Help with a probability problem please

May 2010
1
0
I'm doing practice problems for a test on probability and I'm stuck on this one.

An archer hits a target 1/3 of the time, using binomial theorem, tell the probability that he hits the target 10 times on 50 shots.

Can you please tell me how to get the answer and explain thoroughly please? Sorry, I have a really bad teacher and we never went over how to use binomial theorem with probability.
 
Nov 2009
63
14
I'm doing practice problems for a test on probability and I'm stuck on this one.

An archer hits a target 1/3 of the time, using binomial theorem, tell the probability that he hits the target 10 times on 50 shots.

Can you please tell me how to get the answer and explain thoroughly please? Sorry, I have a really bad teacher and we never went over how to use binomial theorem with probability.
it is called binomial distribution (not binomial theorem)....
the formula:
\(\displaystyle P(X=x) = _{n}C_{x} p^{(x)} q^{(n-x)}\)

\(\displaystyle n\): number of trial
\(\displaystyle p\): probability to success
\(\displaystyle q\): probability to fail
\(\displaystyle x\): total number of success in \(\displaystyle n\) trial
NOTE: \(\displaystyle q=1-p\)

you have to find the value of n, p, q and x first...
===hope it'll help===(Nerd)
 
Dec 2009
3,120
1,342
I'm doing practice problems for a test on probability and I'm stuck on this one.

An archer hits a target 1/3 of the time, using binomial theorem, tell the probability that he hits the target 10 times on 50 shots.

Can you please tell me how to get the answer and explain thoroughly please? Sorry, I have a really bad teacher and we never went over how to use binomial theorem with probability.
If he takes 50 shots, we can write the binomial as

\(\displaystyle (p+q)^n=\left(p+q\right)^{50}\) where p=probability of a hit, q=probability of a miss.

\(\displaystyle =\binom{50}{0}p^0q^{50}+\binom{50}{1}p^1q^{49}+\binom{50}{2}p^2q^{48}+\binom{50}{3}p^3q^{47}+.......\)

sums the probabilities of no hits (which is 50 misses), that's the first term in the expansion,
one hit (49 misses), that's the 2nd term,
two hits (48 misses), this is the third term etc etc

The \(\displaystyle 50c_k\) values are the number of ways that number of "k" hits can happen.

Hence 10 hits includes 40 misses and there are \(\displaystyle \binom{50}{10}=\binom{50}{40}\) ways this can happen.

Hence the probability of exactly 10 hits is

\(\displaystyle \binom{50}{10}\left(\frac{1}{3}\right)^{10}\left(\frac{2}{3}\right)^{40}\)
 
Jul 2007
894
298
New Orleans
I'm doing practice problems for a test on probability and I'm stuck on this one.

An archer hits a target 1/3 of the time, using binomial theorem, tell the probability that he hits the target 10 times on 50 shots.

Can you please tell me how to get the answer and explain thoroughly please? Sorry, I have a really bad teacher and we never went over how to use binomial theorem with probability.
Use these formulas and the problem should be straight forward

\(\displaystyle \sigma = \sqrt{p(1-p)n}\)

\(\displaystyle \mu = np\)

n= 50

p = \(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{X - \mu}{\sigma} = z\)

remember also to use the continuity correction factor(\(\displaystyle X\pm .5\))
 
Dec 2009
3,120
1,342
The binomial expansion will give the probability of "exactly" 10 hits.

The binomial approximation to the Normal Distribution
will give close answers to the probabilities of up to 10 hits
or 10 or more hits.
 
Jul 2007
894
298
New Orleans
The binomial expansion will give the probability of "exactly" 10 hits.

The binomial approximation to the Normal Distribution
will give close answers to the probabilities of up to 10 hits
or 10 or more hits.
O sorry I misread the question. I did not see that he wanted to use the binomial expansion.
 
Dec 2009
3,120
1,342
It was a good idea to bring it in.
Given the teacher is dodgy, could be either one yet!