1. ## Basic probability problem..

Hi,

I am having trouble with this problem I know it might seem really simple but i dont understand how to calculate what the questions ask me.

Q:

The probabilty A hits a target is 1/3

The probabilty A hits a target is 1/5

They both fire at the same time.Find the probability of the following:

(i) A does not hit target
(ii) both A and B hit target
(iii) at least one of them hit target
(iv) Neither hit target.

2. Originally Posted by ColdLamp
Hi,

I am having trouble with this problem I know it might seem really simple but i dont understand how to calculate what the questions ask me.

Q:

The probabilty A hits a target is 1/3

The probabilty A hits a target is 1/5

They both fire at the same time.Find the probability of the following:

(i) A does not hit target
(ii) both A and B hit target
(iii) at least one of them hit target
(iv) Neither hit target.
What problem are you having with (i)?

(ii) is the product of the individual probabilities that A hits and B hits.

(iii) is 1 minus the probability that both miss

CB

3. i just dont understand were or how to begin calculating to get anywere near an answer.I know it might sound like im lazy but ive had to post here because i cannot apply anything i have found online to the questions

4. Originally Posted by ColdLamp
i just dont understand were or how to begin calculating to get anywere near an answer.I know it might sound like im lazy but ive had to post here because i cannot apply anything i have found online to the questions
This question comes down to understanding how combining probabilities work and how likelihood relates to unlikelihood. The probability of A not occurring is 1 minus the probability is does occur. If you want to know the chances of two events both occurring you multiply their chances together. At least one occurring is A or B occurring, so you add them. Neither occurring is the opposite of at least one occurring, so you take 1 minus the previous answer.

These seem difficult at first but it's really just applying a few concepts on combining probabilities.

5. Originally Posted by Jameson
This question comes down to understanding how combining probabilities work and how likelihood relates to unlikelihood. The probability of A not occurring is 1 minus the probability is does occur. If you want to know the chances of two events both occurring you multiply their chances together. At least one occurring is A or B occurring, so you add them. Neither occurring is the opposite of at least one occurring, so you take 1 minus the previous answer.

These seem difficult at first but it's really just applying a few concepts on combining probabilities.

Adding A and B for atleast 1 doesn't work, because they are both separate probabilities, unlike dice or a coin.

If you were looking for at least a 1, 2, or 3 on a dice, you would add all 3 probabilities together, so 1/6 + 1/6 + 1/6 = 1/2

However, this is more like the chance of it raining out, if you have a 75% chance that it will rain on days A and B, there's a (25%*25%) 12.5% chance that it won't rain either day, therefore a (1-.125) 87.5% chance that it will rain at least one day.

6. Originally Posted by Brennan010
Adding A and B for atleast 1 doesn't work, because they are both separate probabilities, unlike dice or a coin.

If you were looking for at least a 1, 2, or 3 on a dice, you would add all 3 probabilities together, so 1/6 + 1/6 + 1/6 = 1/2

However, this is more like the chance of it raining out, if you have a 75% chance that it will rain on days A and B, there's a (25%*25%) 12.5% chance that it won't rain either day, therefore a (1-.125) 87.5% chance that it will rain at least one day.
Yes this is correct. Thank you. Since the two events occur simultaneously the combinations of the event are 1 hit, both hit or neither hit. So at least one hitting it P(A)+P(B)-P(A and B).

7. Originally Posted by Jameson
Yes this is correct. Thank you. Since the two events occur simultaneously the combinations of the event are 1 hit, both hit or neither hit. So at least one hitting it P(A)+P(B)-P(A and B).
Again, the probability that at least one hits is most easily computed as 1 minus the probability that both miss.

CB