# Probability check

• Jan 3rd 2011, 11:17 AM
Mathematicsfan
Probability check
I have a last task to do, can you guide me through please.

- The probability of a sniper shot to strike in a single shot is "P".
It's known that if the sniper fires 3 times then the probability that from the three shots the sniper will strike at least one time and miss at least one time is 0.48.
A- Find "P".

So I was able to get these two equation according to Bernoulli's table:
$3P*(1-P)^2 - 0.48 = 3P^2*(1-P)^1 - 0.48$
Are these correct ? according to the task ? can anyone check it and I'll solve it, and if It's not truth can anyone help me fix it ?
Final equation was :
$-6P^2 +3P=0$
Final answer is : 0.8 / 0.2
• Jan 3rd 2011, 11:55 AM
Quote:

Originally Posted by Mathematicsfan
I have a last task to do, can you guide me through please.

- The probability of a sniper shot to strike in a single shot is "P".
It's known that if the sniper fires 3 times then the probability that from the three shots the sniper will strike at least one time and miss at least one time is 0.48.
A- Find "P".

So I was able to get these two equation according to Bernoulli's table:
$3P*(1-P)^2 - 0.48 = 3P^2*(1-P)^1 - 0.48$
Are these correct ? according to the task ? can anyone check it and I'll solve it, and if It's not truth can anyone help me fix it ?
Final equation was :
$-6P^2 +3P=0$
Final answer is : 0.8 / 0.2

The probability of at "least 1 hit and at least 1 miss" is
the sum of the probabilities of "2 hits and 1 miss" and 1 hit and 2 misses",
which is, from the Binomial expansion of $(p+q)^3$

$\binom{3}{2}p^2(1-p)+\binom{3}{1}p(1-p)^2=0.48$

$\Rightarrow\ 3\left(p^2-p^3\right)+3p\left(1-2p+p^2\right)=0.48$

$\Rightarrow\ 3p^2-3p^3+3p-6p^2+3p^3=0.48$

$\Rightarrow\ -3p^2+3p=0.48\Rightarrow\ p^2-p+0.16=0$

from which $p=0.8,\;0.2$
are obtained.
• Jan 3rd 2011, 12:33 PM
Mathematicsfan
Thank you so much you have helped me way beyond my expectations !!!
I finally understood it deeply.
Thanks again !!!