# Thread: One bag contains 4 white balls

1. ## One bag contains 4 white balls

Question : One bag contains 4 white balls and 2 black balls; another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that (i) both are white , (ii) one is white and one is black.
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Pr(X = white ball from bag 1) = \frac{4}{6}

Pr(X = white ball from bag 2) = \frac{3}{8}

2. Hello, zorro!

One bag contains 4 white balls and 2 black balls.
Another contains 3 white balls and 5 black balls.
If one ball is drawn from each bag, find the probability that
. . (a) both are white
. . (b) one is white and one is black.

We have: .$\displaystyle \boxed{\begin{array}{c}\text{Bag 1} \\ \hline \text{4 White} \\ \text{2 Black} \end{array}} \qquad \boxed{\begin{array}{c}\text{Bag 2} \\ \hline \text{3 White} \\ \text{5 Black}\end{array}}$

(a) Both White.

. . $\displaystyle \begin{array}{ccc}P(\text{W from Bag 1}) &=& \dfrac{4}{6} \\ \\[-3mm] P(\text{W from Bag 2}) &=& \dfrac{3}{8} \end{array}$

Therefore: .$\displaystyle P(\text{both W}) \:=\:\frac{4}{6}\cdot\frac{3}{8} \:=\:\frac{1}{4}$

(b) one W, one B

There are two scenariors:

. . [1] W from Bag 1, B from Bag 2: . $\displaystyle \frac{4}{6}\cdot\frac{5}{8} \:=\:\frac{20}{48}$

. . [2] B from Bag 1, W from Bag 2: . $\displaystyle \frac{2}{6}\cdot\frac{3}{8} \:=\:\frac{6}{48}$

Therefore: .$\displaystyle P(\text{one W, one B}) \:=\:\frac{20}{48} + \frac{6}{48} \:=\:\frac{26}{48} \:=\:\frac{13}{24}$

3. Originally Posted by zorro
Question : One bag contains 4 white balls and 2 black balls; another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that (i) both are white , (ii) one is white and one is black.
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Pr(X = white ball from bag 1) = \frac{4}{6}

Pr(X = white ball from bag 2) = \frac{3}{8}
Draw a tree diagram.