Conditional Probability and proper use of Bayes' Theorem

Question

We're shooting arrows at a target today and we know that $P(\text{bull's eye})=\frac{1}{5}$ and $10$ shots are fired.

We need to find out the probability that

(a) bull's eye is hit at least two times and

(b) the bull's eye is hit at least twice given that its been hit at least once.

For part a) I know that the $P(\text{bull's eye at least twice})=1-P(\text{bull's eye at most twice})$.

Using Bayes' Rule, would my equation look like this?

$$1- \bigg(\frac{4}{25}\bigg){8 \choose 2} \, + \, \bigg(\frac{1}{25}\bigg){8 \choose 2}+\bigg(\frac{4}{25}\bigg) \frac{{8 \choose 2}}{{10 \choose 2}}$$

If not, how can I think of formulating the Bayes' Rule equation here?

For part b), we're looking for $P(\text{bull's eye at least twice}\mid \text{bull's eye at least once})$ and this equals $$P(\text{bull's eye at least once}\mid \text{bull's eye at least twice})+ \frac{P(\text{bull's eye at least twice})}{P(\text{bull's eye once})}.$$

How would you find P(bull's eye at least once|bull's eye at least twice); I'm not sure how this makes sense.

Can someone help me understand this better?

Answer 1

For part a), at least 2 is $1-(P(1 \ \text{hit and 9 misses}) +P( 0 hits)) $

This is just A binomial distribution, I.e. Bernoulli Trials. Unless I am mistaken.

Are you sure that is how part (b) has been asked?

Part b is asking for the probability of at least 2 bulls knowing that there is at least 1 bull.

So the only contradiction to this is if there are just 1 Bull. Using same concept as part a, 1-P(1 bull)

Sorry about poor formatting AI am on my IPhone.

Answer 2

$$P(\text{Hit the bull's eye}) = \frac{1}{5}$$

What is the probability of Hitting the Bull's eye more than twice in 10 Shots

$$P( 2 \text{ or more in }10) = 1 - P(0\text{ in } 10) - P(1 \text{ in } 10) = 1 - \bigg(\frac{4}{5}\bigg)^{10} - {10 \choose 1} \frac{1}{5} \bigg(\frac{4}{5}\bigg)^9$$

For the second part when you say the probability that the bull's eye is hit given that its been hit at least once. this is just

$$P(\text{ Bull's eye is Hit } \mid 1 \text{ or more in }10) = P(1 \text{ or more in }10 \mid 1 \text{ or more in }10) = 1$$

I assume you mean to calculate the probability that the Bull's eye is hit twice given that it is hit at least once. Then this is

$$P(2 \text{ or more in }10 \mid 1 \text{ or more in }10) = \frac{P([2 \text{ or more in }10] \cap [1 \text{ or more in }10])}{P(1 \text{ or more in }10)} = \frac{P([2 \text{ or more in }10] }{P(1 \text{ or more in }10)} = \frac{1 - \bigg(\frac{4}{5}\bigg)^{10} - {10 \choose 1} \frac{1}{5} \bigg(\frac{4}{5}\bigg)^9}{1 - \bigg(\frac{4}{5}\bigg)^{10} }$$

Answer 3

Some Corrections

First of all, this problem does not require Baye's theorem (BT). BT is useful if, for example, we had three archers labeled A,B, and C, each of which have their own probabilities for hitting the bull's eye, and given that an arrow hits the bull's eye, what is the probability that archer C fired the arrow? Ultimately, BT is entirely inappropriate for this problem.

Also, you wrote, "$P(\text{bull's eye at least twice})=1-P(\text{bull's eye at most twice})$." But this is not correct. It should be

$$P(\text{bull's eye at least twice})=1-P(\text{bull's eye at most once})$$

because "at most twice" still includes the value 2.

My Solution

First, let us define the event

$$B = \mathrm{Hitting \enspace the \enspace Bull's \enspace Eye}$$

to promote brevity.

With our new notation and the information given in the problem, we define the probability $P$ that an archer shoots the bull's eye as

$$P(B) \, \, = \, \, \frac{1}{5}$$

Let $X$ be the number of times the archer successfully hits the bull's eye. Furthermore, let each attempt where the archer prepares to shoot the arrow be called a trial. So trial one is the archer's first attempt, trial nine is her ninth attempt, and so on up to ten (since we are firing each arrow ten times).

a) There are two ways to compute this probability.

i) We could iterate through every possible value of $x$ that satisfies the inequality $2\leq x \leq 10$. We first note that the trials are independent: e.g. firing the arrow on my fourth attempt will not affect my chances on my fifth attempt, so the probability of success $P(B)$ is preserved in each trial. We conclude that $P(B)$ is constant. Since the trials are independent, we have to add together the various probabilities for each trial. In the case where $X=4$, which means that the archer successfully hits the bull's eye $4$ out of the $10$ shots, we have $(\frac{1}{5})^4 (\frac{4}{5})^6$. But notice our four successful shots can occur anywhere among the sequence of 10 shots. For example, if $S$ denotes a successful shot and $F$ a failure, then five possible outcomes would be

$$SSSSFFFFFF$$ $$SFSFSFSFFF$$ $$FFFFFFSSSS$$ $$SSFFSSFFFF$$ $$SFFFSSFFFS$$

The total number of subsets consisting of four successes and six failures is $10 \choose 4$

Therefore, the probability that the archer hits the bull's eye $4$ times out of $10$ attempts is $P(X=4) = {10\choose 4} (\frac{1}{5})^4 (\frac{4}{5})^6 \approx 0.0881.$

By extending $x$ to satisfy $2 \leq x \leq 10$, we use sigma notation to deduce

$$P(2 \leq x \leq 10) \, = \, \sum_{x=2}^{10} {10 \choose x} \bigg(\frac{1}{5}\bigg)^x \bigg(\frac{4}{5}\bigg)^{10-x} \approx 0.6242.$$

Therefore, the probability that the archer will hit the bull's eye at least two times out of ten is approximately $0.6242$.

ii) Another way to solve this is using the complement rule. If we have a sample space $S$ and two mutually exclusive events $A$ and $\overline{A}$ such that $S = A \cup \overline{A}$, then $P(S) = 1 = P(A) + P(\overline{A})$, which implies that $P(\overline{A}) = 1 - P(A)$. Similarly,

$$P(X\geq 2) = 1 - P(x < 2) = 1-P(x = 0,1) = 1- P(x=0) - P(x=1)$$

We deduce that $P(X=0) = {10 \choose 0}\Big(\frac{1}{5}\Big)^0 \Big(\frac{4}{5}\Big)^{10} = \Big(\frac{4}{5}\Big)^{10} \approx 0.1074$. Similarly, $P(X=1) \approx 0.2684, which is funny, because the two probabilities are saying that it is more likely for the archer to make one shot among ten attempts than it is her for her miss completely.

Therefore, we have

$$1 - P(X=0) - P(X=1) \approx 0.6242$$

which agrees with our first answer obtained in part i).

b) The point of this part is interesting. Imagine that you've successfully shot the arrows into the bull's eye $4$ times. You wonder if among the next six attempts, your chances of hitting the bull's eye will increase. In other words, what are the chances of you hitting the bull's eye $(x+1)$ more times if you've already hit the bull's eye $x$ more times, where $1 \leq x \leq 10$?

The probability of hitting the bull's eye at least once can be determined using any of the two methods applied in part a. Since the complement rule is easier to apply, we see that

$$P(X\geq 1) = 1 - P(X=0) \approx 0.8926$$

Now we can find $P(X \geq 2 | X \geq 1)$, which is

$$P(X \geq 2 | X \geq 1) = \frac{P(X \geq 2) \enspace and \enspace P(X\geq 1)}{P(X \geq 1)} = \frac{P(X \geq 2)}{P(X\geq 1)} \approx 0.69927 \approx 0.7000$$