SOLUTION: My homework is due tonight and I can't figure out these two problems:
5. A fair coin is flipped 20 times and the sequence of heads and tails is observed. Find the probability th
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Question 465798: My homework is due tonight and I can't figure out these two problems:
5. A fair coin is flipped 20 times and the sequence of heads and tails is observed. Find the probability that the sequence consists of AT LEAST two heads. Answer in fraction reduced to the lowest terms.
The previous answer provided was wrong:
P(2<= x <=20) = 1 - P(0<= x <=1) = 1 - [1/2^20 + 20C2(1/2^20)]
= 1 - [(1+190)/2^20]
= 1 - (191/2^20)
---
== 1048385/1048576
8. Refer to the given graph. Assume one can move only down or to the right. If a route starts at A and ends at C, what is the probability that the route passes through B? Answer in fraction reduced to the lowest terms.
a
[][][]b[][]
[][][][][]
[][][][][]
[][][][][]
[][][][][]c
i said c(4,1) x c(6,2) = 60
But it was wrong
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You are approaching the problem correctly, but I'm not sure you have the formula appropriately used.
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
but you want )
which would require calculating:
\ =\ \sum_{i=2}^{20}\ \left(20\cr\ i\right\)\left(0.5\right)^i\left(1\,-\,0.5\right)^{20\,-\,i})
A daunting task indeed. However )
which can be rendered:
\ =\ 1\ -\ \left\[\left(20\cr\ 0\right\)\left(0.5\right)^0\left(0.5\right)^{20}\ +\ \left(20\cr\ 1\right\)\left(0.5\right)^1\left(0.5\right)^{19}\right\]\ =\ 1\ -\ 21\,\cdot\,\left(0.5\right)^{20}\ >\ 0.999)
Is much easier to calculate. In fact, looking a little deeper, you managed to come up with the correct denominator, but your numerator is off by a touch -- should have been 1048576 minus 21 = 1048555.
You only get one problem per posting. But I'll give you a hint for the second one. Since you can only move down and right, if any of your first three moves are down, you will miss point B. Therefore your first three moves must be to the right. From there you have a limited number of ways to go, so it should be relatively easy to count the number of successful paths.
By the way, I know just by looking at your answer of 60 that it is incorrect without looking at any other part of the problem. Why do you think that is? Hint: You are trying to find a probability.
John
My calculator said it, I believe it, that settles it
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