The probability of a man hitting a target in one fire is $\frac{1}{6} .$ The number of times at least he must fire at the target in order that his chance of hitting the target at least once will exceed $\frac{2}{5}$ will be
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Solution

The correct option is A 3
Probability of hitting target $= \frac{1}{6}$

Probability of missing target $= \frac{5}{6}$

Probability of hitting target as least once = 1 - (Probability of not hitting n times)

$= 1 -^{n} C_{0} p^{0} q^{n}$

$= 1 - {(\frac{5}{6})}^{n}$

$1 - {(\frac{5}{6})}^{n} > \frac{2}{5}$

${(\frac{5}{6})}^{n} < \frac{3}{5}$

$n > 2.8$

$n_{l e a s t} = 3$

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[Kurukshetra CEE 1998]

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The probability of a man hitting a target in one fire is 16. The number of times at least he must fire at the target in order that his chance of hitting the target at least once will exceed 25 will be 3

The correct option is A 3Probability of hitting target =16 Probability of missing target =56 Probability of hitting target as least once = 1 - (Probability of not hitting n times) =1− nC0 p0 qn =1−(56)n 1−(56)n>25 (56)n<35 n>2.8 nleast=3

The probability of a man hitting a target in one fire is $\frac{1}{6} .$ The number of times at least he must fire at the target in order that his chance of hitting the target at least once will exceed $\frac{2}{5}$ will be
3