A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. lf p satisfies the inequality 625p^2 - 175p + 12 <0 then value of n is?

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A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. lf p satisfies the inequality 625p^2 - 175p + 12 <0 then value of n is?

Answer is 3

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Answer 1 · 2018-03-03T16:31:43

$n = 3$ $n=3$

Explanation:

$p (n) = Hitting for the 1st time at the n-th trial$ $p(n) = "Hitting for the 1st time at the n-th trial"$
$\Rightarrow p (n) = {0.8}^{n - 1} \cdot 0.2$ $=> p(n) = 0.8^(n-1) * 0.2$

$Boundary of the inequality 625 p^{2} - 175 p + 12 = 0$ $"Boundary of the inequality "625 p^2 - 175 p + 12 = 0"$
$is the solution of a quadratic equation in p :$ $"is the solution of a quadratic equation in "p" :"$
$disc : 175^{2} - 4 \cdot 12 \cdot 625 = 625 = 25^{2}$ $"disc : "175^2 - 4*12*625 = 625 = 25^2$
$\Rightarrow p = \frac{175 \pm 25}{1250} = \frac{3}{25} or \frac{4}{25}$ $=> p = (175 pm 25)/1250 = 3/25 " or "4/25"$
$So p (n) is negative between those two values.$ $"So "p(n)" is negative between those two values."$

$p (n) = \frac{3}{25} = {0.8}^{n - 1} \cdot 0.2$ $p(n) = 3/25 = 0.8^(n-1) * 0.2$
$\Rightarrow \frac{3}{5} = {0.8}^{n - 1}$ $=> 3/5 = 0.8^(n-1)$
$\Rightarrow log (\frac{3}{5}) = (n - 1) log (0.8)$ $=> log(3/5) = (n-1) log(0.8)$
$=> n = 1 + log(3/5)/log(0.8) = 3.289....$

$p(n) = 4/25 = ...$
$=> n = 1 + log (4/5)/log(0.8) = 2$

$"So " 2 < n < 3.289... => n = 3 " (as n is integer)"$

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A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. lf p satisfies the inequality 625p^2 - 175p + 12 <0 then value of n is?

Answer is 3

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