SOLUTION: Someone flips 15 biased coins once. The coins are weighted such that the probability of a head with any coin is 0.85.
(a) What is the probability of getting exactly 14 heads?
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Question 584299: Someone flips 15 biased coins once. The coins are weighted such that the probability of a head with any coin is 0.85.
(a) What is the probability of getting exactly 14 heads?
(b) What is the probability of getting at least 14 heads?
(c) What is the probability of getting exactly 3 tails?
Would I use the equation p= n!/(r!)(n-r)* p^r q^n - r
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Yep:
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 !})
For part (a) you want:
\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right))
Calculation Hint: \ =\ \left(n\cr 1\right\)\ =\ n)
For part (b) you want the probability of exactly 15 plus the result of part (a):
\ +\ P_{15}(15,0.85)\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right)\ +\ \left(15\cr 15\right\)\left(0.85\right)^{15}\left(0.15\right)^0)
Calculation Hint: \ =\ \left(n\cr 0\right\)\ =\ 1)
and 
, hence:
\ +\ P_{15}(15,0.85)\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right)\ +\ \left(0.85\right)^{15})
Part (c) is just straightforward fill in the numbers, get out the calculator, and crunch noting that the probability of a tail is 1 minus the probability of a head (save for that very, very tiny probability that the coin lands on its edge):
\ =\ \left(15\cr \,3\right\)\left(0.15\right)^{3}\left(0.5\right)^{12})
On the other hand you can save yourself a boatload of arithmetic drugery if you open up a spreadsheet (Excel under Windows or Numbers on a Mac, they both work the same), pick a convenient empty cell, and type in:
Part (a)
=BINOMDIST(14,15,0.85,false)
Part (b)
=BINOMDIST(14,15,0.85,false) + BINOMDIST(15,15,0.85,false)
Alternatively you could enter:
=1 - BINOMDIST(13,15,0.85,true)
Part (c)
=BINOMDIST(3,15,0.15,false)
John
My calculator said it, I believe it, that settles it
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