Question & Answer
QUESTION

A rat maze consists of a straight corridor, at the end of which 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 function of X? Find the probability that at least 9 rats will turn the same way.

ANSWER Verified Verified
Hint: We have 10 trials. Probability of success of an individual trial is $p=\dfrac{1}{2}$, as the rats can either turn left or right. Use binomial distribution, which is: ${}^{n}{{C}_{X}}{{p}^{X}}{{\left( 1-p \right)}^{n-X}}$, to get the probability distribution function. Then find out the required probability from the distribution function.

Complete step-by-step answer:
We know that the binomial distribution function is a probability distribution function of exactly x success on n repeated trials in an experiment which has two possible outcomes. The function is defined as:
${}^{n}{{C}_{x}}{{p}^{x}}{{\left( 1-p \right)}^{n-x}}$, where p is the probability of success.
According to the question the number of trials is 10.
The two possible outcomes are left and right.
It is given in the question that the random variable X denotes the number of right turns taken by the rats.
If we consider the success of the event by turning right of a rat, then the probability of success will be:
$p=\dfrac{1}{2}$ , there are only two possible outcomes.
 Here, X denotes the number of successes and the number of repeated trials is 10. Therefore, the distribution function will be defined as:
${}^{n}{{C}_{x}}{{p}^{x}}{{\left( 1-p \right)}^{n-x}}={}^{10}{{C}_{X}}{{\left( \dfrac{1}{2} \right)}^{X}}{{\left( 1-\dfrac{1}{2} \right)}^{10-X}}={}^{10}{{C}_{X}}{{\left( \dfrac{1}{2} \right)}^{X}}{{\left( \dfrac{1}{2} \right)}^{10-X}}={}^{10}{{C}_{X}}{{\left( \dfrac{1}{2} \right)}^{X}}^{+10-X}={}^{10}{{C}_{X}}{{\left( \dfrac{1}{2} \right)}^{10}}$
Therefore, the probability distribution function will be as follows:


     X
Probability of X or P(X)
     0
 $P\left( X=0 \right)={}^{10}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{10}}$
    1
$P\left( X=1 \right)={}^{10}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{10}}$
    2
$P\left( X=2 \right)={}^{10}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{10}}$
     3
$P\left( X=3 \right)={}^{10}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{10}}$
      4
$P\left( X=4 \right)={}^{10}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{10}}$
      5
$P\left( X=5 \right)={}^{10}{{C}_{5}}{{\left( \dfrac{1}{2} \right)}^{10}}$
     6
$P\left( X=6 \right)={}^{10}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}$
      7
$P\left( X=7 \right)={}^{10}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}$
       8
$P\left( X=8 \right)={}^{10}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}$
       9
$P\left( X=9 \right)={}^{10}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}$
     10
$P\left( X=10 \right)={}^{10}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}$

Now we need to find out the probability that atleast 9 rats will turn the same way.
Therefore,
Probability of at least 9 rats take the same way = probability of at least 9 rats left or right turn = probability of 9 right turns or 10 right turns or 0 right turn or 1 right turn.
$P\left( X=9 \right)+P\left( X=10 \right)+P\left( X=0 \right)+P\left( X=1 \right)$
$={}^{10}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}$
$={{\left( \dfrac{1}{2} \right)}^{10}}\left( {}^{10}{{C}_{0}}+{}^{10}{{C}_{1}}+{}^{10}{{C}_{9}}+{}^{10}{{C}_{10}} \right)$
We know that: ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$
Therefore,
$=\dfrac{1}{{{2}^{10}}}\left( \dfrac{10!}{\left( 10-0 \right)!\times 0!}+\dfrac{10!}{\left( 10-1 \right)!\times 1!}+\dfrac{10!}{\left( 10-9 \right)!\times 9!}+\dfrac{10!}{\left( 10-10 \right)!\times 10!} \right)$
$=\dfrac{1}{{{2}^{10}}}\left( \dfrac{10!}{10!}+\dfrac{10!}{9!}+\dfrac{10!}{9!}+\dfrac{10!}{10!} \right)$ , by putting 0! = 1, 1! = 1.
$=\dfrac{1}{{{2}^{10}}}\left( 1+\dfrac{10\times 9!}{9!}+\dfrac{10\times 9!}{9!}+1 \right)$
$=\dfrac{1}{{{2}^{10}}}\left( 1+10+10+1 \right)$
$=\dfrac{22}{{{2}^{10}}}$
$=0.02148$
Therefore, the probability that at least 9 rats will turn the same way is 0.02148.

Note: Here we can make mistakes while calculating the probability of at least 9 rats will turn the same way. We have to take all the possibilities that can happen. That is 9 right turns or 10 right turns or 9 left turns or 10 left turns. If we miss any one of the possibilities the probability will be wrong.