138- In two boxes there are respectively 20 and 12 bulbs. In the first box 4 bulbs are defective and in the second box 3 bulbs are defective. From the first box 5 bulbs are randomly taken and placed in the second box. With what probability is a randomly selected bulb from the new second box defective?
(1) $\dfrac{5}{24}$ (2) $\dfrac{11}{48}$ (3) $\dfrac{13}{48}$ (4) $\dfrac{7}{24}$
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127-- In a bag there are 16 balls numbered 1 to 16. Two balls are drawn randomly and without replacement. If we know that the number of the second ball is less than the number of the first ball, what is the probability that the number of the first ball is 16?
(1) $\dfrac{1}{16}$ (2) $\dfrac{1}{12}$ (3) $\dfrac{1}{8}$ (4) $\dfrac{1}{4}$
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A bag contains some candies, $\frac { 2 } { 5 }$ of them are made of white chocolate and the remaining $\frac { 3 } { 5 }$ are made of dark chocolate. Out of the white chocolate candies, $\frac { 1 } { 3 }$ are wrapped in red paper, the rest are wrapped in blue paper. Out of the dark chocolate candies, $\frac { 2 } { 3 }$ are wrapped in red paper, the rest are wrapped in blue paper. If a randomly selected candy from the bag is found to be wrapped in red paper, then what is the probability that it is made up of dark chocolate?
(A) $\frac { 2 } { 3 }$
(B) $\frac { 3 } { 4 }$
(C) $\frac { 3 } { 5 }$
(D) $\frac { 1 } { 4 }$

A brand called Jogger's Pride produces pairs of shoes in three different units that are named $U _ { 1 } , U _ { 2 }$ and $U _ { 3 }$. These units produce $10 \% , 30 \% , 60 \%$ of the total output of the brand with the chance that a pair of shoes being defective is $20 \% , 40 \% , 10 \%$ respectively. If a randomly selected pair of shoes from the combined output is found to be defective, then what is the chance that the pair was manufactured in the unit $U _ { 3 }$?
(A) $30 \%$
(B) $15 \%$
(C) $\frac { 3 } { 5 } \times 100 \%$
(D) Cannot be determined from the given data.

Shubhaangi thinks she may be allergic to Bengal gram and takes a test that is known to give the following results:

• For people who really do have the allergy, the test says ``Yes'' $90 \%$ of the time.
• For people who do not have the allergy, the test says ``Yes'' $15 \%$ of the time.

If $2 \%$ of the population has the allergy and Shubhaangi's test says ``Yes'', then the chances that Shubhaangi does really have the allergy are
(A) $1 / 9$
(B) $6 / 55$
(C) $1 / 11$
(D) cannot be determined from the given data.

29. If $\vec { E }$ and $\vec { F }$ are the complementary events of events $E$ and $F$ respectively and if $0 < \mathrm { P } ( \mathrm { F } ) < 1$, then.
(A) $\quad P ( E / F ) + P ( \vec { E } / F ) = 1$
(B) $\quad P ( E / F ) + P ( E / \vec { F } ) = 1$
(C) $\mathrm { P } ( \vec { E } / \mathrm { F } ) + \mathrm { P } ( \mathrm { E } / \vec { F } ) = 1$
(D) $\quad \mathrm { P } ( \mathrm { E } / \vec { F } ) + \mathrm { P } ( \vec { E } / \vec { F } ) = 1$

A is targeting to $\mathrm { B } , \mathrm { B }$ and C are targeting to A . Probability of hitting the target by $\mathrm { A } , \mathrm { B }$ and C are $2 / 3,1 / 2$ and $1 / 3$ respectively. If A is hit then find the probability that B hits the target and C does not.

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{5}$
(D) $\frac{1}{5}$

Let $E^c$ denote the complement of an event $E$. Let $E$, $F$, $G$ be pairwise independent events with $P(G) > 0$ and $P(E \cap F \cap G) = 0$. Then $P(E^c \cap F^c | G)$ equals
(A) $P(E^c) + P(F^c)$
(B) $P(E^c) - P(F^c)$
(C) $P(E^c) - P(F)$
(D) $P(E) - P(F^c)$

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 $B$ 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) $\frac { 9 } { 20 }$

A box $B _ { 1 }$ contains 1 white ball, 3 red balls and 2 black balls. Another box $B _ { 2 }$ contains 2 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.
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
(A) $\frac { 116 } { 181 }$
(B) $\frac { 126 } { 181 }$
(C) $\frac { 65 } { 181 }$
(D) $\frac { 55 } { 181 }$

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 $n _ { 1 } , n _ { 2 } , n _ { 3 }$ 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) $n _ { 1 } = 6 , n _ { 2 } = 12 , n _ { 3 } = 5 , n _ { 4 } = 20$

A 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 $n _ { 1 }$ and $n _ { 2 }$ is(are)
(A) $\quad n _ { 1 } = 4$ and $n _ { 2 } = 6$
(B) $\quad n _ { 1 } = 2$ and $n _ { 2 } = 3$
(C) $n _ { 1 } = 10$ and $n _ { 2 } = 20$
(D) $n _ { 1 } = 3$ and $n _ { 2 } = 6$

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
(A) $\frac{36}{73}$
(B) $\frac{47}{79}$
(C) $\frac{78}{93}$
(D) $\frac{75}{83}$

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) $\frac { 15 } { 256 }$
(B) $\frac { 3 } { 16 }$
(C) $\frac { 5 } { 52 }$
(D) $\frac { 1 } { 8 }$

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 $\frac { 1 } { 2 }$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac { 1 } { 6 }$. Then the probability that the student knows the answer of a randomly chosen question is
(A) $\frac { 1 } { 12 }$
(B) $\frac { 1 } { 7 }$
(C) $\frac { 5 } { 7 }$
(D) $\frac { 5 } { 12 }$

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
$$P ( U ) = \frac { 1 } { 2 } , \quad P ( V ) = \frac { 1 } { 10 } , \quad \text { and } \quad P ( W ) = \frac { 1 } { 12 }$$
then $P ( T )$ is equal to

(A)

$\frac { 13 } { 36 }$

(B)

$\frac { 1 } { 3 }$

(C)

$\frac { 19 } { 60 }$

(D)

$\frac { 1 } { 4 }$

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 $\_\_\_\_$.

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2 , respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is
(1) 0.06
(2) 0.14
(3) 0.2
(4) None of these

Let $A$ and $E$ be any two events with positive probabilities Statement I: $P ( E / A ) \geq P ( A / E ) P ( E )$. Statement II: $P ( A / E ) \geq P ( A \cap E )$.
(1) Both the statements are false
(2) Both the statements are true
(3) Statement-I is false, Statement-II is true
(4) Statement - I is true, Statement - II is false

A box $A$ contains 2 white, 3 red and 2 black balls. Another box $B$ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box $B$ is :
(1) $\frac { 7 } { 8 }$
(2) $\frac { 9 } { 16 }$
(3) $\frac { 7 } { 16 }$
(4) $\frac { 9 } { 32 }$

A box ' $A$ ' contains 2 white, 3 red and 2 black balls. Another box ' $B ^ { \prime }$ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ' $B ^ { \prime }$ is
(1) $\frac { 7 } { 16 }$
(2) $\frac { 9 } { 32 }$
(3) $\frac { 7 } { 8 }$
(4) $\frac { 9 } { 16 }$

Let $A$ and $B$ be two non-null events such that $A \subset B$. Then, which of the following statements is always correct?
(1) $P(A \mid B) \geq P(A)$
(2) $P(A \mid B) = P(B) - P(A)$
(3) $P(A \mid B) \leq P(A)$
(4) $P(A \mid B) = 1$

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is
(1) $\frac { 1 } { 4 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 1 } { 8 }$
(4) $\frac { 1 } { 9 }$

In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35 \% , 20 \%$ and $10 \%$ respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is:
(1) $\frac { 14 } { 45 }$
(2) $\frac { 7 } { 45 }$
(3) $\frac { 8 } { 45 }$
(4) $\frac { 28 } { 45 }$