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Given to events A and B such that odd in favour A are 2 : 1 and odd in favour of $A \cup B$ are 3 : 1. Consistent with this information the smallest and largest value for the probability of event B are given by

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A bag contain different kind of balls in which 5 yellow, 4 black & 3 green balls. If 3 balls are drawn at random then find the probability that no black ball is chosen

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Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ballsis 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 is red, is:

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A computer producing factory has only two plants T_{1} and T_{2} produces 20% and plant T_{2} produces 80% of the total computers produced. 7% of the 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} ). 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

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Let U and V be two events of a sample space S and P(A) denote the probability of an event A. Which of the following statements is true?

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If a man purchases a raffle ticket, he can win a first prize of Rs.5,000 or a second prize of Rs.2,000 with probabilities 0.001 and 0.003 respectively. What should be a fair price to pay for the ticket?

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A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps, he is one step away from the starting point is

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Two numbers $a$ and $b$ are chosen are random from a set of the first 30
natural numbers, then the probability that $a^2 - b^2$ is divisible by
3 is

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If $0{\lt}P(A){\lt}1$ and $0{\lt}P(B){\lt}1$ and $P(A\cap B)=P(A)P(B)$, then

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A four-digit number is formed using the
digits 1, 2, 3, 4, 5 without repetition. The
probability that is divisible by 3 is

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There are two circles in xy −plane whose
equations are $x^2+y^2-2y=0$ and $x^2+y^2-2y-3=0$. A point $(x,y)$ is
chosen at random inside the larger circle.
Then the probability that the point has been
taken from smaller circle is

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Two person A and B agree to meet 20 april 2018 between 6pm to 7pm with understanding that they will wait no longer than 20 minutes for the other. What is the probability that they meet?

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Three numbers a,b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that is divisible by 3 is,

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A and B play a game where each is asked to select a number from 1 to 25. If the two number match, both of them win a prize. The probability that they will not win a prize in a single trial is :

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Probability that they win a prize = 25/625 = 1/25

Thus, the probability that they will not win a prize in a single trial = 1 - 1/25 = 24/25

A and B are independent witness in a case. The chance that A speaks truth is x and B speaks
truth is y. If A and B agree on certain statement, the probability that the statement is true is

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P(B speaks truth) = y

Since, both A and B agree on certain statement.

Hence, Total Probability =P(A)P(B)+P(A')P(B')

=xy + (1-x)(1-y)

If statement is true then it means both A and B speaks truth.

∴ Required Probability =

In an entrance test there are multiple choice questions, with four possible answer to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If the student gets the correct answer to a question, then the probability that he as guessing is

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A man is known to speak the truth 2 out of 3 times. He threw a dice cube with 1 to 6 on its faces and reports that it is 1. Then the probability that it is actually 1 is

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Let A and B be two events such that , and

where stands for the complement of event A. Then the events A and B are

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If E_{1} and E_{2} are two events associated with a random experiment such that P (E_{2}) = 0.35, P (E_{1} or E_{2}) = 0.85 and P (E_{1} & E_{2}) = 0.15 then P(E_{1}) is

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The probability of occurrence of two events E and F are 0.25 and 0.50, respectively. the probability of their simultaneous occurrence is 0.14. the probability that neither E nor F occur is

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If A and B are two events and , the A and B are two events which are

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The probability that a man who is x years old will die in a year is p. Then, amongst n persons $A_1,A_2,\ldots A_n$ each x year old now, the probability that ${{A}}_1$ will die in one year and (be the first to die ) is

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If a number x is selected at random from natural numbers 1,2,…,100, then the probability for $x+\frac{100}{x}{\gt}29$ is

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If three thrown of three dice, the probability of throwing triplets not more than twice is

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A problem in Mathematics is given to 3 students A, B, and C. If the probability of A solving the problem is 1/2
and B not solving it is 1/4
. The whole probability of the problem being solved is 63/64
, then what is the probability of solving it by C?

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A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both win a prize. The probability that they will not win a prize in a single trial is

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Thus, probability of not winning = $1-\frac{1}{25}=\frac{24}{25}$

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 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

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A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely incontradict each other while narrating some incident is

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