Two bad eggs are mixed accidentally with 10 good ones. Three egg

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 Multiple Choice QuestionsShort Answer Type

981. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).
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 Multiple Choice QuestionsLong Answer Type

982. Find the variance of the number obtained on a throw of an unbiased die.
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983. A die is tossed twice. Getting a number greater than 4 is considered a success. Find the variance of the probability distribution of the number of successes.
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984. Two bad eggs are mixed accidentally with 10 good ones. Three eggs are drawn at random without replacement from this lot. Find the mean and variance for the number of bad eggs.


Let X represent the number of bad eggs. X takes the values 0, 1, 2.
 P(X = 0) = P(3 good eggs)
               equals space fraction numerator straight C presuperscript 10 subscript 3 over denominator straight C presuperscript 12 subscript 3 end fraction space equals fraction numerator 10 space cross times 9 space cross times space 8 over denominator 12 space cross times space 11 space cross times 10 end fraction space equals space 12 over 22
P(X = 1) = P(2 good eggs and one bad egg)
               equals space fraction numerator straight C presuperscript 10 subscript 2 space cross times straight C presuperscript 2 subscript 1 over denominator straight C presuperscript 12 subscript 3 end fraction space equals fraction numerator begin display style fraction numerator 10 space cross times space 9 over denominator 1 space cross times space 2 end fraction end style cross times begin display style 2 over 1 end style over denominator begin display style fraction numerator 12 space cross times space 11 space cross times space 10 over denominator 1 space cross times space 2 space cross times 3 end fraction end style end fraction space equals space 9 over 22
P(X = 2) = P(one good egg and two bad eggs)
               equals space fraction numerator straight C presuperscript 10 subscript 1 space cross times space straight C presuperscript 2 subscript 2 over denominator straight C presuperscript 12 subscript 3 end fraction space equals space fraction numerator begin display style 10 over 1 end style cross times 1 over denominator begin display style fraction numerator 12 space cross times space 11 space cross times space 10 over denominator 1 space cross times 2 space cross times space 3 end fraction end style end fraction space equals space 1 over 22
therefore space space we space have



straight mu space equals space sum straight x space straight p subscript straight i space equals space 1 half
straight sigma squared space equals space sum from blank to blank of straight x squared space straight p subscript straight i space minus space straight mu squared space equals space 13 over 22 minus 1 fourth space equals fraction numerator 26 minus 11 over denominator 44 end fraction space equals 15 over 44


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985. Three bad eggs got mixed up with 7 good eggs. If three eggs are drawn (without replacement) from 10 eggs, find the mean and variance for the number of bad eggs among them.
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986. Five defective bulbs are accidentally mixed with twenty good ones. It is not possible to just look at a bulb and tell whether or not a bulb is defective. Four bulbs are drawn at random from this lot. Find the mean number of defective bulbs drawn.
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987. Two cards are drawn simultaneously (or successively, without replacement) from a well-shuffled deck of 52 cards. Compute σ2 for the number of aces.
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988. In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
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 Multiple Choice QuestionsMultiple Choice Questions

989. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
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  • 2

  • 5

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990. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
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  • 5 over 13
  • 1 over 13
  • 1 over 13
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