Calculating Probability of Winning Candy for 10 Students

Click For Summary
In a scenario with 10 students drawing from a bag containing 9 white marbles and 1 black marble, each student does not have an equal chance of winning candy. The probability of the third person winning can be calculated by considering the outcomes of the first two draws. If neither the first nor the second student draws the black marble, the third student has a chance to draw it. The calculation involves multiplying the probabilities of the first two students not drawing the black marble and the third student drawing it. Thus, the third student does have a chance, but it is contingent on the previous draws.
Shanelani2006
Messages
1
Reaction score
0
There are 10 marbles in a bag, 9 white and 1 black. Each day a teacher goes around the room of 10 students and allows each student to pick one marble. The one who draws the black marble gets candy for the day. The students never change their seats and she always distribute the marbles along the same path (the first person is always the first person to draw and the last person is always the last to draw). Does each student have an equal chance to win candy for the day? How would the probability be calculated for third person in line?
I thought every student has an equal chance of 10% an the probability for the third person would be 1/8*9/10*8/9 (the prob. that 3rd person draws it times the prob. that the first person did not draw it times the prob. that the 2nd person didnt draw it) But i don't know if i am making a wrong asumption that the third person even has a chance. please help
 
Physics news on Phys.org
Your analysis sounds fine to me.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K