A bag contains 4 red 6 blue and a second bag contains 4 blue and 6 green balls. A ball is taken out from each bag. find the probablity that one ball is red and other ball is green the probability that an event A happens in one trail of an experiment is 0.4 three independent trails of the experiment are performed. find the probablity that the event A happens at least one
Answers
Answer:
There are 2 possible cases: first ball drawn is red or is black.
∴ P(second ball red | first ball drawn is black) =
4+8
4
=
3
1
∴ P(second ball red | first ball drawn is red) =
6+6
4+2
=
2
1
Hence, the probability of the second ball being red
= P(first ball black)×P(second ball red | first ball black)+P(first ball red)×P(second ball red | first ball red)
=
10
6
×
3
1
+
10
4
×
2
1
=
5
2
This is the required probability.
Answer:
There are total (4 r + 6 b) = 10 balls in the bag . The required event happens if one draw first red then a blue ball that is (r, b) or first blue then a red ball i.e. (b, r) without replacing the first, before the second draw, in both the cases. Hence the required probability
= P{(r, b)} + P {(b, r)} = {(4/10) × (6/9)} + {(6/10) ×(4/9)} = 48/90 = 8/15 .