# Probability: tree diagram

• Dec 8th 2009, 09:32 PM
thcbender
Probability: tree diagram
A bag contains six white marbles, and four blue marbles. Three marbles are drawn in succession. At each draw, if the marble is white, it is replaced, and if its blue it is not replaced.
.
a) draw a tree diagram with this information with all probabiltiies shown (2)
b) find the probability of drawing:
1) no mblue marbles (1)
ii) two blue marbles (1)

another question:

a comittee of 4 people is to be chosen at random from among 10 people.
what is the probability that both James and Sarah will be chosen.
• Dec 9th 2009, 05:38 AM
Hello thcbender
Quote:

Originally Posted by thcbender
A bag contains six white marbles, and four blue marbles. Three marbles are drawn in succession. At each draw, if the marble is white, it is replaced, and if its blue it is not replaced.
.
a) draw a tree diagram with this information with all probabiltiies shown (2)
b) find the probability of drawing:
1) no mblue marbles (1)
ii) two blue marbles (1)

another question:

a comittee of 4 people is to be chosen at random from among 10 people.
what is the probability that both James and Sarah will be chosen.

I attach a partly completed tree diagram. I hope that you can complete it.

(b) (i) The probability of drawing no blue marbles is, as I have shown: $\displaystyle \frac{6}{10}\times\frac{6}{10}\times\frac{6}{10} =$ ... ?

(ii) There are three routes through the probability tree that correspond to two blue marbles: BWW, WBW and WWB. Work out the probabilities of each of them and add the resulting fractions together. I'll start you off with BBB:
$\displaystyle \frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}= \frac{1}{10}$
Can you complete it?

Question 2. (On an administrative note, you're asked not to post two different questions in the same thread. See Rule 14.)

There are $\displaystyle {^{10}C_{4}}$ ways of choosing a committee of $\displaystyle 4$ people from $\displaystyle 10$.

If James and Sarah have been chosen, we need to choose $\displaystyle 2$ from the remaining $\displaystyle 8$ to complete the committee. In how many ways can this be done?

Divide the second number by the first to get the probability that James and Sarah are on a committee chosen at random.

I make the answer $\displaystyle \frac{2}{15}$. Do you agree?