Conditional probabiltiy

• Mar 5th 2006, 11:08 AM
d.darbyshire
Conditional probabiltiy
Suppose that a bag contains 6 red marbles and 5 yellow marbles. A single marble is picked at random out of the bag and is not put back in the bag; then a second marble is drawn at random out of the bag. Find
a. the probability that both marbles are yellow
b. the probabilty that the second marble is red
c. the probability that the second marble is red given that the first is red
d. the probabilty that the second marble is red given that the first is yellow.

I have to also attach a tree diagram with this problem and don't understand the multipliang and adding acrooss the diagram.
• Mar 5th 2006, 02:19 PM
ThePerfectHacker
Quote:

Originally Posted by d.darbyshire
Suppose that a bag contains 6 red marbles and 5 yellow marbles. A single marble is picked at random out of the bag and is not put back in the bag; then a second marble is drawn at random out of the bag. Find
a. the probability that both marbles are yellow
b. the probabilty that the second marble is red
c. the probability that the second marble is red given that the first is red
d. the probabilty that the second marble is red given that the first is yellow.

I have to also attach a tree diagram with this problem and don't understand the multipliang and adding acrooss the diagram.
1]There are $_5C_2=10$ ways to chose to yellow marbles. There are $_{11}C_2=55$ ways to chose two marbles. Thus, the ratio is the probability which is $\frac{10}{55}=\frac{2}{11}$.
2]Thus, either RR or YR. Probability of chosing a red first is $\frac{6}{11}$ and then another red is $\frac{5}{10}=\frac{1}{2}$ thus, the probability of RR is $\frac{6}{11}\cdot\frac{1}{2}=\frac{3}{11}$. The probability of chosing a yellow first is $\frac{5}{11}$ and then choosing a red is $\frac{6}{10}=\frac{3}{5}$ thus the probability of YR is $\frac{5}{11}\cdot\frac{3}{5}=\frac{3}{11}$. Thus, the probability of RR or YR is
$\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$.