A rat maze consists of a straight corridor, at the end of which the rats take either right or left turn. If $10$ rats are placed in the maze one at a time and the random variable $X$ denotes the number of right turns taken by the rats then what is the probability distribution of $X$ ? Find the probability that atleast $9$ rats will turn the same way.

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Solution

Consider $x$ be the number of right turns taken by the rats

Therefore, $p (s u c e s s) = p = \frac{1}{2}, q = \frac{1}{2}, n = 10$

$p (a t l e a s t 9 r a t s t a k e t h e s a m e t u r n) = p (a t l e a s t 9 r a t s t a k e t h e l e f t o r r i g h t t u r n) = p (9 r i g h t t u r n s o r 10 r i g h t t u r n s o r 0 r i g h t t u r n o r 1 r i g h t t u r n)$

So,

$\begin{matrix} = p (x = 0) + p (x = 1) + p (x = 9) + p (x = 10) =^{10} c_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{10} +^{10} c_{1} + {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{9} +^{10} c_{9} {(\frac{1}{2})}^{9} {(\frac{1}{2})}^{1} +^{10} c_{10} {(\frac{1}{2})}^{10} {(\frac{1}{2})}^{0} = (^{10} c_{0} +^{10} c_{1} +^{10} c_{9} +^{10} c_{10}) = \frac{1}{2^{10}} = (1 + 10 + 10 + 1) \times \frac{1}{1024} = \frac{22}{1024} = \frac{11}{512} = 0.0214 \end{matrix}$

Hence, The probability of at-least $9$ rats will return the same way is $0.0214$

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