Question 382011: i really need help please help me
1. Consider an experiment of picking a ball twice in a box with 4 different balls numbering from 1 to 4. The picked ball in the first try should be returned to the box for the second try.
1) Specify all the possible events from the experiment.
2) Find the probability of happening the event A, “the sum of the numbers from the two balls is 3”.
3) Find the probability of happening the event B, “the number of the 2nd ball is 2”.
4) Find the probability of happening the event of A and B.
5) Find the probability of happening the event of A after the event B happens.
6) Can we say that event A and B are mutually exclusive? If yes, why? or if not ,why not?
7) Can we say that event A and B are independent? If yes, why? or if not ,why not?
Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. Consider an experiment of picking a ball twice in a box with 4 different balls numbering from 1 to 4. The picked ball in the first try should be returned to the box for the second try.
1) Specify all the possible events from the experiment.
There are 4*4=16 possible events ranging from 1/1, to 4/4
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2) Find the probability of happening the event A, “the sum of the numbers from the two balls is 3”.
Events: 1/2, 2,2, 2/1
P(sum is 3) = 3/16
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3) Find the probability of happening the event B, “the number of the 2nd ball is 2”.
Events: 1/2,2/2,3/2,4/2
P(2nd ball is 2) = 4/16
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4) Find the probability of happening the event of A and B.
Event: 1/2 is in both event A and event B
P(A and B) = 1/16
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5) Find the probability of happening the event of A after the event B happens.
Event: P(sum is 3|2nd ball is 2) = 1/4
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6) Can we say that event A and B are mutually exclusive? If yes, why? or if not ,why not?
No, there is an element (1/2) that appears in both A and B.
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7) Can we say that event A and B are independent? If yes, why? or if not ,why not?
P(A|B) = 1/4
P(A)*P(B) = (3/16)(4/16) is not equal to P(A|B)
So A and B are not independent.
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Cheers,
Stan H.
Answer by edjones(8007)  (Show Source):
You can put this solution on YOUR website! 1) There are 16 possible events:
11, 12, 13, 14
21, 22, 23, 24
31, 32, 33, 34
41, 42, 43, 44
.
2) 2/16
.
3) 4/16
.
4) 1/16
.
5) 4/16 * 2/16 = 1/32
.
6) No, 4) would have to be zero.
.
7) No, P(A)<>P(A|B)
.
Ed
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