# Probability check

• January 6th 2011, 08:58 AM
Mathematicsfan
Probability check
the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
Find P.

here is the equation I got:

$6p^4-12P^3+6P^2-27/128$

FINAL EQUATION:

$6P^2-12P+5/101/128=0$

FINAL RESULT IS 3\4

i have to do it according to bernouli's principle.
• January 6th 2011, 09:15 AM
alexmahone
Quote:

Originally Posted by Mathematicsfan
the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
Find P.

here is the equation I got:

$6p^4-12P^3+6P^2-27/128$

FINAL EQUATION:

$6P^2-12P+5/101/128=0$

FINAL RESULT IS 3\4

i have to do it according to bernouli's principle.

$6P^2(1-P)^2=\frac{27}{128}$

$P^2(1-P)^2=\frac{9}{256}$

$P(1-P)=\frac{3}{16}$ (The negative value is rejected because both factors on the LHS are positive.)

$P-P^2=\frac{3}{16}$

$16P^2-16P+3=0$

$16P^2-4P-12P+3=0$

$4P(4P-1)-3(4P-1)=0$

$(4P-1)(4P-3)=0$

$P=\frac{1}{4}$ or $P=\frac{3}{4}$

But $P>\frac{1}{2}$ (Why?)

Therefore, $P=\frac{3}{4}$.
• January 6th 2011, 09:44 AM
Quote:

Originally Posted by Mathematicsfan
the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
Find P.

here is the equation I got:

$6p^4-12P^3+6P^2-27/128$

FINAL EQUATION:

$6P^2-12P+5/101/128=0$

FINAL RESULT IS 3\4

i have to do it according to bernouli's principle.

Binomial Expansion:

$\displaystyle\ (p+q)^4=p^4+\binom{4}{1}p^3q+\binom{4}{2}p^2q^2+\b inom{4}{3}pq^3+q^4$

The probability of exactly two successes is

$\displaystyle\binom{4}{2}p^2(1-p)^2=6p^2(1-p)^2=\frac{27}{128}$

$\Rightarrow\displaystyle\ p^2q^2=\frac{9}{256}\Rightarrow\ pq=\frac{\sqrt{9}}{\sqrt{256}}=\frac{3}{16}$

$pq=p(1-p)=p-p^2\Rightarrow\ 16p-16p^2=3\Rightarrow\ 16p^2-16p+3=0$

$(4p)^2-4(4p)+3=0\Rightarrow\ (4p-1)(4p-3)=0$

$p>(1-p)\Rightarrow\ 4p=3$
• January 6th 2011, 09:45 AM
Mathematicsfan
thanks a lot you guys 1
• January 6th 2011, 09:56 AM
Mathematicsfan
I'm totally confused, about something why did you guy use root on 9\256
? can you explain it to me cause I didn't study to use it, and hence I wasn't able to solve this.
• January 6th 2011, 10:05 AM
$\displaystyle\ p^2q^2=(pq)^2=\frac{9}{256}$
so we need the square root of that fraction for $pq$
$\displaystyle\ pq=\sqrt{(pq)^2}=\sqrt{\frac{9}{256}}=\frac{\sqrt{ 9}}{\sqrt{256}}$