I've just solved an exercise related to discrete random variables and maybe to the binomial distribution as well. I would like to know if my solution is correct, so here goes the problem statement followed by my answer:


In a justice court, $12$ juries have to dictaminate the sentence of an accused that was taken to court. This person will be convicted if at least $8$ of the $12$ juries vote to find him/her guilty. Suppose the accused has probability $\alpha$ of having commited the crime and that each jury decides his/her vote independently, with probability $\theta$ of making the right decision. Which is the probability of the accused being convicted?

My solution

Let's define the binomial random variable $X=\{\text{number of successes}\}$ where a success is defined to be a "positive vote" (the accused is found guilty), and define $Y=\{0,1\}$ where $0$ is means "the accused is indeed guilty of the crime" and $1$ is the opposite.

We want to calculate $P(X\geq 8)$. Applying total probability, we have $$P(X\geq 8)=P(X \geq 8|Y=0)P(Y=0)+P(X \geq 8|Y=1)P(Y=1)$$$$=\sum_{i=8}^{12} P(X=i|Y=0)P(Y=0)+P(X=i|Y=1)P(Y=1)$$$$=\sum_{i=8}^{12} \left[(\binom{12}{i}{\theta}^i(1-\theta)^{12-i})\alpha+(\binom{12}{i}(1-\theta)^i{\theta}^{12-i})(1-\alpha) \right].$$

I would appreciate if someone could check my answer, thanks in advance.

  • 2
    $\begingroup$ $\color{green}{\checkmark}$ Yes. Your working seems correct. Though technically, if those summation signs are mean to span both terms you should bracket them. $\endgroup$ – Graham Kemp Sep 3 '14 at 9:50

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