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BITSAT 2012 Paper

Section: Mathematics

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Question : 118 of 150

Marks: +1, -0

A boy is throwing stones at a target. The probability of hitting the target at any trial is

\frac{1}{2}

. The probability of hitting the target 5th time at the 10th throw is :

$\frac{5}{2^{10}}$
$\frac{63}{2^{9}}$
$\frac{\,{}^{10}C_5}{2^{10}}$
None

Solution:

The probability of hitting the target 5th time at the 10th throw = P(the probability of hitting the target 4 times in the first 9 throws) × P(the probability of hitting the target at the 10th throw) =

\left[\,{}^{9}C_4 \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^5\right] \left(\frac{1}{2}\right)

=

\frac{9!}{4!5!}

×

\left(\frac{1}{2}\right)^{10}

=

\frac{63}{2^{9}}

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