Question

A bag contains 6 red balls, 4 green balls, and 3 blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

Answer 1

Answer:

Probability of first ball Will be green is 4/13.

Probability of second will be red is 6/12 that is equal to1/2

May it help you

Answer 2

p(first ball)=4/13

p(second ball)=6/12=1/2

Step-by-step explanation:

p(first ball)=no.of green balls/total no.of balls

=4/13

one ball is removed then total no.of balls=12

p(second ball)=no.of red balls/total no.of balls

=6/12

=1/2