- #1
Solve Probability Question #6: Find the Correct Answer Now!
- Thread starter Ronaldo95163
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Homework Help Overview
The discussion revolves around a probability question, specifically focusing on the calculations and reasoning involved in determining the correct answer to a problem labeled as number 6. Participants are also referencing related questions, particularly question 11, which introduces additional complexity regarding repeated digits and specific conditions for a four-figure number.
Discussion Character
- Mixed
Approaches and Questions Raised
- One participant shares their calculation for question 6, while another confirms agreement with that answer, detailing the reasoning behind their probability calculations. Other participants express confusion regarding the requirements of question 11, particularly the relationship between its parts and how to approach the calculations needed to find the probability of a specific digit appearing.
Discussion Status
The discussion is active, with participants sharing their attempts and questioning the clarity of the problem statements. Some have provided calculations, while others are seeking clarification on how to proceed with their reasoning, particularly in relation to question 11.
Contextual Notes
There appears to be confusion regarding the phrasing of question 11, specifically the connection between its parts and the implications for the calculations required. Participants are encouraged to clarify their approaches and provide more detailed explanations of their reasoning.
- #2
kiritee Gak
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yes i got the same answer.It is (4/10)*(8/11)+(6/10)*(4/11)=28/55.(The first term for getting a red from first bag and again a red from the second bag and second term for picking a black in first bag and again picking a black in second bag)
- #3
Ronaldo95163
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Ahh thanks man
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- #5
HallsofIvy
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11 (c) says "There is at least one repeated digit, but no digit appears more than twice in a number. Find the probability that a four figure number chose at random from the from the set of numbers in case (a) above contains at least one 6".
I find that very confusing. The first sentence does not appear to have anything to do with the problem posed in the second sentence!
Problem 11(a) you calculated how many such numbers there were. Now calculate how many of them do NOT have a "6". The answer to the question asked in 11(c) is 1 minus (the number that have no "6" divided by the answer to (6)).
I find that very confusing. The first sentence does not appear to have anything to do with the problem posed in the second sentence!
Problem 11(a) you calculated how many such numbers there were. Now calculate how many of them do NOT have a "6". The answer to the question asked in 11(c) is 1 minus (the number that have no "6" divided by the answer to (6)).
- #6
Ray Vickson
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Ronaldo95163 said:
Please try to respect PF standards, by typing up the questions and your answers. Your solution image comes out sideways, and I am unable to rotate it.
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I assume the last sentence in 11c is supposed to be a completely separate question.Ronaldo95163 said:
Your attached working is impossible to decipher. Please explain your approach to 11c (first part).
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