What Is the Probability That the Least Number Drawn Is 5?

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Homework Help Overview

The problem involves drawing three chits numbered 1 to 7 with replacement and determining the probability that the least number drawn is 5. Participants are exploring the implications of the conditions set by the problem statement.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the requirement for the number 5 to be present at least once among the drawn chits and the condition that no number should be less than 5. There is also a consideration of different interpretations of the problem statement regarding the least number drawn.

Discussion Status

The discussion is ongoing, with some participants clarifying their understanding of the problem's requirements. There is recognition of different interpretations, particularly concerning the conditions for the drawn numbers.

Contextual Notes

Participants are navigating the nuances of the problem's wording, particularly the phrase "the least number on any selected chit is 5," which has led to varying interpretations of the probability calculation.

zorro
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Homework Statement



Seven chits are numbered 1 to 7. Three are drawn one by one with replacements. The probability that the least number on any selected chit is 5, is?

The Attempt at a Solution



We require 5 to be present on atleast one of the three chits and no number should be less than 5.

So required probability = P(three times 5) + P(two times 5) + P(one time 5)
= (1/7)3 + 1/7*1/7*2/7*3 + 1/7*2/7*2/7*3=19/343

The answer given is
(3/7)3
 
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Abdul Quadeer said:

Homework Statement



Seven chits are numbered 1 to 7. Three are drawn one by one with replacements. The probability that the least number on any selected chit is 5, is?

The Attempt at a Solution



We require 5 to be present on atleast one of the three chits and no number should be less than 5.

So required probability = P(three times 5) + P(two times 5) + P(one time 5)
= (1/7)3 + 1/7*1/7*2/7*3 + 1/7*2/7*2/7*3=19/343

The answer given is
(3/7)3

If you interpret the problem to ask for the probability for each set of chit to have at least one number to be 5, and no number less than 5, then your result is correct.

But looking at the problem, I think it asks for the probability such that each number in the set is no less than 5: "The probability that the least number on any selected chit is 5".
 


Okay, that makes sense now. Thanks!
 


Since you replace the chit after each drawing, you want the probability that you draw 5,6 or 7 on the first draw, 5,6 or 7 on the second draw and 5,6 or 7 on the third draw.

R.G. Vickson
 

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