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

AI Thread Summary
The problem involves drawing three chits numbered 1 to 7 with replacement, and the goal is to find the probability that the least number drawn is 5. To achieve this, at least one chit must show the number 5, while no chits can show numbers less than 5. The initial calculation presented was incorrect, as it misinterpreted the requirement for the least number. The correct approach is to calculate the probability of drawing either 5, 6, or 7 for all three chits, leading to a different probability outcome. The clarification emphasizes the importance of understanding the problem's phrasing to arrive at the correct solution.
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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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