Probabilitys I know how to use the calculator not formulas

clatie

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Nov 30, 2006
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I understand how to use the calculator TI-83 on these problems but can you assist me on the formula and how it is worked out?

roll exactly 2 threes in 5 rolls of a die?
binompdf(5,1/6,2) = .16075

.......................k........n-k
P(X=K)=(n/k)p (1-p)

...........................2.............5-2
P(X=2)=(5/2)(1/6) (1-1/6) ????


More than 3 heads in 6 flips of coin?

binompdf(6, .5,3)=.65625 1-.65625=.34375
............................................i...........n-i
P(X<or equal to K)=sum(n/i)p (1-p)

................................................ i............6-i
P(X<or equal to 3)= (6/i).5 (1-.5) ??????
 
Re: Probabilitys I know how to use the calculator not formul

Hello, clatie!

Your formula is basically correct.

. . P(x=k)  =  (nk)pk(1p)nk\displaystyle P(x=k)\;=\;{n\choose k}p^k(1-p)^{n-k}


Roll exactly 2 threes in 5 rolls of a die?
p(3)=16,  P(not 3)=56\displaystyle p(3)\,=\,\frac{1}{6},\;P(\text{not }3)\,=\,\frac{5}{6}

\(\displaystyle P(k=2)\;=\;{\5\choose2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3 \;=\;0.160751029\)


More than 3 heads in 6 flips of coin

P(head)=12,  P(tail)=12\displaystyle P(\text{head}) \,=\,\frac{1}{2},\;P(\text{tail})\,=\,\frac{1}{2}

"More than 3 heads" means: 4 heads or 5 heads or 6 heads.
We must calculate these separately and then add them.

. . \(\displaystyle P(k=4)\:=\:{\6\choose4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^2 \;=\;\frac{15}{64}\)

. . P(k=5)  =(65)(12)5(12)1  =  664\displaystyle P(k=5)\;=\:{6\choose5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^1\;=\;\frac{6}{64}

. . P(k=6)  =  (66)(12)6(12)0  =  164\displaystyle P(k=6)\;=\;{6\choose6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^0 \;= \;\frac{1}{64}


Therefore: P(k>3)  =  1564+664+164  =  2264  =  0.34375\displaystyle \,P(k\,>\,3)\;=\;\frac{15}{64}\,+\,\frac{6}{64}\,+\,\frac{1}{64}\;=\;\frac{22}{64} \;=\;0.34375

 
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