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

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
 
on Phys.org


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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