# Thread: Tree diagram, probability.

1. ## Tree diagram, probability.

At the start of a gameshow there are 10 contestants of which 6 are female. In each round of
the game, one contestant is eliminated. All of the contestants have the same chance of
progressing to the next round each time.
(a) Show that the probability that the first two contestants to be eliminated are
both male is 2/15

(b) Find the probability that more females than males are eliminated in the first three
rounds of the game.

(c) Given that the first contestant to be eliminated is male, find the probability that the next
two contestants to be eliminated are both female.
Hi,

Can someone please take a look at my tree diagram and point out whats 'wrong' with it, becasue I dont seem to get the right answer.

btw- 'E'- stands for elimnated
NE- stands for not elminated.

($\displaystyle \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$ )+
($\displaystyle \frac{6}{10} \times \frac{4}{9} \times \frac{5}{8}$ )+
($\displaystyle \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$) +
($\displaystyle \frac{4}{10} \times \frac{6}{9} \times \frac{5}{8}$)

2. Hello, Tweety!

Your tree diagram should indicate the gender of the person eliminated.

At the start of a gameshow there are 10 contestants of which 6 are female.
In each round of the game, one contestant is eliminated.
All contestants have the same chance of progressing to the next round each time.

(a) Show that the probability that the first two contestants to be eliminated
are both male is 2/15
Probability that the first eliminated is Male: .$\displaystyle \frac{4}{10}$
Probability that the second eliminated is Male: .$\displaystyle \frac{3}{9}$

Hence: .$\displaystyle P(\text{1st two Male}) \:=\:\frac{4}{10}\cdot\frac{3}{9} \:=\:\frac{2}{15}$

(b) Find the probability that more females than males
are eliminated in the first three rounds of the game.
I won't even try to draw a tree diagram . . .

. . $\displaystyle \begin{array}{cccccc}P(MFF) &=& \frac{4}{10}\cdot\frac{6}{9}\cdot\frac{5}{8} &=& \frac{120}{720} \\ \\[-3mm] P(FMF) &=& \frac{6}{10}\cdot\frac{4}{9}\cdot\frac{5}{8} &=& \frac{120}{720} \\ \\[-3mm] P(FFM) &=& \frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{9} &=& \frac{120}{720} \\ \\[-3mm] P(FFF) &=& \frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8} &=& \frac{120}{720} \end{array}$

Therefore: .$\displaystyle P(F > M) \:=\:4\cdot\frac{120}{720} \:=\:\frac{2}{3}$

(c) Given that the first contestant to be eliminated is male,
find the probability that the next two eliminated are both female.

Since a Male has been eliminated, there remain: 3 Males and 6 Females.

. . $\displaystyle \begin{array}{ccc}P(\text{2nd is F}) &=& \frac{6}{9} \\ \\[-3mm]P(\text{3rd is F}) &=& \frac{5}{8}\end{array}$

Therefore: .$\displaystyle P(\text{2nd and 3rd are F}) \:=\:\frac{6}{9}\cdot\frac{5}{8} \:=\:\frac{5}{12}$