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I have searched the forum for a solution to this problem and didn't see one. I can get to the answer mentally but am having trouble figuring out how to show it mathematically. Any help would be appreciated.
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The figure shown represents a board with four rows of pegs, and at the bottom of the board are four cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has probablility 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end up in cell 2?
Yes this is easier to figure in your head than put into mathematical formulas
Once passing through the original two points, the ball goes either left or right.
50% says it goes left, 50% goes right
After going left, it could either go left then right, or right then left, to get into section 2. meaning that there are two chances (at 50% of 50% each) or two quarters, or a 50% chance that, once going left, the ball will end up in section two.
If it goes right, however, it must then go left twice (50% of 50%) to get to section 2, so there is a 25% chance that, once going right, the ball will end up in section two.
so of the 50% chance of it going left, 50% of the time it will end up in section two, so the probability of this is 25%, or 1/4.
of the 50% chance that it goes right, 25% of the time it will end up in section two, so the probability of this is 12.5%, or 1/8.