conditional probability that both answers are correct given that at least one is correct I came across this question whiles doing conditional probability and need a vivid explanation. A person answers each of two multiple choice questions at random. if there are four possible choices on each of the question, what is the conditional probability that both answers are correct given that at least one is correct
 A: Call the questions A and B. Given that at least one of these questions was answered correctly, we must be in one of the following seven situations:
$$\begin{array}{c|c}
\text{Question A}&\text{Question B}\\ \hline
\text{right answer}&\text{right answer}\\
\text{right answer}&\text{first wrong answer}\\
\text{right answer}&\text{second wrong answer}\\
\text{right answer}&\text{third wrong answer}\\
\text{first wrong answer}&\text{right answer}\\
\text{second wrong answer}&\text{right answer}\\
\text{third wrong answer}&\text{right answer}
\end{array}$$
These seven situations are equally likely, and only one of them has both answers correct, so the desired conditional probability is $\frac17$.
(The other nine possible outcomes involve combining one of the three wrong answers to Question A with one of the three wrong answers to Question B; this can be done in $3\cdot3=9$ different ways.)
A: An alternative to Brian's solution.
Let $C$ and $D$ denote the events that questions $A$ and $B$ respectively were
answered correctly. Then, the event that both are answered correctly is
$C\cap D$ while the event that at least one is answered correctly is
$C\cup D$.
We are asked to find the conditional probability that both questions were
answered correctly given that at least one was answered correctly, that is,
$P((C \cap D) \mid (C \cup D))$. Since the event $C \cap D$ is a subset
of the event $C \cup D$, we get from the definition of conditional
probabiility that
$$P((C \cap D) \mid (C \cup D))
= \frac{P((C \cap D) \cap (C \cup D))}{P(C \cup D)}
= \frac{P(C \cap D)}{P(C \cup D)}.$$
If we are given only the numerical values of $P(C \cap D)$ and 
$P(C \cup D)$, then we can simply substitute into the
above formula and complete the solution. But we are told that
the questions are answered at random by choosing one of the
four choices in each case with equal probability, that is
$C$ and $D$ are independent events
of probability $\frac{1}{4}$ each. Thus, we can readily
compute
$$\begin{align}
P(C\cap D) &= P(C)P(D)& &= \frac{1}{4}\times\frac{1}{4}& &= \frac{1}{16}\\
P(C\cup D) &= P(C) + P(D) - P(C\cap D)& &= \frac{1}{4}+\frac{1}{4}-\frac{1}{16}&
&= \frac{7}{16}\\
P((C \cap D) \mid (C \cup D))
&= \frac{P(C \cap D)}{P(C \cup D)}& &= \frac{1/16}{7/16}& &= \frac{1}{7}.
\end{align}$$
A: Brian has given a lovely (better!) answer. Here is how I would have done it.
Let $A$ represent the event that both are correct, and $B$ the event that at least one is correct. Then
$$
\begin{align}
P(A | B) &= P(A \cap B)/P(B) \\
 &= P(A)/P(B)
\end{align}
$$
Now $P(A)=1/16$ clearly. $P(B)$ is a little harder, but with a little counting (ala Brian's answer) you can see $P(B) = 7/16$. Thus $P(A|B) = 1/7$. 
A: Let $A$ be the event of getting  both answer are correct and $B$ be the event at least one answer are correct. Now the probability of both answers begin correct is $P(A)= \frac{1}{4} \cdot \frac{1}{4}$ (because given one multiple choice question with 4 choices, the probability for choosing one option is right $1/4$)
Now we need to find $P(A | B)= \frac{P(A \text{ and } B)}{P(B)}$. Now $A \text{ and }B$ is both answer are right, then $P(A \text{ and }B)= P(A)$. But then we get $P(A | B)= \frac{P(A)}{P(B)}$. To find $P(B)$, there are three possible cases:

*

*Question 1 answer is right and question 2 answer is wrong.

*Question 1 answer is wrong and question 2 answer is right.

*Question 1 and question 2 both answer are right.

So $P(B)= \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4}= \frac{7}{16}$.
From the fact that $P(A | B)= \frac{P(A)}{P(B)}$, $P(A | B)= \frac{1/16}{7/16}= \frac{1}{7}$, at least, I think.
