Thread: Probability question - finding the number of attempts

1. 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

2. 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.

3. Re: Probability question - finding the number of attempts

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

4. Re: Probability question - finding the number of attempts 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...

5. Re: Probability question - finding the number of attempts 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".

6. 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?

7. Re: Probability question - finding the number of attempts 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.

8. Re: Probability question - finding the number of attempts

Ok thx man

attempts, finding, number, probability, question 