+4 There are 9 b***s in a sac. There are 3 reds b***s, 2 white b***s and 4 black b***s. If three b***s are drawn from the sac without replacement.
i. find the probability that the 1st ball is red and the last ball is black
ii. find the probabilty that the 1st ball is black and the last ball is either white or black
+118723 There are 9 b***s in a sac. There are 3 reds b***s, 2 white b***s and 4 black b***s. If three b***s are drawn from the sac without replacement.
i. find the probability that the 1st ball is red and the last ball is black
I wasn't really sure but I did it 2 ways and got the same answer so I feel confident reasonably.
I knew the second way was fine but this is the easier way.
It would be the same as if the first ball was red and the second one was black and the last anything.
(3/9)*(4/8)*(7/7)=1/6
ii. find the probabilty that the 1st ball is black and the last ball is either white or black
(4/9)*(6/8)*(7/7) = 1/3
+118723 There are 9 b***s in a sac. There are 3 reds b***s, 2 white b***s and 4 black b***s. If three b***s are drawn from the sac without replacement.
i. find the probability that the 1st ball is red and the last ball is black
I wasn't really sure but I did it 2 ways and got the same answer so I feel confident reasonably.
I knew the second way was fine but this is the easier way.
It would be the same as if the first ball was red and the second one was black and the last anything.
(3/9)*(4/8)*(7/7)=1/6
ii. find the probabilty that the 1st ball is black and the last ball is either white or black
(4/9)*(6/8)*(7/7) = 1/3
