Box 1 (RRGG) Box 2 (RRRGG)
A ball is drawn at random out of box 1 and placed in box 2 without its color being observed.
A. If a ball is then drawn out of box 2, what is the probability that it is green?
I have equal probabilities of the ball drawn from box 1 is either R (1/2) or G (1/2)
if R is drawn from box 1 (Box 2 will now have RRRRGG)
p(G) 1/2* 1/3= 1/6
if G is drawn from box 1 (Box 2 will now have RRRGGG)
p(G) 1/2 * 1/2 = 1/4
A/ p(G) = 1/6 + 1/4 = 5/12
B. First, a box is chosen at random, and then a ball is drawn at random out of the chosen box. What is the probability that this ball is red?
p( box 1)= 1/2 * 2/4 = 2/8 = 1/4
p( box 2)= 1/4 * 3/5 = 3/10
p( red)= 1/4 + 3/10 = 11/20
Is this the correct way to calculate it? Thanks.
A ball is drawn at random out of box 1 and placed in box 2 without its color being observed.
A. If a ball is then drawn out of box 2, what is the probability that it is green?
I have equal probabilities of the ball drawn from box 1 is either R (1/2) or G (1/2)
if R is drawn from box 1 (Box 2 will now have RRRRGG)
p(G) 1/2* 1/3= 1/6
if G is drawn from box 1 (Box 2 will now have RRRGGG)
p(G) 1/2 * 1/2 = 1/4
A/ p(G) = 1/6 + 1/4 = 5/12
B. First, a box is chosen at random, and then a ball is drawn at random out of the chosen box. What is the probability that this ball is red?
p( box 1)= 1/2 * 2/4 = 2/8 = 1/4
p( box 2)= 1/4 * 3/5 = 3/10
p( red)= 1/4 + 3/10 = 11/20
Is this the correct way to calculate it? Thanks.