LUANDATO wrote:A certain board game has a row of squares numbered 1 to 100. If a game piece is placed on a random square and then moved 7 consecutive spaces in a random direction, what is the probability the piece ends no more than 7 spaces from the square numbered 49?
A. 7%
B. 8%
C. 14%
D. 15%
E. 28%
The OA is D.
I'm really confused by this PS question. Experts, any suggestion please? Thanks in advance.
In order to be within 7 spaces of the 49th space, the piece has to end up no farther away than the 42nd or 56th space. This is after the initial placement and the random move of 7 spaces.
So consider the lower bound of 42. The farthest down the initial placement can be is 35, since you only have the possibility of a 7 space move up. For example, if the initial placement was at 34 and you moved 7 spaces up, you would end up at 41, outside the lower bound of 42.
Now consider the upper bound of 56. By the same logic above, the farthest to the "right" the initial placement can be is 63, considering the potential 7 move left down towards 56.
So the total span for the initial placement is from 35 to 63 if you consider the potential 7 space move following.
Now let's look at the probabilities.
The probability of the initial placement being between 35 and 48 inclusive is 14/100. The only moves that keep the piece within 7 of 49 are to the right with a 50% probability.. So the total probability is (14/100) *.5 = 7/100.
Likewise, the probability of being between 50 and 63 inclusive is also 14/100, with the only random 7 space moves being to the left with a 50% probability. So the total probability is also 7/100.
Notice that we skipped the potential for the initial placement on 49. That has a probability of 1/100. However, if this piece is then moved either left or right, it will still remain within 7, so that probability is 1. So for 49th position the probability is 1/100.
Adding up the three probabilities, 7/100 + 7/100 + 1/100 = [spoiler]15/100, 15%, D[/spoiler]
