Choose the correct option from the given alternatives: The probability of a shooter hitting a target is $\frac{3}{4}$ How many minimum numbers of times must he fire so that the probability of hitting the target at least once is more than 0·99?

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KollegeApply · Accepted Answer

4 Hint: P(X $\geq$ 1) > 0.99 $\therefore$ 1 - P(X = 0) > 0.99 $\therefore$ P(X = 0) < 0.01 = $\frac{1}{100}$ $\therefore$ `"^nC_0 ($\frac{3}{4}$)^0 ($\frac{1}{4}$)^n < $\frac{1}{100}$ $\therefore$ ($\frac{1}{4}$)^n < $\frac{1}{100}$` $\therefore$ n = 4

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