There are 6 black counters and 4 white counters in bag A.
There are 7 black counters and 3 white counters in bag B.
There are 5 black counters and 5 white counters in bag C.

Bernie takes at random a counter from bag A and puts the counter in bag B. He then takes at random a counter from bag B and puts the counter in bag C.

Find the probability that there are now more black counters than white counters in bag C.

The probability of an event in an experiment is calculated as the number of desired outcomes in the context of the experiment divided by the number of total outcomes in the context of the experiment.

In this problem, the desired probability is the probability that a black counter is moved to bag C, and there are two possible cases:

Black from A to B(6/10 = 3/5 probability) and Black from B to C(8/11 probability).

White from A to B(4/10 = 2/5 probability) and black from B to C(7/11 probability).

Hence the probability is:

p = 3/5 x 8/11 + 2/5 x 7/11 = 24/55 + 14/55 = 38/55 = 0.6909 = 69.09%.

More can be learned about probabilities at https://brainly.com/question/14398287

#SPJ1