Math Help
Hi, trying to help my younger sister with math hw problems and I got stump. Help is needed!! I have 2 questions:
1. The probability that a bomber hits its target on any particular mission is .80. Four bombers are sent after the same target. What is the probability:
a. They all hit the target? Ans: (0.8) ^ 4 = 0.4096
b. None hit the target? Ans: (0.2) ^ 4 = 0.0016
c. At least one hits the target? Ans: 1
1. The probability that a bomber hits its target on any particular mission is .80. Four bombers are sent after the same target. What is the probability:
a. They all hit the target? Ans: (0.8) ^ 4 = 0.4096
b. None hit the target? Ans: (0.2) ^ 4 = 0.0016
c. At least one hits the target? Ans: 1
For #1:
You need to include P(hit)*P(nohit)*P(hit)*p(hit), p(hit)*p(nohit)*p(hit)*p(nohit), etc. This is because p(hit)*p(nohit)*p(hit)*p(nohit) is a different event than p(hit)*p(hit)*p(nohit)*p(nohit) even though both result in 2 hits and 2 misses. Ie, the order matters. Alternatively, it's MUCH easier to simply take 1 minus the probability that no bombers hit. The only outcomes are no hits, 1 hit, 2 hits, 3 hits, or 4 hits and they must sum to 1.
So, Pr(x=0) + Pr(x=1)+Pr(x=2)+Pr(x=3)+Pr(x=4) = 1
Pr (x=1,2,3,4) + Pr(x=0) =1 {Just rewrote the above}
Pr (x=1,2,3,4) = 1 - Pr(x=0)
For #2
I have no idea what your asking. You calculated the mean salary earlier. It's a sort of stupid question to begin with because statistically you can't determine the upper limit for the last class. It's very unlikely that anyone in the sample population would have a salary bigger than 1,000,000 but you never know... maybe you've got a Bill Gates. You could determine a 95 or 99 confidence interval that no one was above a certain salary either by approximating the distribution as normal (its not), or using a stats program like SHAZAM... but you aren't at that level.
If you're asking how to calculate the mean... take the median number for each range is all you can really do. That or take the mean using the low end, mid point, and high end of each range and then average the three means you get. Problem with that point is wtf do you use for the high point of the last income bracket? That or I'm totally wrong on what they want you to do...
You need to include P(hit)*P(nohit)*P(hit)*p(hit), p(hit)*p(nohit)*p(hit)*p(nohit), etc. This is because p(hit)*p(nohit)*p(hit)*p(nohit) is a different event than p(hit)*p(hit)*p(nohit)*p(nohit) even though both result in 2 hits and 2 misses. Ie, the order matters. Alternatively, it's MUCH easier to simply take 1 minus the probability that no bombers hit. The only outcomes are no hits, 1 hit, 2 hits, 3 hits, or 4 hits and they must sum to 1.
So, Pr(x=0) + Pr(x=1)+Pr(x=2)+Pr(x=3)+Pr(x=4) = 1
Pr (x=1,2,3,4) + Pr(x=0) =1 {Just rewrote the above}
Pr (x=1,2,3,4) = 1 - Pr(x=0)
For #2
I have no idea what your asking. You calculated the mean salary earlier. It's a sort of stupid question to begin with because statistically you can't determine the upper limit for the last class. It's very unlikely that anyone in the sample population would have a salary bigger than 1,000,000 but you never know... maybe you've got a Bill Gates. You could determine a 95 or 99 confidence interval that no one was above a certain salary either by approximating the distribution as normal (its not), or using a stats program like SHAZAM... but you aren't at that level.
If you're asking how to calculate the mean... take the median number for each range is all you can really do. That or take the mean using the low end, mid point, and high end of each range and then average the three means you get. Problem with that point is wtf do you use for the high point of the last income bracket? That or I'm totally wrong on what they want you to do...
Originally Posted by S2KANDRE,May 1 2007, 11:17 PM
pm magician
[QUOTE=nwk00,May 1 2007, 10:32 PM]Hi, trying to help my younger sister with math hw problems and I got stump. Help is needed!! I have 2 questions:
1. The probability that a bomber hits its target on any particular mission is .80. Four bombers are sent after the same target. What is the probability:
a. They all hit the target? Ans: (0.8) ^ 4 = 0.4096
b. None hit the target? Ans: (0.2) ^ 4 = 0.0016
c. At least one hits the target? Ans: 1
1. The probability that a bomber hits its target on any particular mission is .80. Four bombers are sent after the same target. What is the probability:
a. They all hit the target? Ans: (0.8) ^ 4 = 0.4096
b. None hit the target? Ans: (0.2) ^ 4 = 0.0016
c. At least one hits the target? Ans: 1
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