The probability of a man hitting a target is | vedclass

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Question

(A) Let $p$ be the probability of hitting the target,$p = \frac{2}{5}$.Then the probability of missing the target is $q = 1 - p = 1 - \frac{2}{5} = \f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let $p$ be the probability of hitting the target,$p = \frac{2}{5}$.Then the probability of missing the target is $q = 1 - p = 1 - \frac{2}{5} = \frac{3}{5}$.The probability of hitting the target at least once in $k$ trials is given by $1 - P(	ext{hitting zero times}) = 1 - q^k$.We are given that this probability is greater than $\frac{7}{10}$:$1 - (\frac{3}{5})^k > \frac{7}{10}$Subtracting $1$ from both sides:$-(\frac{3}{5})^k > \frac{7}{10} - 1$$-(\frac{3}{5})^k > -\frac{3}{10}$Multiplying by $-1$ (and reversing the inequality sign):$(\frac{3}{5})^k For $k=1$: $(\frac{3}{5})^1 = 0.6$,which is not less than $0.3$.For $k=2$: $(\frac{3}{5})^2 = \frac{9}{25} = 0.36$,which is not less than $0.3$.For $k=3$: $(\frac{3}{5})^3 = \frac{27}{125} = 0.216$,which is less than $0.3$.Thus,the minimum value of $k$ is $3$.

The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times ($k$ is a given number). The minimum value of $k$ such that the probability of hitting the target at least once is more than $\frac{7}{10}$ is:

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