Louise Johnson
Junior Member
- Joined
- Jan 21, 2007
- Messages
- 103
Question:
Three boxes contain marbles with the contents of each box indicated below. Andrea randomely selects a box, and then randomly selects a marble from that box.
Box#1 has 3 red, 5 blue Box#2 has 2 red, 4 yellow Box#3 has 4 red ,2 blue
a) Find the probability that Andrea selected a red marble.
My answer
\(\displaystyle \L\\\begin{array}{l}
\frac{1}{3} \times \frac{3}{8} + \frac{1}{3} \times \frac{2}{6} + \frac{1}{3} \times \frac{4}{6} \\
= \frac{1}{8} + \frac{1}{9} + \frac{2}{9} \\
= \frac{{11}}{{24}} \\
\end{array}\)
b) Find the probability that Andrea selected a red marble, given that she did not select box #3.
my answer
\(\displaystyle \L\\\begin{array}{l}
\frac{1}{2} \times \frac{3}{8} + \frac{1}{2} \times \frac{2}{6} \\
= \frac{3}{{16}} + \frac{1}{6} \\
= \frac{{17}}{{48}} \\
\end{array}\)
I am feeling rather confident with these as I had a few good examples to work from. It would be great if someone could let me know how I did?
Thank you
Louise
Three boxes contain marbles with the contents of each box indicated below. Andrea randomely selects a box, and then randomly selects a marble from that box.
Box#1 has 3 red, 5 blue Box#2 has 2 red, 4 yellow Box#3 has 4 red ,2 blue
a) Find the probability that Andrea selected a red marble.
My answer
\(\displaystyle \L\\\begin{array}{l}
\frac{1}{3} \times \frac{3}{8} + \frac{1}{3} \times \frac{2}{6} + \frac{1}{3} \times \frac{4}{6} \\
= \frac{1}{8} + \frac{1}{9} + \frac{2}{9} \\
= \frac{{11}}{{24}} \\
\end{array}\)
b) Find the probability that Andrea selected a red marble, given that she did not select box #3.
my answer
\(\displaystyle \L\\\begin{array}{l}
\frac{1}{2} \times \frac{3}{8} + \frac{1}{2} \times \frac{2}{6} \\
= \frac{3}{{16}} + \frac{1}{6} \\
= \frac{{17}}{{48}} \\
\end{array}\)
I am feeling rather confident with these as I had a few good examples to work from. It would be great if someone could let me know how I did?
Thank you
Louise
