What is the probability that the ball was labeled

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Discussion Overview

The discussion revolves around calculating probabilities related to selecting balls from a tank containing different types of balls, specifically focusing on the probability of selecting labeled frosted balls under various conditions. The questions include scenarios of single and multiple selections, with or without replacement.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants calculate the probability that a randomly selected frosted ball is labeled as 10/17.
  • For the second question, participants discuss the probabilities of selecting a frosted ball followed by a mild ball, with calculations depending on whether the first ball is returned to the tank or not.
  • One participant suggests that if the first ball is not returned, the probability of selecting a frosted ball first and a mild ball second is (17/50)(33/49), while if the first ball is returned, it is (17/50)(33/50).
  • In the third question, participants agree that the probability of selecting two frosted balls without replacement is (17/50)(16/49).
  • Some participants seek confirmation of their calculations and express uncertainty about the correctness of their results.
  • There is a mention of presenting results as single fractions rather than decimal forms, indicating a preference for a specific format in presenting probabilities.

Areas of Agreement / Disagreement

Participants generally agree on the calculations for the probabilities, but there is some uncertainty regarding the interpretation of the second question and the format of presenting results. Some participants seek confirmation of their answers, indicating a lack of complete consensus.

Contextual Notes

There are unresolved assumptions regarding the conditions of the selections, particularly whether the first ball is returned to the tank or not, which affects the probabilities calculated. Additionally, there is a discussion about the preferred format for presenting the results, which may reflect different expectations in educational contexts.

Jason000000
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A tank contains 50 balls. 10 are frosted (F) and labeled (L), 23 are mild (M) and unlabeled, 7 are frosted and not labelled (N), 10 are mild and labeled.

1. A ball is randomly selected from the tank and it was frosted. What is the probability that the ball was labeled?
2. The 1st ball was frosted and the 2nd one was mild (not frosted) What is the probability if we would do two trials?
3. To select a ball was frosted and this ball was not returned to the tank, and select the 2nd one frosted again. What is the probability?
 
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I finally figured it out .. but I would appreciate if anyone can confirm the answers .. thanx

for Q1. P(L|F) = 10/17

for Q2. P(F) x P(M) = 17/50 x 33/50 = 0.34 x 0.66 = 0.2244

for Q3. P(F1) = 17/50 & P(F2) = 16/49

P(F1|F2) = 17/50 x 16/49 = 0.34 x 0.326 = 0.111
 
Jason000000 said:
A tank contains 50 balls. 10 are frosted (F) and labeled (L), 23 are mild (M) and unlabeled, 7 are frosted and not labelled (N), 10 are mild and labeled.

1. A ball is randomly selected from the tank and it was frosted. What is the probability that the ball was labeled?
The ball is one of the 17 frosted balls of which 10 are labeled. The probability this frosted ball is labeled is 10/17.

2. The 1st ball was frosted and the 2nd one was mild (not frosted) What is the probability if we would do two trials?
What is the probability of what? Are you asking "if we take two balls what is the probability that the first is frosted and the second is not"? If so then is the first ball returned to the tank or not?
There are initially 50 balls, 17 of which are frosted. The probability the first ball selected is frosted is 17/50. If that ball is not returned to the tank, there are then 49 balls, 33 of which are not frosted. The probability that the second ball selected is not frosted is 33/49. In this situation the probability the first ball is frosted and the second ball is not is (17/50)(33/49). If the first ball is returned to the tank, the probability second ball is not frosted is (33/50) so the probability of "frosted, not frosted", in that order, is (17/50)(33/50).

3. To select a ball was frosted and this ball was not returned to the tank, and select the 2nd one frosted again. What is the probability?
Initially there are 50 balls in the tank, 17 of which are frosted. The probability the first ball is frosted is 17/50. Since that ball is not returned there are 49 balls, 16 of which are frosted. The probability the second ball is also frosted is 16/49. The probability of two frosted balls, without replacement, is (17/50)(16/49).
 
Country Boy said:
The ball is one of the 17 frosted balls of which 10 are labeled. The probability this frosted ball is labeled is 10/17.What is the probability of what? Are you asking "if we take two balls what is the probability that the first is frosted and the second is not"? If so then is the first ball returned to the tank or not?
There are initially 50 balls, 17 of which are frosted. The probability the first ball selected is frosted is 17/50. If that ball is not returned to the tank, there are then 49 balls, 33 of which are not frosted. The probability that the second ball selected is not frosted is 33/49. In this situation the probability the first ball is frosted and the second ball is not is (17/50)(33/49). If the first ball is returned to the tank, the probability second ball is not frosted is (33/50) so the probability of "frosted, not frosted", in that order, is (17/50)(33/50).Initially there are 50 balls in the tank, 17 of which are frosted. The probability the first ball is frosted is 17/50. Since that ball is not returned there are 49 balls, 16 of which are frosted. The probability the second ball is also frosted is 16/49. The probability of two frosted balls, without replacement, is (17/50)(16/49).
can u confirm above the answers I provided before your reply? many thanx!
 
Jason000000 said:
can u confirm above the answers I provided before your reply? many thanx!
Why are you asking CBoy that? His answers are SAME as yours!
 
Wilmer said:
Why are you asking CBoy that? His answers are SAME as yours!

because I have extra to his answer:-

P(F) x P(M) = 17/50 x 33/50 = 0.34 x 0.66 = 0.2244
P(F1|F2) = 17/50 x 16/49 = 0.34 x 0.326 = 0.111

is it correct to multiply and get these results?
 
Jason000000 said:
because I have extra to his answer:-

P(F) x P(M) = 17/50 x 33/50 = 0.34 x 0.66 = 0.2244
P(F1|F2) = 17/50 x 16/49 = 0.34 x 0.326 = 0.111

is it correct to multiply and get these results?
Depends on what your teacher expects.
Usually shown as single fraction: 561/2500 and 136/1225.
 

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