# Conditional probability exercise - apples and oranges

We have two crates, crate 1 and crate 2. Crate 1 has 2 oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2.

The first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from crate 1 was an orange(which is $\frac{2}{6}$) and then saying that I have 3 fruits in crate 2, $1+\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\%$ probability that the fruit we take from crate 2 was an orange. The probability I got seems reasonable, but I don't know for sure if what I did was correct.

Anyway, point 2 of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A: Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\:P\left(A|B\right)$ I think, which means "Probability that A happens if we know B happened", but I'm not so sure about this.

Could anyone give me a hint on how to go about solving this problem?

• Your answer to the first question is correct. For the second question, $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ where $A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. – N. F. Taussig Jan 29 '17 at 12:18
• But what is $P\left(A\cap B\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? – MikhaelM Jan 29 '17 at 12:32
• It is the probability that you put an orange from the first crate into the second, and having done that then took an orange from the second . $$\mathsf P(A\cap B)= \mathsf P(A)~\mathsf P(B\mid A)$$ – Graham Kemp Jan 29 '17 at 12:34
• Doing that I get $\:P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(A\right)\cdot P\left(B|A\right)}{P\left(B\right)}=\frac{\frac{2}{6}\cdot \frac{2}{3}}{\frac{4}{9}}=\frac{1}{2}$ so $50\%$. Is this correct or am I making a mistake somewhere? – MikhaelM Jan 29 '17 at 12:45
• The answer you obtained in your comment is correct. – N. F. Taussig Jan 29 '17 at 13:15

Let $A,B$ be the events of removing an orange from the first and second crates, respectively.

You have calculated $\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \begin{align}\mathsf P(B) ~&=~\mathsf P(A)~\mathsf P(B\mid A)+\mathsf P(A^\complement)~\mathsf P(B\mid A^\complement) \\ &=~ \tfrac 26\cdot\tfrac 23+\tfrac 46\cdot\tfrac 13 & =&~ \frac{\tfrac 2 6+1}3 \\ &=~ \tfrac 49 \end{align}

Where $\mathsf P(A)$ is the probability of taking an orange from cart 1, $\mathsf P(A^\complement)$ is that of taking an apple from cart 1, $\mathsf P(B\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange to that cart, and $\mathsf P(B\mid A^\complement)$ is the probability of taking a orange from cart 2 when given that you have added an apple to that cart.

Now you just need to calculate $\mathsf P(A\mid B)$ the probability of having taken an orange from cart 1 when given that you took an orange from cart 2.

Use Bayes' Rule.

Yes for second part solve P(A|B).

Probability of A happens if we know B happened.