SOLUTION: A coin is tossed 5 times. Find P(at most 4 tails.)
a. 3/16
b. 13/16
c. 1/32
d. 31/32
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Question 331633: A coin is tossed 5 times. Find P(at most 4 tails.)
a. 3/16
b. 13/16
c. 1/32
d. 31/32
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
Your problem is that you want the probability of "at most 4" tails in 5 flips. That means you need to calculate the probability of 0 tails in 5 flips and add that to the probability of 1 tail in 5 flips, 2 tails in 5...you get the picture. Each flip has a probability of 0.5 for a tail, so your calculation will look like this:
\ =\ \sum_{i=0}^4\left(5\cr i\right\)(0.5)^i(0.5)^{5\,-\,i})
Which calculation is guaranteed to provide all the entertainment value that can be obtained watching paint dry.
But all is not lost. Notice that the only alternative to "at most 4 tails" in 5 flips is all 5 of the flips coming up tails. That means we can simply find the probability of 5 tails out of 5 flips and subtract from 1 to get our answer.
\ =\ 1\ -\ P_5(5,0.5)\ =\ 1\ -\ \left(5\cr 5\right\)(0.5)^5(0.5)^{0})
And this looks MUCH better, particularly when you realize that \ =\ 1\ \forall\ n\ \in\ \mathbb{N})
and 
, so really this reduces to:
\ =\ 1\ -\ P_5(5,0.5)\ =\ 1\ -\ (0.5)^5\ =\ 1\ -\ \frac{1}{32}\ \approx\ 97%)
John
My calculator said it, I believe it, that settles it
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