Q37 / 90 · JEE Main 2019 (10 Jan Shift 2) · Mathematics

If the probability of hitting a target by a shooter, in any shot is $\frac{1}{3},$ then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than $\frac{5}{6},$ is:

A
$4$
Correct answer
B
$5$
C
$6$
D
$3$

Solution

P (hitting target) = \frac{1}{3}

Let number of trials

= n

P (h i t t i n g t a r g e t a t l e a s t o n c e)

= 1 - P

(not hit)

= 1 - {(\frac{2}{3})}^{n} > \frac{5}{6}

\Rightarrow (\frac{1}{6}) > {(\frac{2}{3})}^{n}

So minimum number of trials

= 5

Hence, the correct answer is option: 'B'.

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