Question on Conditional Probability and independent events

 Question on Conditional Probability and independent events

 

 

 

question upon conditional probability and independent events

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A fair coin is tossed twice , Find the probability of getting a 4, 5

           or 6 on the first toss and 1,2 ,3 or 4 on the second toss

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Answer

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For the 'First Toss , we have 6 ways of getting a different numbers

on the dice.

 

For the 'Second Toss , we have 6 ways of getting different numbers

on the dice .

 

So , considering the two tosses , there are overall 6*6=36  equally

likely possibilities of obtaining outcomes on the dice.

 

Let , A be the event of getting {4,5,6} on the throw of the dice .

Let , B be the event of getting a {1,2,3,4} on the throw of the dice .

 

Each of the 3 ways in which a '1' can occur over the dice can also

be associated with each of the 4 ways in which an event a2 can occur

to give 3*4 ways in which both A1 and A2 both can occur .

 

P(A1 n A2) = 12/36 = 1/3

 

A1 and A2 being independent events and none of the

event being dependent over the other , we can get

P(A1 n A2)

= P(A1) * P(A2)

= 3/6 * 4/6

=12/36

=1/3