Hi!

Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.

Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?

Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!

Thanks alot!!

There are 6 balls (2 of them are yellow)

so P("picking a yellow ball with the first try") = 2/6

P("not picking a yellow ball with the first try") = 4/6

A) We need to calculate

P("picking a yellow ball with the first try and picking a non-yellow-ball with the second try")

This is equal to 2/6 * 4/5 (this is because of "without replacement" after picking a ball, there are 5 remaining)

Anyway there are more events:

Most obvious one: picking a non-yellow ball and then a yellow ball

P((non-yellow, yellow)) = 4/6*2/5

And another possible event:

P("Picking a yellow ball with the first try and picking a yellow ball with the second try")

After picking a yellow ball (probability equals 2/6 ) there are 5 balls remaining, one of them is yellow

picking another yellow ball -> 1/5

So

P("Picking a yellow ball with the first try and picking a yellow ball with the second try") = P(yellow, yellow ) = 2/6 *1/5

And therefor the solution is

P(yellow, non-yellow) + P(non-yellow, yellow) + P(yellow, yellow)

Any questions?

Rapha

Edit: This is a more complicated way. It is better/easier to solve it like e^(i*pi)'s.

He calculated

1- P(non-yellow, non-yellow) = P(at least one yellow ball)

The solution should be the same.