A die is tossed twice. Getting a number greater than 4 is consid

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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.


The number of successes is a random variable. Let us denote it by X. X can take the values 0, 1, 2.
The number greater than 4 is 5 or 6. Let S denote the successes in getting 5 or 6.
therefore space space space straight P left parenthesis straight S right parenthesis space equals space straight P left parenthesis 5 space or space 6 right parenthesis space equals space 2 over 6 space equals space 1 third
space space space space space space space space straight P open parentheses straight S with bar on top close parentheses space equals space 1 minus 1 third space equals space 2 over 3
therefore space space space space straight P left parenthesis straight X space equals space 0 right parenthesis space equals space straight P left parenthesis straight S with bar on top space. space straight S with bar on top right parenthesis space equals space 2 over 3 cross times 2 over 3 space equals space 4 over 9
space space space space space space space space straight P left parenthesis straight X space equals space 1 right parenthesis space equals space straight P left parenthesis straight S. space straight S with bar on top right parenthesis space plus space straight P left parenthesis straight S with bar on top space. space straight S right parenthesis space equals space 1 third cross times 2 over 3 plus 2 over 3 cross times 1 third space equals 4 over 9
space space space space space space space space space straight P left parenthesis straight X space equals space 2 right parenthesis space equals space straight P left parenthesis straight S. space straight S right parenthesis space equals space 1 third cross times 1 third space equals space 1 over 9
therefore      X takes the values 0, 1, 2 with probabilities 4 over 9 comma space 4 over 9 space and space 1 over 9 respectively.


straight mu space equals space sum straight x space straight p subscript straight i space equals space 2 over 3
therefore space space space space straight sigma squared space equals space sum space straight x squared space straight p subscript straight i space minus space straight mu squared space equals space 8 over 9 minus 4 over 9 space equals 4 over 9
therefore space space space variance space equals space 4 over 9.

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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.
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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
  • 1

  • 2

  • 5

  • 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
  • 37 over 221
  • 5 over 13
  • 1 over 13
  • 1 over 13
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