The probability of India winning a test match against West Indies is $1/2$. Assuming independence from match to match the probability that in a $5$ match series India's second win occurs at third test is
Let $A, B, C$ be three mutually independent events. Consider the two statements ${S_1}$ and ${S_2}$
${S_1}\,:\,A$ and $B \cup C$ are independent
${S_2}\,:\,A$ and $B \cap C$ are independent
Then,
An unbiased die with faces marked $1,2,3,4,5$ and $6$ is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than $2$ and the maximum face value is not greater than $5,$ is then:
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting, points $0,$ $1$ and $2$ are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are independent, the probability of India getting at least $7$ points is
One hundred identical coins, each with probability, $p,$ of showing up heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is
A student appears for tests, $I$, $II$ and $III$. The student is successful if he passes either in tests $I$ and $II$ or tests $I$ and $III$. The probabilities of student passing in tests $I$, $II$ and $III$ are $p, q$ and ${1 \over 2}$ respectively. If the probability that the student is successful is ${1 \over 2}$, then
The probability that at least one of the events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.2,$ then $P\left( {\overline A } \right) + P\left( {\overline B } \right)$ is
A box contains $24$ identical balls of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4$th time on the $7$th draw is
Fifteen coupons are numbered $1, 2 ........15,$ respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is $9,$ is
If $A$ and $B$ are two events such that $P\left( A \right) > 0,$ and $P\left( B \right) \ne 1,$ then $P\left( {{{\overline A } \over {\overline B }}} \right)$ is equal to
A.
$1 - P({A \over B})$ (Here $\overline A $ and $\overline B $ are complements of $A$ and $B$ respectively).
B.
$1 - P({{\overline A } \over B})$ (Here $\overline A $ and $\overline B $ are complements of $A$ and $B$ respectively).
C.
${{1 - P\left( {A \cup B} \right)} \over {P\left( {\overline B } \right)}}$ (Here $\overline A $ and $\overline B $ are complements of $A$ and $B$ respectively).
D.
${{P\left( {\overline A } \right)} \over {P\left( {\overline B } \right)}}$ (Here $\overline A $ and $\overline B $ are complements of $A$ and $B$ respectively).
Two events $A$ and $B$ have probabilities $0.25$ and $0.50$ respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$. Then the probability that neither $A$ nor $B$ occurs is
The probability that an event $A$ happens in one trial of an experiment is $0.4.$ Three independent trials of the experiment are performed. The probability that the event $A$ happens at least once is
Two fair dice are tossed. Let $x$ be the event that the first die shows an even number and $y$ be the event that the second die shows an odd number. The two events $x$ and $y$ are:
A factory has a total of three manufacturing units, $M_1, M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1, M_2$, and $M_3$ produce bulbs in the proportions of $2: 2: 1$, respectively. It is known that $20 \%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1, 15 \%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$.
If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is __________.
Correct Answer: 0.27TO0.33
Explanation:
$H_1$: The bulb is produced by unit $M_1$.
$H_2$: The bulb is produced by unit $M_2$.
$H_3$: The bulb is produced by unit $M_3$.
$E$: The bulb is defective.
The unit production proportions and known probabilities are:
A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i=1,2,3$, let $W_i, G_i$, and $B_i$ denote the events that the ball drawn in the $i^{\text {th }}$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P\left(W_1 \cap G_2 \cap B_3\right)=\frac{2}{5 N}$ and the conditional probability $P\left(B_3 \mid W_1 \cap G_2\right)=\frac{2}{9}$, then $N$ equals ________.
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is _____________.
Let $X$ be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For example, 22240 is in $X$ while 02244 and 44422 are not in $X$. Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5 . Then the value of $38 p$ is equal to :
Correct Answer: 31
Explanation:
Number of elements in $\mathrm{X}$ which are multiple of 5
Among these 38 elements, let us calculate when element is not divisible by 20
Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i=0,1,2,3,4$. Let $X$ be a random variable such that for $i=0,1,2,3,4$, the probability $P(X=i)=p_i$. Then the value of $7 E(X)$ is :
Correct Answer: 24
Explanation:
Number of points having 0 friend $=0$
Number of points having 1 friend $=0$
Number of points having 2 friends $=4$
Number of points having 3 friends $=5 \times 4=20$
Number of points having 4 friends $=49-24=25$
$\mathrm{P}_{\mathrm{i}}=$ Probability that randomly selected points has friends
Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{49}$. Let $p$ be the probability that they are friends. Then the value of $7 p$ is :
Correct Answer: 0.50
Explanation:
Total number of ways of selecting 2 persons $={ }^{49} \mathrm{C}_2$
Number of ways in which 2 friends are selected $=6 \times 7 \times 2=84$
A number of chosen at random from the set {1, 2, 3, ....., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is __________.
Correct Answer: 214
Explanation:
Given, set = {1, 2, 3, ...., 2000}
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$\therefore$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
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 ${{625} \over 4}{p_1}$ is ___________.
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 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
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
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 $=q=\frac{1}{2}$
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 $\mathrm{E}_3$.
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?
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 ${3 \over 10}$.
Correct Answer: A,C
Explanation:
It is given that there are three bags B1, B2 and B3 and probabilities of being chosen B1, B2 and B3 are respectively.
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 ${n_2} = 6$
Correct Answer: C,D
Explanation:
$\therefore$ P (drawing red ball from B1) = ${1 \over 3}$
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)
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.
$P\left( {{X^c} \cap Y} \right) = {1 \over 3}$
Correct Answer: A,B
Explanation:
Let's analyze the given conditions and evaluate which options are correct.
Given:
$ P(X|Y) = \frac{1}{2} $
$ P(Y|X) = \frac{1}{3} $
$ P(X \cap Y) = \frac{1}{6} $
Now, let's proceed step by step through each option.
Option A: $ P(X \cup Y) = \frac{2}{3} $
We can use the formula for the union of two events:
$ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) $
To find $ P(X) $ and $ P(Y) $, we use the definitions of conditional probability:
$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $
$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $
$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $
Similarly,
$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $
$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $
$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $
Now, substituting these values into the union formula:
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 $\left. {|X} \right] = {7 \over 8}$
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.
$P\left( E \right) = {3 \over 5},P\left( F \right) = {4 \over 5}$
Correct Answer: A,D
Explanation:
Let $P(E) = e$ and $P(F) = f$.
$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$
$ \Rightarrow e + f - 2ef = {{11} \over {25}}$ ...... (1)
$P(\overline E \cap \overline F ) = {2 \over {25}}$
$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$
$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$ ...... (2)
From Eqs. (1) and (2), we get
$ef = {{12} \over {25}}$ and $e + f = {7 \over 5}$
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?
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
A.
$P\left( {B/A} \right) = P\left( B \right) - P\left( A \right)$
$E$ 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.
$\,P\left( E \right) = 1/4,P\left( F \right) = 1/3$
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.
$A$ and $B$ are two independent events. $C$ is even in which exactly one of $A$ or $B$ occurs. Prove that $P\left( C \right) \ge P\left( {A \cup B} \right)P\left( {\overline A \cap \overline B } \right)$
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">
$A$ 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.
