{E_2}$ or ${E_3}$ occurs satisfy the equations $\left( {\alpha -2\beta } \right)p = \alpha \beta $ and $\left( {\beta - 3\gamma } \right)p = 2\beta \gamma .$ All the given probabilities are assumed to lie in the interval $(0, 1)$.
Then ${{\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_1}} \over {\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_3}}}$
Correct Answer: 6
Explanation:
Given, three independent events $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$.
Probability that only
$\mathrm{E}_1 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\alpha$
Probability that only
$\mathrm{E}_2 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\beta$
Probability that only
$\mathrm{E}_3 \text { occurs }=P\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)=\gamma$
Probability that none of $E_1, E_2$ or $E_3$ occurs
$=P\left(\overline{\mathrm{E}}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\rho$
$\begin{aligned} & \text { Let } \mathrm{P}\left(\mathrm{E}_1\right)=x, \mathrm{P}\left(\mathrm{E}_2\right)=y \text { and } \mathrm{P}\left(\mathrm{E}_3\right)=z \\ & \alpha=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \alpha=x(1-y)(1-z) \quad \text{... (i)}\\ & \beta=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \beta=(1-x) \cdot(y)(1-z) \quad \text{... (ii)}\\ & \gamma=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\mathrm{E}_3\right) \\ & \Rightarrow \quad \gamma=(1-x)(1-y) z \quad \text{... (iii)}\\ & \rho=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \rho=(1-x)(1-y)(1-z) \quad \text{... (iv)}\\ \end{aligned}$
Given $(\alpha-2 \beta) \rho=a \beta$ and $(\beta-3 \gamma) \rho=2 b \gamma$
$\begin{aligned} \Rightarrow \quad & {[x(1-y)(1-z)-2(1-x) y(1-z)](1-x) } \\ & (1-y)(1-z) \\ & =x(1-y)(1-z) \cdot(1-x) y(1-z) \text { and } \\ & {[(1-x) y(1-z)-3(1-x)(1-y) z](1-x) } \\ & (1-y)(1-z) \\ & =2(1-x) y(1-z) \cdot(1-x)(1-y) z \\ \Rightarrow \quad & {[x-x y-2 y+2 x y](1-x)(1-y)(1-z)^2 } \\ & =x y(1-x)(1-y)(1-z)^2 \text { and } \\ & {[y-y z-3 z+3 y z](1-x)^2(1-y)(1-z) } \\ & =2 y z(1-x)^2(1-y)(1-z) \\ \Rightarrow \quad & (x-2 y+x y)=x y \text { and }(y-3 z+2 y z)=2 y z \\ \Rightarrow \quad & x=2 y \text { and } y=3 z \\ \Rightarrow \quad & \frac{x}{2}=y=3 z \\ \Rightarrow \quad & \frac{x}{Z}=6 \end{aligned}$
$\Rightarrow \frac{\text { Probability of occurrence of } E_1}{\text { Probability of occurrence of } E_3}=\frac{x}{Z}=6$
Hints :
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are independent events, then probability of occurrence of only event A
$=\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})$.
2021
JEE Advanced
MSQ
JEE Advanced 2021 Paper 1 Online
Let E, F and G be three events having probabilities $P(E) = {1 \over 8}$, $P(F) = {1 \over 6}$ and $P(G) = {1 \over 4}$, and let P (E $\cap$ F $\cap$ G) = ${1 \over {10}}$. For any event H, if Hc denotes the complement, then which of the following statements is (are) TRUE?
A.
$P(E \cap F \cap {G^c}) \le {1 \over {40}}$
B.
$P({E^c} \cap F \cap G) \le {1 \over {15}}$
C.
$P(E \cup F \cup G) \le {{13} \over {24}}$
D.
$P({E^c} \cup {F^c} \cup {G^c}) \le {5 \over {12}}$
2019
JEE Advanced
MSQ
JEE Advanced 2019 Paper 1 Offline
There are three bags B1, B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, B2 and B3 have probabilities ${3 \over {10}}$, ${3 \over {10}}$ and ${4 \over {10}}$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
A.
Probability that the chosen ball is green, given that the selected bag is B3, equals ${3 \over 8}$.
B.
Probability that the selected bag is B3, given that the chosen ball is green, equals ${5 \over 13}$.
C.
Probability that the chosen ball is green equals ${39 \over 80}$.
D.
Probability that the selected bag is B3 and the chosen ball is green equals ${3 \over 10}$.
2017
JEE Advanced
MSQ
JEE Advanced 2017 Paper 1 Offline
Let X and Y be two events such that $P(X) = {1 \over 3}$, $P(X|Y) = {1 \over 2}$ and $P(Y|X) = {2 \over 5}$. Then
A.
$P(Y) = {4 \over {15}}$
B.
$P(X'|Y) = {1 \over 2}$
C.
$P(X \cup Y) = {2 \over 5}$
D.
$P(X \cap Y) = {1 \over 5}$
2015
JEE Advanced
MSQ
JEE Advanced 2015 Paper 2 Offline
Let ${n_1}$ and ${n_2}$ be the number of red and black balls, respectively, in box ${\rm I}$. Let ${n_3}$ and ${n_4}$ be the number of red and black balls, respectively, in box ${\rm I}{\rm I}.$
A ball is drawn at random from box ${\rm I}$ and transferred to box ${\rm I}$${\rm I}.$ If the probability of drawing a red ball from box ${\rm I},$ after this transfer, is ${1 \over 3},$ then the correct option(s) with the possible values of ${n_1}$ and ${n_2}$ is(are)
A.
${n_1} = 4$ and ${n_2} = 6$
B.
${n_1} = 2$ and ${n_2} = 3$
C.
${n_1} = 10$ and ${n_2} = 20$
D.
${n_1} = 3$ and ${n_2} = 6$
2015
JEE Advanced
MSQ
JEE Advanced 2015 Paper 2 Offline
Let ${n_1}$ and ${n_2}$ be the number of red and black balls, respectively, in box ${\rm I}$. Let ${n_3}$ and ${n_4}$ be the number of red and black balls, respectively, in box ${\rm I}{\rm I}.$
One of the two boxes, box ${\rm I}$ and box ${\rm I}{\rm I},$ 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 ${\rm I}{\rm I}$ is ${1 \over 3},$ then the correct option(s) with the possible values of ${n_1}$ ${n_2},$ ${n_3}$ and ${n_4}$ is (are)
A.
${n_1} = 3,{n_2} = 3,{n_3} = 5,{n_4} = 15$
B.
${n_1} = 3,{n_2} = 6,{n_3} = 10,{n_4} = 50$
C.
${n_1} = 8,{n_2} = 6,{n_3} = 5,{n_4} = 20$
D.
${n_1} = 6,{n_2} = 12,{n_3} = 5,{n_4} = 20$
2012
JEE Advanced
MSQ
IIT-JEE 2012 Paper 2 Offline
Let $X$ and $Y$ be two events such that $P\left( {X|Y} \right) = {1 \over 2},$ $P\left( {Y|X} \right) = {1 \over 3}$ and $P\left( {X \cap Y} \right) = {1 \over 6}.$ Which of the following is (are) correct ?
A.
$P\left( {X \cup Y} \right) = {2 \over 3}$
B.
$X$ and $Y$ are independent
C.
$X$ and $Y$ are not independent
D.
$P\left( {{X^c} \cap Y} \right) = {1 \over 3}$
2012
JEE Advanced
MSQ
IIT-JEE 2012 Paper 1 Offline
A ship is fitted with three engines ${E_1},{E_2}$ and ${E_3}$. The engines function independently of each other with respective probabilities ${1 \over 2},{1 \over 4}$ and ${1 \over 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 Let ${X_1},{X_2}$ and ${X_3}$ denote respectively the events that the engines ${E_1},{E_2}$ and ${E_3}$ are functioning. Which of the following is (are) true?
A.
$P\left[ {X_1^c|X} \right] = {3 \over {16}}$
B.
$P$ [exactly two engines of the ship are functioning $\left. {|X} \right] = {7 \over 8}$
C.
$P\left[ {X|{X_2}} \right] = {5 \over {16}}$
D.
$P\left[ {X|{X_1}} \right] = {7 \over {16}}$
2011
JEE Advanced
MSQ
IIT-JEE 2011 Paper 2 Offline
Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\,{{11} \over {25}}$ and the probability of none of them occurring is $\,{{2} \over {25}}$. If $P(T)$ denotes the probability of occurrence of the event $T,$ then
A.
$P\left( E \right) = {4 \over 5},P\left( F \right) = {3 \over 5}$
B.
$P\left( E \right) = {1 \over 5},P\left( F \right) = {2 \over 5}$
C.
$P\left( E \right) = {2 \over 5},P\left( F \right) = {1 \over 5}$
D.
$P\left( E \right) = {3 \over 5},P\left( F \right) = {4 \over 5}$
1999
JEE Advanced
MSQ
IIT-JEE 1999
The probabilities that a student passes in Mathematics, Physics and Chemistry are $m, p$ and $c,$ respectively. Of these subjects, the student has a $75%$ chance of passing in at least one, a $50$% chance of passing in at least two, and a $40$% chance of passing in exactly two. Which of the following relations are true?
A.
$p+m+c=19/20$
B.
$p+m+c=27/20$
C.
$pmc=1/10$
D.
$pmc=1/4$
1998
JEE Advanced
MSQ
IIT-JEE 1998
If $\overline E $ and $\overline F $ are the complementary events of events $E$ and $F$ respectively and if $0 < P\left( F \right) < 1,$ then
A.
$P\left( {E/F} \right) + P\left( {\overline E /F} \right) = 1$
B.
$P\left( {E/F} \right) + P\left( {E/\overline F } \right) = 1$
C.
$P\left( {\overline E /F} \right) + P\left( {E/\overline F } \right) = 1$
D.
$P\left( {E/\overline F } \right) + P\left( {\overline E /\overline F } \right) = 1$
1995
JEE Advanced
MSQ
IIT-JEE 1995 Screening
Let $0 < P\left( A \right) < 1,0 < P\left( B \right) < 1$ and
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( A \right)P\left( B \right)$ then
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( A \right)P\left( B \right)$ then
A.
$P\left( {B/A} \right) = P\left( B \right) - P\left( A \right)$
B.
$P\left( {A' - B'} \right) = P\left( {A'} \right) - P\left( {B'} \right)$
C.
$P\left( {A \cup B} \right)' = P\left( {A'} \right) - P\left( {B'} \right)$
D.
$P\left( {A/B} \right) = P\left( A \right)$
1993
JEE Advanced
MSQ
IIT-JEE 1993
$E$ and $F$ are two independent events. The probability that both $E$ and $F$ happen is $1/12$ and the probability that neither $E$ nor $F$ happens is $1/2.$ Then,
A.
$\,P\left( E \right) = 1/3,P\left( F \right) = 1/4$
B.
$\,P\left( E \right) = 1/2,P\left( F \right) = 1/6$
C.
$\,P\left( E \right) = 1/6,P\left( F \right) = 1/2$
D.
$\,P\left( E \right) = 1/4,P\left( F \right) = 1/3$
1991
JEE Advanced
MSQ
IIT-JEE 1991
For any two events $A$ and $B$ in a simple space
A.
$P\left( {A/B} \right) \ge {{P\left( A \right) + P\left( B \right) - 1} \over {P\left( B \right)}},P\left( B \right) \ne 0$ is always true
B.
$P\left( {A \cap \overline B } \right) = P\left( A \right) - P\left( {A \cap B} \right)\,\,$ does not hold
C.
$P\left( {A \cup B} \right) = 1 - P\left( {\overline A } \right)P\left( {\overline B } \right),$ if $A$ and $B$ are independent
D.
$P\left( {A \cup B} \right) = 1 - P\left( {\overline A } \right)P\left( {\overline B } \right),$ if $A$ and $B$ are disjoint.
1989
JEE Advanced
MSQ
IIT-JEE 1989
If $E$ and $F$ are independent events such that $0 < P\left( E \right) < 1$ and $0 < P\left( F \right) < 1,$ then
A.
$E$ and $F$ are mutually exclusive
B.
$E$ and ${F^c}$ (the complement of the event $F$) are independent
C.
${E^c}$ and ${F^c}$ are independent
D.
$P\left( {E|F} \right) + P\left( {{E^c}|F} \right) = 1.$
1988
JEE Advanced
MSQ
IIT-JEE 1988
For two given events $A$ and $B,$ $P\left( {A \cap B} \right)$
A.
not less than $P\left( A \right) + P\left( B \right) - 1$
B.
not greater than $P\left( A \right) + P\left( B \right)$
C.
equal to $P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\,\,$
D.
$P\left( A \right) + P\left( B \right) + P\left( {A \cup B} \right)\,\,$
1984
JEE Advanced
MSQ
IIT-JEE 1984
If $M$ and $N$ are any two events, the probability that exactly one of them occurs is
A.
$P\left( M \right) + P\left( N \right) - 2P\left( {M \cap N} \right)$
B.
$P\left( M \right) + P\left( N \right) - P\left( {M \cap N} \right)$
C.
$P\left( {{M^c}} \right) + P\left( {{N^c}} \right) - 2P\left( {{M^c} \cap {N^c}} \right)$
D.
$P\left( {M \cap {N^c}} \right) + P\left( {{M^c} \cap N} \right)$
2005
JEE Advanced
Numerical
IIT-JEE 2005
A person goes to office either by car, scooter, bus or train, the probability of which being ${1 \over 7},{3 \over 7},{2 \over 7}$ and ${1 \over 7}$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is ${2 \over 9},{1 \over 9},{4 \over 9}$ and ${1 \over 9}$ respectively. Given that he reached office in time, then what is the probability that he travelled by a car.
Correct Answer: $${1 \over 7}$$
2004
JEE Advanced
Numerical
IIT-JEE 2004
$A$ and $B$ are two independent events. $C$ is even in which exactly one of $A$ or $B$ occurs. Prove that $P\left( C \right) \ge P\left( {A \cup B} \right)P\left( {\overline A \cap \overline B } \right)$
Correct Answer: Solve it.
2004
JEE Advanced
Numerical
IIT-JEE 2004
A box contains $12$ red and $6$ white balls. Balls are drawn from the box one at a time without replacement. If in $6$ draws there are at least $4$ white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)
Correct Answer: <img class="question-image" src="https://imagex.cdn.examgoal.net/r3yIiW32oClH1Ip2G/QNz79hshQVDYMLKOiHGajvVEQt8Nz/41RpeyHlHrSzgzx2n7SUoe/uploadfile.jpg" loading="lazy" alt="IIT-JEE 2004 Mathematics - Probability Question 54 English Answer">
2003
JEE Advanced
Numerical
IIT-JEE 2003
$A$ is targeting to $B, B$ and $C$ are targeting to $A.$ Probability of hitting the target by $A,B$ and $C$ are ${2 \over 3},{1 \over 2}$ and ${1 \over 3}$ respectively. If $A$ is hit then find the probability that $B$ hits the target and $C$ does not.
Correct Answer: $${1 \over 2}$$
2003
JEE Advanced
Numerical
IIT-JEE 2003
For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is $p.$ If he fails in one of the exams then the probability of his passing in the next exam is ${p \over 2}$ otherwise it remains the same. Find the probability that he will qualify.
Correct Answer: $$2{p^2} - {p^3}$$
2002
JEE Advanced
Numerical
IIT-JEE 2002
A box contains $N$ coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1/2$, while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. what is the probability that the coin drawn is fair?
Correct Answer: $${{9m} \over {m + 8N}}$$
2001
JEE Advanced
Numerical
IIT-JEE 2001
An unbiased die, with faces numbered $1,2,3,4,5,6,$ is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that, among the numbers $1,2,3,4,5,6,$ only three numbers appear in this list?
Correct Answer: $${{6{c_3}\left[ {{3^n} - 3\left( {{2^n}} \right) + 3} \right]} \over {{6^n}}}$$
2001
JEE Advanced
Numerical
IIT-JEE 2001
An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?
Correct Answer: $$\,{m \over {m + n}}$$
2000
JEE Advanced
Numerical
IIT-JEE 2000
A coin has probability $p$ of showing head when tossed. It is tossed $n$ times. Let ${p_n}$ denote the probability that no two (or more) consecutive heads occur. Prove that ${p_1} = 1,{p_2} = 1 - {p^2}$ and ${p_n} = \left( {1 - p} \right).\,\,{p_{n - 1}} + p\left( {1 - p} \right){p_{n - 2}}$ for all $n \ge 3.$
Correct Answer: Solve it.
1999
JEE Advanced
Numerical
IIT-JEE 1999
Eight players ${P_1},{P_2},.....{P_8}$ play a knock-out tournament. It is known that whenever the players ${P_i}$ and ${P_j}$ play, the player ${P_i}$ will win if $i < j.$ Assuming that the players are paired at random in each round, what is the probability that the player ${P_4}$ reaches the final?
Correct Answer: $$4/35$$
1998
JEE Advanced
Numerical
IIT-JEE 1998
Three players, $A,B$ and $C,$ toss a coin cyclically in that order (that is $A, B, C, A, B, C, A, B,...$) till a head shows. Let $p$ be the probability that the coin shows a head. Let $\alpha ,\,\,\,\beta $ and $\gamma $ be, respectively, the probabilities that $A, B$ and $C$ gets the first head. Prove that $\beta = \left( {1 - p} \right)\alpha $ Determine $\alpha ,\beta $ and $\gamma $ (in terms of $p$).
Correct Answer: $$\alpha = {p \over {1 - {{\left( {1 - p} \right)}^3}}},$$ $$\beta = {{\left( {1 - p} \right)p} \over {1 - {{\left( {1 - p} \right)}^3}}},$$ $$\gamma = {{p{{\left( {1 - p} \right)}^2}} \over {1 - {{\left( {1 - p} \right)}^3}}}\,\,$$
1998
JEE Advanced
Numerical
IIT-JEE 1998
Let ${C_1}$ and ${C_2}$ be the graphs of the functions $y = {x^2}$ and $y = 2x,$ $0 \le x \le 1$ respectively. Let ${C_3}$ be the graph of a function $y=f(x),$ $0 \le x \le 1,$ $f(0)=0.$ For a point $P$ on ${C_1},$ let the lines through $P,$ parallel to the axes, meet ${C_2}$ and ${C_3}$ at $Q$ and $R$ respectively (see figure.) If for every position of $P$ (on ${C_1}$ ), the areas of the shaded regions $OPQ$ and $ORP$ are equal, determine the function$f(x).$
Correct Answer: $$f\left( x \right) = {x^3} - {x^2}$$
1997
JEE Advanced
Numerical
IIT-JEE 1997
If $p$ and $q$ are chosen randomly from the set $\left\{ {1,2,3,4,5,6,7,8,9,10} \right\},$ with replacement, determine the probability that the roots of the equation ${x^2} + px + q = 0$ are real.
Correct Answer: $$0.62$$
1996
JEE Advanced
Numerical
IIT-JEE 1996
In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, $3$ in the front and $4$ at the back? How many seating arrangements are possible if $3$ girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of $3$ girls sitting together in a back row on adjacent seats?
Correct Answer: $$7\left( {13!} \right),12!,1/9!$$
1994
JEE Advanced
Numerical
IIT-JEE 1994
An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2, 3,4,.....12$ is picked and the number on the card is noted. What is the probability that the noted number is either $7$ or $8$?
Correct Answer: $$0.2436$$
1993
JEE Advanced
Numerical
IIT-JEE 1993
Numbers are selected at random, one at a time, from the two- digit numbers $00, 01, 02 ......, 99$ with replacement. An event $E$ occurs if only if the product of the two digits of a selected number is $18$. If four numbers are selected, find probability that the event $E$ occurs at least $3$ times.
Correct Answer: $${{97} \over {{{\left( {25} \right)}^4}}}$$
1992
JEE Advanced
Numerical
IIT-JEE 1992
A lot contains $50$ defective and $50$ non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events $A, B, C$ are defined as
$A=$ (the first bulbs is defective)
$B=$ (the second bulbs is non-defective)
$C=$ (the two bulbs are both defective or both non defective)
Determine whether
(i) $\,\,\,\,\,$ $A, B, C$ are pairwise independent
(ii)$\,\,\,\,\,$ $A, B, C$ are independent
$A=$ (the first bulbs is defective)
$B=$ (the second bulbs is non-defective)
$C=$ (the two bulbs are both defective or both non defective)
Determine whether
(i) $\,\,\,\,\,$ $A, B, C$ are pairwise independent
(ii)$\,\,\,\,\,$ $A, B, C$ are independent
Correct Answer: $$A, B, C$$ are pairwise independent but $$A, B, C$$ are dependent.
1991
JEE Advanced
Numerical
IIT-JEE 1991
In a test an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he make a guess is $1/3$ and the probability that he copies the answer is $1/6$. The probability that his answer is correct given that he copied it, is $1/8$. Find the probability that he knew the answer to the questions given that he correctly answered it.
Correct Answer: $$24/29$$
1990
JEE Advanced
Numerical
IIT-JEE 1990
A is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P.$ $A$ subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.
Correct Answer: $${\left( {{3 \over 4}} \right)^n}$$
1989
JEE Advanced
Numerical
IIT-JEE 1989
Suppose the probability for A to win a game against B is $0.4.$ If $A$ has an option of playing either a "best of $3$ games" or a "best of $5$ games" match against $B$, which option should be choose so that the probability of his winning the match is higher ? (No game ends in a draw).
Correct Answer: best of $$3$$ games
1988
JEE Advanced
Numerical
IIT-JEE 1988
A box contains $2$ fifty paise coins, $5$ twenty five paise coins and a certain fixed number $N\,\,\left( { \ge 2} \right)$ of ten and five paise coins. Five coins are taken out of the box at random. Find the probability that the total value of these $5$ coins is less than one rupee and fifty paise.
Correct Answer: $$\,1 - {{10\left( {N + 2} \right)} \over {N + 7{c_5}}}$$
1987
JEE Advanced
Numerical
IIT-JEE 1987
A man takes a step forward with probability $0.4$ and backwards with probability $0.6$ Find the probability that at the end of eleven steps he is one step away from the starting point.
Correct Answer: $$0.37$$
1986
JEE Advanced
Numerical
IIT-JEE 1986
A lot contains $20$ articles. The probability that the lot contains exactly $2$ defective articles is $0.4$ and the probability that the lot contains exactly $3$ defective articles is $0.6$. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelth testing.
Correct Answer: $$99/1900$$
1985
JEE Advanced
Numerical
IIT-JEE 1985
In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidate decides to tick the answers at random, If he is allowed upto three chances to answer the questions, find the probability that he will get marks in the questions.
Correct Answer: $$1/5$$
1984
JEE Advanced
Numerical
IIT-JEE 1984
In a certain city only two newspapers $A$ and $B$ are published, it is known that $25$% of the city population reads $A$ and $20$% reads $B$ while $8$% reads both $A$ and $B$. It is also known that $30$% of those who read $A$ but not $B$ look into advertisements and $40$% of those who read $B$ but not $A$ look into advertisements while $50$% of those who read both $A$ and $B$ look into advertisements. What is the percentage of the population that reads an advertisement?
Correct Answer: $$13.9$$%
1983
JEE Advanced
Numerical
IIT-JEE 1983
$A, B, C$ are events such that
$P\left( A \right) = 0.3,P\left( B \right) = 0.4,P\left( C \right) = 0.8$
$P\left( {AB} \right) = 0.08,P\left( {AC} \right) = 0.28;\,\,P\left( {ABC} \right) = 0.09$
$P\left( A \right) = 0.3,P\left( B \right) = 0.4,P\left( C \right) = 0.8$
$P\left( {AB} \right) = 0.08,P\left( {AC} \right) = 0.28;\,\,P\left( {ABC} \right) = 0.09$
If $P\left( {A \cup B \cup C} \right) \ge 0.75,$ then show that $P$ $(BC)$ lies in the interval $0.23 \le x \le 0.48$
Correct Answer: Solve it.
1983
JEE Advanced
Numerical
IIT-JEE 1983
Cards are drawn one by one at random from a well - shuffled full pack of $52$ playing cards until $2$ aces are obtained for the first time. If $N$ is the number of cards required to be drawn, then show that ${P_r}\left\{ {N = n} \right\} = {{\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \over {50 \times 49 \times 17 \times 13}}$ where $2 \le n \le 50$
Correct Answer: Solve it.
1982
JEE Advanced
Numerical
IIT-JEE 1982
$A$ and $B$ are two candidates seeking admission in $IIT.$ The probability that $A$ is selected is $0.5$ and the probability that both $A$ and $B$ are selected is atmost $0.3$. Is it possible that the probability of $B$ getting selected is $0.9$ ?
Correct Answer: No.