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Solution
a. Within 100ns, the number of movements = 100ns / 10ns = 10
So in these 10 movements, net position should be 6 positions left.
This is possible when it has travelled total 8 positions left and 2 positions right, irrespective of the order.
p for left = p for right = 0.5
P(left 6) = 10C8*(0.5)8 *(0.5)2 = 0.0439
b. Most likely place is the same place where it starts as at each and every movement, there is an equal chance of moving left or right, hence, on an avergae we can expect the position to remain the same as:
Expected change = -1*0.5 + 1*0.5 = 0
Probability = Moving 5 steps left and 5 steps right in 10 steps: 10C5 * 0.55 * 0.55 = 0.2461
c. It does not mean that it will never meet the target as the above event is just an expectation and reality may differ from expectation.
