1. ## Probability Tricky Qsts

A bag contains 5 red cubes and 3 black cubes. Three cubes are chosen at random.
Find the probability of at least 2 reds being chosen, given that the first cube was red:
a
if the cubes are replaced after each draw

b if the cubes are not replaced after each draw.

Thanks.

2. Hello ilovemaths
Originally Posted by ilovemaths
A bag contains 5 red cubes and 3 black cubes. Three cubes are chosen at random.
Find the probability of at least 2 reds being chosen, given that the first cube was red:
a
if the cubes are replaced after each draw

b if the cubes are not replaced after each draw.

Thanks.
(a) Given that the first cube was red, we have to find the probability of drawing at least one more red cube, when 2 cubes are drawn from the bag that contains 5 red and 3 black, where the cube is replaced after the first draw.

This is easiest if we find the probability that neither cube is red, and subtract from 1.

So, the probability that the first cube is black is $\frac38$; and, since the cube is replaced, the probability that the second cube is black is also $\frac38$. So the probability that both cubes are black is $\frac38\times\frac38=\frac{9}{64}$.

Therefore the probability that at least one of the cubes is red $= 1-\frac{9}{64}=\frac{55}{64}$.

(b) In this case, we have to find the probability of drawing at least one more red cube, when 2 cubes are drawn from the bag that contains 4 red and 3 black, where the cube is not replaced after the first draw.

I'll start it for you, and leave you to complete it. Again, it's easiest if we find the probability that neither cube is red, and subtract from 1.

So, the probability that the first cube is black is $\frac37$. If the first cube is black, there are then 2 black and 4 red left in the bag. So the probability that the second cube is also black is $\frac26$. So the probability that both cubes are black is ...?

Therefore the probability that at least one of the cubes is red = 1 - ... Can you complete it?