Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X.
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An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colours is
Correct Answer - C `(8)/(15)`
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`E_(P)` of a good is -3 . At a price of Rs. 8 per unit a consumer busy 160 units of the good . How many units of the good will the consumer buy when p
Correct Answer - 280 units
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An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 m s^-2.
1. Assuming absence of stars in the vicinity, the only gravitational force exerted on astronaut is by the spaceship. 2. But this force is negligible. 3. Hence, once astronaut is out of...
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Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement. Let X be the number of bad eggs drawn. Find the mean a
Correct Answer - `"Mean "=(6)/(4), "variance "=(30)/(49)`
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(i) A box contins 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect
Correct Answer - ` (i) 1/10 (ii) (a) 9/10 (b) 1/5`
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An urn contains 9 red, 7 white and 4 black balls. A ball is drawn at random. Find the probability that the ball drawn is (i) red (ii) white (iii) red
Correct Answer - `(i) 9/20 (ii) 7/20 (iii) 4/5 (iv) 11/20 (v) 13/20`
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A card is drawn at random from a well shuffled deck of 2 cards. Find the probability of its being a spade or a king.
Correct Answer - `4/13`
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The digit in the ones place of a 2 digit number is twice the digit at the tens place. What is the number?
Let the digit at ten's place be x Since we are given that the digit in the ones place of a 2 digit number is twice the digit at the tens...
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A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag
Let X be the random variable defined as the number of red balls. Then X = 0, 1 P(X = 0) = \(\frac{3}{4}\times\frac{2}{3}=\frac{6}{12}=\frac{1}{2}\) P(X = 1) = \(\frac{1}{4}\times\frac{3}{3}+\frac{3}{4}\times\frac{1}{3}=\frac{6}{12}=\frac{1}{2}\) Probability Distribution Table: X 0 1 P(X) \(\frac{1}{2}\) \(\frac{1}{2}\)
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Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card
The required probability = P((The first is a red jack card and The second is a jack card) or (The first is a red non-jack card and The second is...
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