Each of the members of a $7$-judge panel independently makes a correct decision with probability .$7$. If the panel’s decision is made by majority rule, what is the probability that the panel makes the correct decision? Given that $4$ of the judges agreed, what is the probability that the panel made the correct decision?

Given in the question that, the 7 judge panel independently makes a correct decision with probability.

We have to find the probability that the panel makes the correct decision, if the panel's decision is made by majority rule.

We need to find the probability that the panel makes the correct decision, if 4 of the judges agreed.

Define random variable $x$that count how many of the judges had made the right decision .

We are given that $X~\mathrm{Binom}(7,0.7)$

The panel will make the correct decision if and only if $x\ge 4$.

Hence, the probability that the panel makes the correct decision is

$P(X\ge 4)=\sum _{k=4}^7 \left(\begin{array}{l}7\\ k\end{array}\right){0.7}^k{0.3}^{7-k}$

$=0.873964$

Suppose that exactly four judges have agreed, there exist two options.

They could have agreed to make the right decision and they could have agreed to make the wrong decision.

Therefore, we could write the two events as $X=4\text{and}X=3$.

So the required conditional probability is :

$P(X\ge 4\mid X=4\cup X=3)=\frac{P\left(\right(X\ge 4)\cap (X=4\cup X=3\left)\right)}{P(X=4\cup X=3)}$

$=\frac{P(X=4)}{P(X-4)+P(X=3)}=\frac{\left(\begin{array}{l}7\\ 4\end{array}\right){0.7}^4{0.3}^3}{\left(\begin{array}{l}7\\ 4\end{array}\right){0.7}^4{0.3}^3+\left(\begin{array}{l}7\\ 3\end{array}\right){0.7}^3{0.3}^4}$

$=0.7$

The probability that the panel makes the correct decision is $0.873964$.

If 4 of the judges agreed, then the probability that the panel made the correct decision is$0.7$