There is a bag filled with 3 blue and 4 red marbles. A marble is taken at random from the bag, the colour is noted and then it is not replaced. Another marble is taken at random. What is the probability of getting 2 reds?

2/7

Altogether there're 7 marbles. Let,probability of getting blue and red marbles be 'B' and 'R' respectively.

In 1st pick

Probability of getting red marble is 4/7

So, there's the probability that the ball in 1st pick may be either blue or red.

So,if the marble is red then we've to eliminate that red marble ( cause,question says so)

Then,in 2nd pick

Probability of getting red ball will be 3/6(if we eliminate that red ball then the total number of marbles will also be 6)

Now,

Probability of getting two red will be

(4/7)×(3/6)

=2/7

It would be more easier to do by tree diagram. Since,I didn't find it convenient here,it may be not so much helpful.But, I tried my best. Hope you are clarified.

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64