Five defective bulbs are accidentally mixed with twenty good one

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 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).
88 Views

 Multiple Choice QuestionsLong Answer Type

982. Find the variance of the number obtained on a throw of an unbiased die.
128 Views

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.
215 Views

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.
400 Views

Advertisement
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.
237 Views

Advertisement

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.


Let us denote by X, the number of defective bulbs. Clearly X can take the values 0, 1, 2, 3, 4.
P(X = 0) = (no defective bulb) = P(all 4  goods ones)
                                              equals space fraction numerator straight C presuperscript 20 subscript 4 over denominator straight C presuperscript 25 subscript 4 end fraction space equals space fraction numerator 20 cross times 19 cross times 18 cross times 17 over denominator 25 cross times 24 cross times 32 cross times 22 end fraction space equals space 969 over 2530
P(X = 1) = P(1 defective and 3 good ones)

              equals space fraction numerator straight C presuperscript 5 subscript 1 cross times straight C presuperscript 20 subscript 3 over denominator straight C presuperscript 25 subscript 4 end fraction space equals space fraction numerator begin display style 5 over 1 end style cross times begin display style fraction numerator 20 cross times 19 cross times 18 over denominator 1 cross times 2 cross times 3 end fraction end style over denominator begin display style fraction numerator 25 cross times 24 cross times 23 cross times 22 over denominator 1 cross times 2 cross times 3 cross times 4 end fraction end style end fraction space equals space 1140 over 2530

P(X = 2) = P(2 defective and 2  good ones)

               equals space fraction numerator straight C presuperscript 5 subscript 2 cross times straight C presuperscript 20 subscript 2 over denominator straight C presuperscript 25 subscript 4 end fraction space equals space fraction numerator begin display style fraction numerator 5 cross times 4 over denominator 1 cross times 2 end fraction end style cross times begin display style fraction numerator 20 cross times 19 over denominator 1 cross times 2 end fraction end style over denominator begin display style fraction numerator 25 cross times 24 cross times 23 cross times 22 over denominator 1 cross times 2 cross times 3 cross times 4 end fraction end style end fraction space equals space 380 over 2530

P(X  = 3) = P(3 defective and one good one)
               equals space fraction numerator straight C presuperscript 5 subscript 3 cross times straight C presuperscript 20 subscript 1 over denominator straight C presuperscript 25 subscript 4 end fraction space equals space fraction numerator begin display style fraction numerator 5 cross times 4 over denominator 1 cross times 2 end fraction end style cross times begin display style 20 over 1 end style over denominator begin display style fraction numerator 25 cross times 24 cross times 23 cross times 22 over denominator 1 cross times 2 cross times 3 cross times 4 end fraction end style end fraction space equals 40 over 2530
P(X = 4) = P(all 4 defective)
                      equals space fraction numerator straight C presuperscript 5 subscript 4 over denominator straight C presuperscript 25 subscript 4 end fraction space equals space fraction numerator begin display style 5 over 1 end style over denominator begin display style fraction numerator 25 cross times 24 cross times 23 cross times 22 over denominator 1 cross times 2 cross times 3 cross times 4 end fraction end style end fraction space equals 1 over 2530
∴ Probability distribution table is


therefore space mean space sum straight x space straight p subscript straight i space equals space 4 over 5
835 Views

Advertisement
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.
82 Views

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.
138 Views

Advertisement

 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

96 Views

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
77 Views

Advertisement