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Three Numbers Are Chosen from 1 to 20. the Probability that They Are Not Consecutive is

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प्रश्न

Three numbers are chosen from 1 to 20. The probability that they are not consecutive is

विकल्प

  • \[\frac{186}{190}\]

     

  • \[\frac{187}{190}\]

     

  • \[\frac{188}{190}\]

     

  • \[\frac{18}{^{20}{}{C}_3}\]

     

MCQ
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उत्तर

Number of ways to choose three numbers from 1 to 20 = \[^{20}{}{C}_3\] = 1140

Now, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ...., (18, 19, 20).
So, the number of ways to choose three numbers from 1 to 20 such that they are consecutive is 18.
P(three numbers choosen are consecutive) =\[\frac{\text{ Number of ways to choose three consecutive numbers from 1 to 20} }{\text{ Number of ways to choose three numbers from 1 to 20} } = \frac{18}{^{20}{}{C}_3} = \frac{18}{1140} = \frac{3}{190}\]

∴ P(three numbers choosen are not consecutive) = 1 − P(three numbers choosen are consecutive) = \[1 - \frac{3}{190} = \frac{187}{190}\]

 

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Concept of Probability - Probability of 'Not', 'And' and 'Or' Events
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अध्याय 33: Probability - Exercise 33.6 [पृष्ठ ७३]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 33 Probability
Exercise 33.6 | Q 36 | पृष्ठ ७३

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