StudiosTom wrote:
chetan2u wrote:
gmatt1476 wrote:
In the figure above, X and Y represent locations in a district of a certain city where the streets form a rectangular grid. In traveling only north or east along the streets from X to Y, how many different paths are possible?
A. 720
B. 512
C. 336
D. 256
E. 56
PS61551.01
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2019-09-21_1421.png
If you look at the streets, you have to move 5 streets in the east and 3 streets to the north..
Thus, the total streets are 5+3 and the 3 N or 5 E can be chosen out of these in any order..
So, total ways = 8C3=\(\frac{8!}{5!3!}=\frac{8*7*6}{3*2}=8*7=56\). Hence 56 ways.
E
chetan2u : I am always confused about when to multiply probability and when to add probability. What is the logic behind this in probability problems
Hi,
It will depend on what the question is asking..
So, in very simple terms, I would say
Multiply..Say you have 10 different things and we are looking for probability to pick a particular item on three different times when we pick one item..
If you see the probability should be DECREASING.. Picking that item is 1/10 in each pick so it becomes 1/10*1/10*1/10
A coin- getting tails on two continuous throws or tails in first throw and heads in second throw---- 1/2*1/2
So, generally we are looking at
multiple events and probability of a particular/different item each time
That is we have AND .. x and y happening simultaneouslyAdd..Say you have 10 different things and we are looking for probability to pick any of two different item when we pick one item..
If you see the probability should be INCREASING, as now we have options... Picking each item is 1/10 in each pick so picking any of the two becomes 1/10+1/10
A coin- getting tails or heads on a throw ---- 1/2+1/2=1, as it is sure that we will get heads or tails in that throw
So, generally we are looking at
a single events and probability of more than one item.
That is we have OR.. option of either of x or y happening