Question

the probability of a hitting a target is 4/5 the probability of b hitting is 3/4 and the probability of c missing the target is 1/3 what is the probability of the being hit at leat twice
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Answer 1

Step-by-step explanation:

probability of target getting hit atleast twice =p(e)

= probability of twice + probability of hitting thrice

= abc'+ab'c+a'bc+abc

let p(a) is prob. of A hitting target,

p(a') is missing the target

p(b) is prob. of b hitting target, p(b') b missing target

p(c) is probability of c hitting target, p(c') missing target

p(a)=4/5, p(a')=1-4/5=1/5

p(b)=3/4, p(b')= 1-3/4=1/4

p(c')=1/3, p(c)=1-1/3=2/3

p(e) = 4/5×3/4×1/3 +

4/5×1/4×2/3+

1/5×3/4×2/3 +

4/5×3/4×2/3

= 1/5 + 2/15 + 1/10 +2/5

=(6+4+3+12)/30

=25/30=5/6