Probability question - finding the number of attempts

Qu:John is shooting at a target. His probabiltiy of hitting the target is 0.6. What is the minimum number of shots needed for the probability of John hitting the target exactly 5 times to be more then 25%.

ive got no idea how to even approach this. Thx in advance

Re: Probability question - finding the number of attempts

If he shoots at the target 5 the probability of hitting the target all 5 times is 0.6^5.

If he shoots at it 6 times the probability of haviung exactly 5 hits is C(6,5) x 0.6^5 x 0.4^1

If he shoots 7 times the probability of hitting it 5 times is C(7,5) x 0.6^5 0.4^2

In general if shooting N times the probability of hitting it 5 times is C(N,5) x 0.6^5 x 0.4^(N-5). This is the binomial probability formula. So try some values of N and find what value leads to a result >0.25.

Re: Probability question - finding the number of attempts

Ok cool thx. Is there anyway of solving out without trial and error?

Re: Probability question - finding the number of attempts

Quote:

Originally Posted by

**JellyOnion** Qu:John is shooting at a target. His probabiltiy of hitting the target is 0.6. What is the minimum number of shots needed for the probability of John hitting the target exactly 5 times to be more then 25%.

ive got no idea how to even approach this. Thx in advance

If $\displaystyle \displaystyle \begin{align*} X \end{align*}$ is the distribution "number of successful shots", then the probability of getting exactly five successful shots after "n" shots is given by $\displaystyle \displaystyle \begin{align*} Pr \left( X \geq 5 \right) &= \sum_{r = 5}^n{{n \choose{r} } \left( 0.6 \right) ^r \left( 0.4 \right) ^{ n - r }} \end{align*}$. You want this to be at least 0.25, so you are left trying to solve $\displaystyle \displaystyle \begin{align*} \sum_{r = 5}^n{{n\choose{r}} \left( 0.6 \right) ^r \left( 0.4 \right) ^{n - r}} \geq 0.25 \end{align*}$, no easy task...

Re: Probability question - finding the number of attempts

Quote:

Originally Posted by

**JellyOnion** Ok cool thx. Is there anyway of solving out without trial and error?

The method ebaines suggests is NOT "trial and error".

Re: Probability question - finding the number of attempts

But he said to try some values for N until i find one that lets the probability equal something greater then 25. Isn't that trial and error?

Re: Probability question - finding the number of attempts

Quote:

Originally Posted by

**JellyOnion** Isn't that trial and error?

Yes, I agree it's probably easiest to try N=5, then 6, and depending on how quickly the value is appoaching 0.25 then then make a new guess, and zero in that way. It'll probably take about 2 minutes to get to the answer.

Re: Probability question - finding the number of attempts