A bag soccer balls contains 8 balls with black design and 7 balls with red design with no other types of soccer balls. Coach graves reaches into the bag and randomly pulls out two soccer balls.
please check my answer
a) explain these events are either dependent or independent.
i think this is independent because the coach just randomly chooses the ball.
b) what is the propability both of the balls will be the ame design color?
is it 98/15.....
see
http://www.jiskha.com/display.cgi?id=1232417324
The result of the second reach into the bag depends on the result of the first reach.
For example if the coach pulled a black one first, then for the second there would be 7 black and 7 red.
probability first one black = 8/15
then probability second one black = 7/14
so 8/15 * 7/14 for both black
probability first one not black = 7/15
probability second one not black = 6/14
so 7/15 * 6/14 for both white
add those two.
8/15 * 7/14 + 7/15 * 6/14
= (1/15)(56/14 + 42/14) = (1/15)(98/14)
= 98/210 = 49/105
Oh, 49/105 = 7/15
a) You are correct, the events of pulling out two soccer balls from the bag are independent. This is because the act of pulling out one soccer ball does not affect the probability of pulling out a specific type of the second soccer ball. In this case, the coach randomly chooses the balls without replacement, meaning that once a ball is removed, it is not put back into the bag.
b) Let's calculate the probability of both balls being the same color.
The first ball has a 15/30 chance (since there are 15 balls in total) of being any color. Once the first ball is removed from the bag, there are now 14 balls remaining, with 7 balls of the same color as the first ball. Therefore, the probability of drawing a second ball of the same color is 7/14.
To find the probability of both balls being the same color, we multiply the probabilities of each event occurring:
(15/30) * (7/14) = (1/2) * (1/2) = 1/4
So, the probability of both balls being the same color is 1/4, not 98/15 as you mentioned.