The probability that all 4 will fire is 1/6 and at most 2 will not fire is 39/49.
From a lot of 10 missiles, 4 are selected at ran- dom and fired.
Sample space = 10
the lot contains 3 defective missiles that will not fire
Non - defective missiles = 7
probability that all 4 will fire
P(all four will fire) = [tex]\frac{7}{10}.\frac{6}{9} .\frac{5}{8}.\frac{4}{7}[/tex]
P(all four will fire) = 840/5040
P(all four will fire) = 1/6
Atmost two will not fire = P(all fire) + P(1 will not fire) + P(two will not fire)
= [tex]\frac{4}{7} + \frac{3}{7}.\frac{1}{3} + \frac{2}{7}.\frac{2}{7}[/tex]
= 4/7 + 3/21 + 4/49
= 39/49
Therefore, the probability that all 4 will fire is 1/6 and at most 2 will not fire is 39/49.
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