Billl has some counters in a bag
3 of the counters are red.
7 of the counters are blue.
The rest are yellow.
Bill takes at random a counter from the bag.
The probability that he takes a tellow counter is 2/7
How many yellow counters are in the bag before bill takes a counter?
Please explain all answers.
here are the numbers: 3 red, 7 blue, X yellow
X yellow = 2 / 7
3 red + 7 blue must equl the other 5 /7
3 + 7 = 10, 5 * 2 = 10
3 red has 3 / 20 chance
7 blue has 7 /20 chance
X yellow has 10 / 20 chance
there are 10 yellow