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MISCELLANEOUS EXERCISE 8 (I) · Q21

Q.The probability of a shooter hitting a target is 34\dfrac{3}{4}. How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 0.990.99? (A) 22 (B) 33 (C) 44 (D) 55

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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With p=34p=\dfrac34 per shot, q=14q=\dfrac14. 'At least one hit in nn shots' is the complement of 'no hit at all': P(X≥1)=1−qn=1−(14)nP(X\ge1)=1-q^n=1-\left(\dfrac14\right)^n.

We need 1−(14)n>0.991-\left(\dfrac14\right)^n>0.99, i.e. (14)n<0.01\left(\dfrac14\right)^n<0.01. …

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