Shooting at a target probabilities.

Z

Zelda22

Junior Member
Joined
Mar 30, 2022
Messages
136
Arthur and Bert hit a target with probabilities of 4/5 and 7/10 respectively.
Find the probability that;

1. they both hit the target
2. they both miss the target
3. at LEAST one of them hits the target.

Please help, I don't know how to solve it, here is what I tried.

(hit,hit)
(4/5 * 7/10) = 14/25

(miss,miss)
(1/5 * 3/10) = 3/50

(hit, miss)
(4/5 * 3/10) = 6/25

(miss,hit)
(1/5 * 7/10) = 7/50


I'm lost. Please help. Thanks.
 
Last edited:
What you did at the end is correct. Good work!
My problem is that you have three questions yet you have four answers on your last line.
Again, your work so far is good, now answer the questions!

p(they both hit the target) = ?
p(they both miss the target)=?
p(at LEAST one of them hits the target) = ?
 
For part c) it is one minus both miss. WHY?
 
You still have not answered all the questions!!!
Can you answer p(at LEAST one of them hits the target) using your work?

Do you understand that p(at LEAST one of them hits the target) can NOT have two answers??

I would normally recommend that you solve the problem as pka suggested, however you need to know how to do the problem using your method since you are doing it wrong.
 
Steven G said:
What you did at the end is correct. Good work!
My problem is that you have three questions yet you have four answers on your last line.
Again, your work so far is good, now answer the questions!

p(they both hit the target) = ?
p(they both miss the target)=?
p(at LEAST one of them hits the target) = ?
Click to expand...
p(they both hit the target) = 14/25
p(they both miss the target)= 3/50
p(at LEAST one of them hits the target) = (here, I'm not sure) I'm thinking about the complementary event. maybe 1-both miss?
 
@zelda, You are correct in using complements.

You could also get the answer by adding the probabilities: P(hit,hit) + P(hit, miss) + P(miss hit)

because these are the 3 ways of getting " at LEAST one hit".

Do that addition to see that you do, in fact, get the same answer.

It is important to know both ways, because usually one will be more efficient.
 
Steven G said:
You need to know how to do it using your method! Being able to do the method without complements is extremely important.
Click to expand...
I was trying but it didn't work, I was thinking the probability of (hit,hit) is 1/4 , (miss, miss) is 1/4, (hit, miss) 1/4 and (miss, hit) 1/4, then I multiply each outcome by 1/4, add them, but didn't work, :( , can you explain to me, please.
 
Harry_the_cat said:
@zelda, You are correct in using complements.

You could also get the answer by adding the probabilities: P(hit,hit) + P(hit, miss) + P(miss hit)

because these are the 3 ways of getting " at LEAST one hit".

Do that addition to see that you do, in fact, get the same answer.

It is important to know both ways, because usually one will be more efficient.
Click to expand...
Thank you.
 
Zelda22 said:
I was trying but it didn't work, I was thinking the probability of (hit,hit) is 1/4 , (miss, miss) is 1/4, (hit, miss) 1/4 and (miss, hit) 1/4, then I multiply each outcome by 1/4, add them, but didn't work, :( , can you explain to me, please.
Click to expand...
No, even though there are 4 possibilities ie HH, HM, MH, MM, they are not equally likely, so 1/4 doesn't come into it. You have calculated the prob of each one correctly in your first post. Just take the three that meet the "at least one hit" criteria and add their probs together.
 
Last edited:
This might help too. Notice that the probs of the 4 outcomes you calculated in your first post add up to 1. This should always be the case if you have accounted for all outcomes.

So P(HH) + P(HM) + P(MH) + P(MM) = 1

Rearranging this, you get

1 - P(MM) = P(HH) + P(HM) + P(MH)

The LHS uses complements, the RHS uses addition of probs of relevant outcomes. They are equal!!
 
Harry_the_cat said:
This might help too. Notice that the probs of the 4 outcomes you calculated in your first post add up to 1. This should always be the case if you have accounted for all outcomes.

So P(HH) + P(HM) + P(MH) + P(MM) = 1

Rearranging this, you get

1 - P(MM) = P(HH) + P(HM) + P(MH)

The LHS uses complements, the RHS uses addition of probs of relevant outcomes. They are equal!!
Click to expand...
Thank you so much. I understand now ?
 
You must log in or register to reply here.
Top