A man wins if he hits the target. The chance of hitting the target in one shoot is $20 %$ . He stops when he hits the target. Let $p$ be the probability that he wins in $n^{th}$ attempt, where $20 p^{2} - 13 p + 2 \leq 0,$ then the possible value of $n$ is

2

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Solution

The correct option is A $2$
Probability of hitting the target is $\frac{1}{5}$ .
Probability of missing the target is $\frac{4}{5}$ .

$20 p^{2} - 13 p + 2 \leq 0$
$\Rightarrow p \in [\frac{1}{4}, \frac{2}{5}]$

$\frac{1}{4} \leq \frac{1}{5} + (\frac{4}{5}) \frac{1}{5} + {(\frac{4}{5})}^{2} \frac{1}{5} + \dots + {(\frac{4}{5})}^{n - 1} \frac{1}{5} \leq \frac{2}{5}$
$\Rightarrow \frac{1}{4} \leq 1 - {(\frac{4}{5})}^{n} \leq \frac{2}{5}$
$\Rightarrow - \frac{3}{4} \leq - {(\frac{4}{5})}^{n} \leq - \frac{3}{5}$
$\Rightarrow \frac{3}{4} \geq {(\frac{4}{5})}^{n} \geq \frac{3}{5}$
$\Rightarrow n = 2$

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A man wins if he hits the target. The chance of hitting the target in one shoot is 20%. He stops when he hits the target. Let p be the probability that he wins in nth attempt, where 20p2−13p+2≤0, then the possible value of n is

The correct option is A 2Probability of hitting the target is 15. Probability of missing the target is 45. 20p2−13p+2≤0 ⇒p∈[14,25] 14≤15+(45)15+(45)215+⋯+(45)n−115≤25 ⇒14≤1−(45)n≤25 ⇒−34≤−(45)n≤−35 ⇒34≥(45)n≥35 ⇒n=2

A man wins if he hits the target. The chance of hitting the target in one shoot is $20 %$ . He stops when he hits the target. Let $p$ be the probability that he wins in $n^{th}$ attempt, where $20 p^{2} - 13 p + 2 \leq 0,$ then the possible value of $n$ is