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anonymous

  • 4 years ago

Four missiles are fired at a target. If the probabilities of hitting the target are 0.3, 0.4, 0.5 and 0.6 respectively and if the missiles are fired independently then what will be the probability i) That all the missiles hit the target? ii) That at least one of the four hits the target? iii) That exactly one hits the target? iv) That exactly 2 hit the target?

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  1. dumbcow
    • 4 years ago
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    multiply independent probabilities i) .3 * .4 * .5 *.6 = .0036

  2. dumbcow
    • 4 years ago
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    oops it should be .036

  3. dumbcow
    • 4 years ago
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    P(none) = .7*.6*.5*.4 = .084 1 - .084 = .916

  4. dumbcow
    • 4 years ago
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    iii) There are 4 distinct cases, in each case 1 hits target while rest miss add up probabilities of each case iv) There are 6 distinct cases where 2 hit while other 2 miss add up probabilities of each case

  5. anonymous
    • 4 years ago
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    what is none please describe the answe in format so i can understand...

  6. dumbcow
    • 4 years ago
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    prob that none of missiles hit target

  7. anonymous
    • 4 years ago
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    Q.Four missiles are fired at a target. If the probabilities of hitting the target are 0.3, 0.4, 0.5 and 0.6 respectively and if the missiles are fired independently then what will be the probability multiply independent probabilities solution. i) 0.3 x 0.4 x 0.5 x 0.6 = 0.036 ii) P(none) = 0.7 x 0.6 x 0.5 x 0.4 = 0.084 P(none) =1 - 0.084 = 0.916 where 0.7, 0.6, 0.5 and 0.4 comes from ? iii) There are 4 distinct cases, in each case 1 hits target while rest miss add up probabilities of each case iv) There are 6 distinct cases where 2 hit while other 2 miss add up probabilities of each I'll try to formalize this. Q.Four missiles are fired at a target. If the probabilities of hitting the target are 0.3, 0.4, 0.5 and 0.6 respectively and if the missiles are fired independently then what will be the probability I'll put a name to each missile. A, B, C, and D respectly. Then: P(A)=0.3 (the first missile hit probability) P(B)=0.4 (the second) P(C)=0.5 (third) P(D)=0.6 (forth) If an event has x chance to happen, then it has 1-x chances to happen, remember, 1 is the total of the all possible probabilities. I'll name ~X (where X is any event) to the probability of X not happening. So: P(~A)=1-0.3=0.7 P(~B)=1-0.4=0.6 P(~C)=1-0.5=0.5 P(~D)=1-0.6=0.4 Then you'll have 4 distinct cases one and just one missile missing ~A, B, C, D A, ~B, C, D A, B, ~C, D A, B, C, ~D Probability of the intersection (two or more happening together) of independent events is the multiplication of the independent probabilities of each one. So: P(~A intersection B intersection C intersection D) = P(~A)*P(B)*P(C)*P(D)=0,7*0,4*0,5*0,6=0,084 P(A intersection ~ B intersection C intersection D) = P(A)*P(~B)*P(C)*P(D)= 0,3*0,6*0,5*0,6=0,054 P(A intersection B intersection ~C intersection D) = P(A)*P(B)*P(~C)*P(D)= 0,3*0,4*0,5*0,6=0,036 P(A intersection B intersection C intersection ~D) = P(A)*P(B)*P(C)*P(~D)= 0,3*0,4*0,5*0,4=0,024 The different ways of two and only two missiles hitting is: The combinatory number 4:2=4!/(2!*2!)=6, because you're taking two elements of a set of four. They're: ~A, ~B, C, D ~A, B, ~C, D ~A, B, C, ~D A, ~B, ~C, D A, ~B, C, ~D A, B, ~C, ~D You can done the probability calculation for this now. The next time try to post your approach so we can help you in a more precise and efficient maner.

  8. anonymous
    • 4 years ago
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    Ah, I answered III and IV All missiles (I) are easy is P(A)*P(B)*P(C)*P(D). Probability of all missing (P(~A intersection ~B intersection ~C intersection ~D) is P(~A)*P(~B)*P(~C)*P(~D) If they not miss the four together, then at least one of them hit the target. So this is: P(~(~A intersection ~B intersection ~C intersection ~D))= 1-(P(~A intersection ~B intersection ~C intersection ~D)=1- P(~A)*P(~B)*P(~C)*P(~D)

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