Probabilitys I know how to use the calculator not formulas

[clatie]

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) ??????

 

[soroban]

Re: Probabilitys I know how to use the calculator not formul

Hello, clatie!

Your formula is basically correct.

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

Roll exactly 2 threes in 5 rolls of a die?

\(\displaystyle p(3)\,=\,\frac{1}{6},\;P(\text{not }3)\,=\,\frac{5}{6}\)

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

More than 3 heads in 6 flips of coin

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

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

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

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

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


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