Bayes' theorem

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\dfrac25$, $\dfrac15$ and $\dfrac25$. The probabilities that the candidate reaches late at the examination centre are $\dfrac15$, $\dfrac13$ and $\dfrac14$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

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 $\dfrac25$.

If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is _____.

A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\dfrac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\dfrac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is

Suppose that

• Box-I contains 8 red, 3 blue and 5 green balls,
• Box-II contains 24 red, 9 blue and 15 green balls,
• Box-III contains 1 blue, 12 green and 3 yellow balls,
• Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\dfrac6{11}$, then $n$ is equal to ______.

Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

Consider three sets $E_1 = \{1, 2, 3\}$, $F_1 = \{1, 3, 4\}$ and $G_1 = \{2, 3, 4, 5\}$. Two elements are chosen at random, without replacement, from the set $E_1$, and let $S_1$ denote the set of these chosen elements. Let $E_2 = E_1 - S_1$ and $F_2 = F_1 \cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.

Let $G_2 = G_1 \cup S_2$. Finally, two elements are chosen at random, without replacement, from the set $G_2$ and let $S_3$ denote the set of these chosen elements.

Let $E_3 = E_2 \cup S_3$. Given that $E_1 = E_3$, let $p$ be the conditional probability of the event $S_1 = \{1, 2\}$. Then the value of $p$ is

There are three bags $B_1$, $B_2$ and $B_3$. The bag $B_1$ contains 5 red and 5 green balls, $B_2$ contains 3 red and 5 green balls, and $B_3$ contains 5 red and 3 green balls. Bags $B_1$, $B_2$ and $B_3$ have probabilities $\dfrac3{10}$, $\dfrac3{10}$ and $\dfrac4{10}$ 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 computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces 20% and plant $T_2$ produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that

$P$(computer turns out to be defective given that it is produced in plant $T_1$) $=10P$(computer turns out to be defective given that it is produced in plant $T_2$),

where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

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 $\dfrac13$, then the correct option(s) with the possible values of $n_1$, $n_2$, $n_3$ and $n_4$ is(are)

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 $B_2$ is

Given that the drawn ball from $U_2$ is white, the probability that head appeared on the coin is

A signal which can be green or red with probability $\dfrac45$ and $\dfrac15$ respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is $\dfrac34$. If the signal received at station B is green, then the probability that the original signal was green is

Let $H_1,H_2,\ldots,H_n$ be mutually exclusive and exhaustive events with $P(H_i)>0$, $i=1,2,\ldots,n$. Let $E$ be any other event with $0<P(E)<1$.

STATEMENT-1: $P(H_i\mid E)>P(E\mid H_i)\cdot P(H_i)$ for $i=1,2,\ldots,n$.

because

STATEMENT-2: $\displaystyle\sum_{i=1}^nP(H_i)=1$.