tree diagram

Oct 2008
897
3
Bristol, England
hi this is my question:

A bag contains 20 coins.
There are 6 gold coins and the rest are silver.

A coin is taken at random from the bag.
The type of coin is recorded and the coin is then returned to the bag.
A second coin is then taken at random from the bag.

A) The tree diagram shows all the ways in which two coins can be taken from the bag.
Write the probabilities on a copy of the diagram.

(G for gold and S for silver)

zzzz.jpg

i have done that, but im stuck on this next part:

Use you tree diagram to calculate the probability that one coin is gold and one coin is silver.

i do, 6/20 X 14/20, but i don't know how to get the answer? would i do 6X14 to get 84 and 20X20 to get 400 so i have 84/400= 42/200= 21/100?

and the same on the next one but, 14/20 x 6/20 and the rest the same?

if that is right so far would i do 21/100 x 21/100= 441/10000 to get the answer???

if that is wrong could someone please tell me where im going wrong and how to correct it, thanks
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
hi this is my question:

A bag contains 20 coins.
There are 6 gold coins and the rest are silver.

A coin is taken at random from the bag.
The type of coin is recorded and the coin is then returned to the bag.
A second coin is then taken at random from the bag.

A) The tree diagram shows all the ways in which two coins can be taken from the bag.
Write the probabilities on a copy of the diagram.

(G for gold and S for silver)

View attachment 17037

i have done that, but im stuck on this next part:

Use you tree diagram to calculate the probability that one coin is gold and one coin is silver.

i do, 6/20 X 14/20, but i don't know how to get the answer? would i do 6X14 to get 84 and 20X20 to get 400 so i have 84/400= 42/200= 21/100?

and the same on the next one but, 14/20 x 6/20 and the rest the same?

if that is right so far would i do 21/100 x 21/100= 441/10000 to get the answer???

if that is wrong could someone please tell me where im going wrong and how to correct it, thanks
\(\displaystyle Pr(G \cap S) = Pr(G) \times Pr(S)\)

\(\displaystyle = \frac{6}{20} \times \frac{14}{20}\)

\(\displaystyle = \frac{21}{100}\).


\(\displaystyle Pr(S \cap G) = Pr(S) \times Pr(G)\)

\(\displaystyle = \frac{14}{20} \times \frac{6}{20}\)

\(\displaystyle = \frac{21}{100}\).


So that means that the probability of either situation is

\(\displaystyle \frac{21}{100} + \frac{21}{100} = \frac{21}{50}\).
 
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