Question

How to solve extended response question using probability?

Can someone please explain to me how to do question 11? Thanks!

Answer 1

A) `1/60`
B) `1/5`
C) `3/5`
D) `6/13`

Explanation:

Take the problem in turns:

Part A

The probability that all three shots hit the target is the probability `P(A nn B nn C)` where A, B, and C refer to the events of each respective shooter's shot hitting the target.

Since we assume that each shooter is independent of each other, this will be `P(A) * P(B) * P(C)`, or `1/5*1/4*1/3 = 1/60`.

Part B

The probability that only C's shot will hit the target is `P(bar(A) nn bar(B) nn C)`, where `bar(A)` and `bar(B)` refer to the event of A's shot not hitting the target (and the same for B).

Again, since we assume that each shooter is independent of each other, this will be `P(bar(A)) * P(bar(B)) * P(C)`, or `4/5 * 3/4 * 1/3 = 12/60 = 1/5`

Part C

The probability of at least 1 of A, B, or C hitting the target may seem tricky to solve, but we have to consider a clever alternate way to solve this. Since the sum of all probabilities in a problem space is 1, we could find the probability that none of the shots hit the target. If we find the complement of that (ie, we subtract that from 1), we are left with all of the probabilities that result in some combination of shots hitting the target; it could be just A, or just B, or just C, or A and B, or A and C, or B and C, or all three. It's far easier to calculate none hitting and subtract from 1 than each of the 7 I just listed and adding them together.

`P(bar(A) nn bar(B) nn bar(C)) = P(bar(A)) * P(bar(B)) * P(bar(C))`

`= 4/5 * 3/4 * 2/3 = 24/60 = 2/5`

Now, the probability of at least 1 hit: `1 - 2/5 = 3/5`

Part D

The probability of C being the shot to hit the target, given that only 1 shot hit is calculated a little differently. This is an example of a conditional probability, or one in which you are finding the probability of a certain event happening given that you know ahead of time that another event happened.

In this case, we are told that a single shot hit the target. (At this point it could be only A, or only B, or only C). Let's say O represents only one shot hitting. The probability question for this question can be represented as `P(C | O)`, which can be read as "the probability of C given O", or "the probability of C hitting the target given only one shot hit the target".

For conditional probability questions, we have a formula we can use:

`P(C | O) = (P(C nn O))/(P(O))`

The top value `P(C nn O)` stands for "the probability of C hitting the target AND only one shot hitting the target", while `P(O)` stands for "the probability of only one shot hitting the target". Let's find each, and then divide to get the answer.

To find `P(C nn O)`, we need to think in terms of the intersection of both events; in other words, a situation where only one shot hit the target and it happened to be C's shot. If you look back to Part B, you'll notice we were asked to find the probability that only C's shot hit the target. This is the exact same probability being sought here. Thus `P(C nn O) = 1/5`

To find `P(O)`, where only 1 shot hits the target, we have to realize that there are 3 scenarios under which that might happen: A is the only one to hit, B is the only one to hit, or C is the only one to hit. (We know this last one is `1/5`). The other two we can calculate the same way as we did for C in Part B:

`P("only" A) = P(A) * P(bar(B)) * P(bar(C)) = 1/5 * 3/4 * 2/3 = 6/60 = 1/10`

`P("only" B) = P(bar(A)) * P(B) * P(bar(C)) = 4/5 * 1/4 * 2/3 = 8/60 = 2/15`

Thus, `P(O) = P("only" A) + P("only" B) + P("only" C)`

`= 1/10 + 2/15 + 1/5 = 3/30 + 4/30 + 6/30 = 13/30`

Lastly, we can now use the conditional formula:

`P(C | O) = (P(C nn O))/(P(O)) = (1/5)/(13/30) = 1/5 *30/13 = 6/13`