Test Index

BITSAT 2016 Paper

Section: Mathematics

Question : 125 of 150

Marks: +1, -0

Solution:

0.4+0.6=1

, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure.

Then, the probability of success in one step

p=0.4=\frac{2}{5}

The probability of failure in one step

=q=0.6=\frac{3}{5}

In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1.

So, the number of successes

=6

The number of failures

=5

or the number of successes

=5

The number of failures

=6

∴ the required probability

=^{11}C_{6}p^{6}q^{5}+^{11}C_{5}p^{5}q^{6}

=^{11}C_{6}\left(\frac{2}{5}\right)^{6}\left(\frac{3}{5}\right)^{5}+^{11}C_{5}\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{6}

=\frac{11!}{6!5!}\cdot\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{5}\left\{\frac{2}{5}+\frac{3}{5}\right\}

=\frac{11\cdot10\cdot9\cdot8\cdot7}{120}\cdot\frac{2^{5}\cdot3^{5}}{5^{10}}=462\times\left(\frac{6}{25}\right)^{5}

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