Dtyjlui

We have 3 cups with balls in them.

• First cup: 3black, 4white


• Second Cup: 2black, 6white


• Third cup: 5black, 1white

We throw a dice. If we throw:

• 1, we take one ball from first cup


• 2 or 3, we take one ball from second cup


• Otherwise we take one ball from third cup


• List item

What is probability of drawing a black ball?

After we draw a black ball, what is probability that we draw it from second cup?

For second question should just be 0.33 as it only depends on the dice, right? As for first question goes, I am totally lost. I tried as $3/7 + 2/8 + 5/6$ but that gives me more then 1 so it can't be right.

This is a problem testing your knowledge of conditional probability

For the first:

Let $B$ be the event that we took a black ball.

Let $X,Y,Z$ be the event that we took a ball from the first, second, or third cup respectively. Note that $X,Y,Z$ create a partition of the sample space.

$Pr(B) = Pr(Bcap (Xcup Ycup Z)) = Pr(Bcap X)+Pr(Bcap Y)+Pr(Bcap Z)$

$=Pr(X)Pr(Bmid X) + Pr(Y)Pr(Bmid Y)+Pr(Z)Pr(Bmid Z)$

$=frac{1}{6}cdot frac{3}{7}+frac{2}{6}cdot frac{2}{8}+frac{3}{6}cdot frac{5}{6}$

For the second, apply Bayes' Theorem.

We wish to calculate $Pr(Ymid B)$. Using Bayes' Theorem we continue as:

$Pr(Ymid B) = dfrac{Pr(Bmid Y)Pr(Y)}{Pr(B)}$, each term of which we should know from the earlier problem part or from quick observations from the problem statement.

For the first question you proceed as follows:
$$P(Choosing Black Ball)$$
$$=P(cup 1)P(Choosing Black Ball|cup 1)+P(cup 2)P(Choosing Black Ball|cup 2)+P(cup 3)P(Choosing Black Ball|cup 3)$$
$$=(1/6)(3/7)+(2/6)(2/8)+(3/6)(5/6)=4/7$$

For the second part you need $P(cup 2|Black Ball Chosen)=frac{P(Black Ball Chosen|cup 2)P(cup 2)}{P(Black Ball Chosen)}=7/48$

We have 3 cups with balls in them.
First cup: 3black, 4white
Second Cup: 2black, 6white
Third cup: 5black, 1white
We throw a die. If we throw:
1, we take one ball from first cup

The probability of a 1 is 1/6. Give that the probability of a black ball is 3/7. The probability of rolling a 1 and drawing a black ball is (3/7)(1/6)= 1/14.

2 or 3, we take one ball from second cup

The probability of a 2 or 3 is 2/6= 1/3. Given that the probability of a black ball is 2/8= 1/4. The probability of rolling a 2 or a 3 and drawing a black ball is (1/4)(1/3)= 1/12.

Otherwise we take one ball from third cup

The probability we get anything other than a 1, 2,, or 3 (i.e. 4, 5, or 6) is 3/6= 1/2. Given that the probability of a black ball is 5/6. The probability of rolling anything other than a 1, 2, or 3 [b]and[/b] drawing a black ball is (1/2)(5/6)= 5/12.

Over all, the probability of drawing a black ball is 1/14+ 1/12+ 5/12= 6/84+ 7/84+ 35/84= 43/84.

Let us denote by $O_{k}$ the output of the die. Moreover, let us designate by $B$ the event "a black ball has been drawn". Therefore we are interested in the event
begin{align*}
textbf{P}(B) & = textbf{P}(Bcap(O_{1}cup O_{2}cup O_{3}cup O_{4}cup O_{5}cup O_{6}))\
& = sum_{k=1}^{6}textbf{P}(Bcap O_{k}) = sum_{k=1}^{6}textbf{P}(B|O_{k})textbf{P}(O_{k})
end{align*}

Therefore the sought result is expressed by
begin{align*}
textbf{P}(B) = frac{1}{6}left[frac{C(3,1)}{C(7,1)} + 2timesfrac{C(2,1)}{C(8,1)} + 3timesfrac{C(5,1)}{C(6,1)}right] = frac{1}{6}left(frac{3}{7} + frac{1}{2} + frac{5}{2}right) = frac{4}{7}
end{align*}

Once we have the probability $textbf{P}(B)$, we are able to answer the second question. Precisely speaking, we are interested in the following probability

begin{align*}
textbf{P}(O_{2}cup O_{3}|B) & = frac{textbf{P}(O_{2}cap B) + textbf{P}(O_{3}cap B)}{textbf{P}(B)} = frac{textbf{P}(B|O_{2})textbf{P}(O_{2}) + textbf{P}(B|O_{3})textbf{P}(O_{3})}{textbf{P}(B)}\
& = frac{displaystylefrac{1}{4}timesfrac{1}{6} + frac{1}{4}timesfrac{1}{6}}{displaystylefrac{4}{7}} = frac{1}{12}timesfrac{7}{4} = frac{7}{48}
end{align*}

Hope this helps.