Conditional probability

Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green.

Then which of the following statements is (are) TRUE ?

The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is :

A bag contains $(N+1)$ coins − N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\dfrac9{16}$, then N is equal to:

Three students $S_1, S_2$, and $S_3$ are given a problem to solve. Consider the following events:

• $U$: At least one of $S_1, S_2$, and $S_3$ can solve the problem,
• $V$: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem,
• $W$: $S_2$ can solve the problem and $S_3$ cannot solve the problem,
• $T$: $S_3$ can solve the problem.

For any event $E$, let $P(E)$ denote the probability of $E$. If

$$P(U) = \frac12,\qquad P(V) = \frac{1}{10},\qquad \text{and}\qquad P(W) = \frac{1}{12},$$

then $P(T)$ is equal to

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

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 $P(W_1 \cap G_2 \cap B_3) = \dfrac{2}{5N}$ and the conditional probability $P(B_3 \mid W_1 \cap G_2) = \dfrac29$, then $N$ equals _____.

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 $38p$ is equal to _____.

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

A random variable $X$ has the following probability distribution:
$$\begin{array}{|c|c|c|c|c|c|}\hline X&0&1&2&3&4\\\hline P(X)&k&2k&4k&6k&8k\\\hline\end{array}$$
The value of $P(1<X<4\mid X\le2)$ is equal to :

Let $E_1$ and $E_2$ be two events such that the conditional probabilities $P(E_1|E_2)=\dfrac12$, $P(E_2|E_1)=\dfrac34$ and $P(E_1\cap E_2)=\dfrac18$. Then :

If A and B are two events such that $P(A)=\dfrac13$, $P(B)=\dfrac15$ and $P(A\cup B)=\dfrac12$, then $P(A|B')+P(B|A')$ 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 :

Let A and B be two events such that $P(B|A)=\dfrac25$, $P(A|B)=\dfrac17$ and $P(A\cap B)=\dfrac19$. Consider

(S1) $P(A'\cup B)=\dfrac56$,

(S2) $P(A'\cap B')=\dfrac1{18}$

Then

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

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 ______

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?

Let $S$ be the sample space of all $3\times3$ matrices with entries from the set $\{0, 1\}$. Let the events $E_1$ and $E_2$ be given by

$E_1 = \{A\in S : \det A = 0\}$ and

$E_2 = \{A\in S : \text{sum of entries of } A \text{ is } 7\}$.

If a matrix is chosen at random from $S$, then the conditional probability $P(E_1|E_2)$ equals ______

Let $X$ and $Y$ be two events such that $P(X) = \frac{1}{3}$, $P(X|Y) = \frac{1}{2}$ and $P(Y|X) = \frac{2}{5}$. Then

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