I am working on the following problem:
We have 2 bags that each contain 3 yellow, 4 blue, 5 red, 6 green and
2 black balls. In a simultaneous draw what is the chance of getting
at least 1 red or 1 green ball?
My approach:
Find probability of getting at least 1 green ball:
$(\frac{14}{20})^2 = (\frac{7}{10})^2= \frac{49}{100}$ is the probability of not getting a green ball in the draw hence $1 – \frac{49}{100} = \frac{51}{100}$ is the probability of getting at least 1 green ball.
Find probability of getting at least 1 red ball:
$(\frac{15}{20})^2 = (\frac{3}{4})^2 = \frac{9}{16}$ is the probability of not getting a red ball in the draw hence $1 – \frac{9}{16} = \frac{7}{16}$ is the probability of getting at least 1 red ball.
Therefore what we are looking for is the sum of these probabilities i.e.
$\frac{51}{100} + \frac{7}{16} = \frac{816 + 700}{1600} =\frac{1516}{1600} = \frac{379}{400}$
But my notes say $\frac{319}{400}$
Is this a typo or is my solution wrong?
Best Answer
Your solution is wrong because you forgot to subtract the joint probability of drawing a red ball and a green one. However, here's the best way to do this problem.
The probability of at least $1$ red or green ball is $1$ minus the probability of no red or green ball on either draw. Thus:
$1-(\frac{9}{20}\cdot\frac{9}{20}) = 1-\frac{81}{400}=\frac{319}{400}$
Does that make sense?