A pair of dice is tossed. Event A means that the sum is divisible by 2 and event B means that the sum is greater than 6. Find P(A|B).

Answer 1

P(A/B)=3/7. Intuition might suggest 1/2, but P(B/A)=1/2 not P(A/B)

There are 11 possibilities for 2 dice: 2,3,4,5,6,7,8,9,10,11 or 12

There is only 1 way get 2, one on each die

2 ways to get 3, two on first die, one on the second

3 ways to get 4, one on 1st die, three on the 2nd, three on the first, one on the 2nd, 2 on each

4 ways to get 5, 1&4, 4&1, 2&3, 3&2,

5 ways to get 6, 1&5, 5&1, 2&4, 4&2, 3&3

6 ways to get 7, 1&6, 6&1, 2&5, 5&2, 4&3, 3&4

5 ways to get 8 2&6, 6&2, 3&5, 5&3, 4&4

4 ways to get 9 3&6, 6&3, 4&5, 5&4

3 ways to get 10 4&6, 6&4, 5&5

2 ways to get 11 6&5, 5&6

1 way to get 12 6&6

Altogether there are 36 possible outcomes, 1+2+3+4+5+6+5+4+3+2+1=36

There are 6+5+4+3+2+1=21 ways to get >6

21/36=7/12=P(B)

P(A)=probability of an even outcome is one-half

There are 36 possible outcomes. Of them, even outcomes are 2,4,6,8,10, and 12

They happen 1+3+5+5+3+1 times or 18 times. 18/36=1/2=P(A)

P(B/A)=the probability of getting greater than 6, given an even outcome. 2,4 and 6 are as equally likely as 8,10 and 12. If you have an even outcome, then P(8,10 or 12)=1/2=P(B/A)

P(A/B)= the probability of getting an even outcome if the outcome is greater than 6.

Use Bayes Theorem: P(A/B)=[P(B/A)P(A)]/P(B)=[(1/2)(1/2)]/(7/12)=(1/4)/(7/12)=(1/4)(12/7)=3/7

P(A/B)=3/7

Or, more directly, P(A/B)=probability of even, if >6

If the outcome is greater than 6, it is 7, 8, 9, 10, 11, or 12. You might be tempted to

say that half are even so chances are 1/2, but the 8, 10 and 12 do not come up as often as

7, 9, and 11. So P(A/B) is slightly less than 1/2 3/7<1/2 by 1/14

7 occurs most, 12 the least 7 happens 6 ways, 8 happens 5 ways, 9 happens 4 ways

10 happens 3 ways, 11 two ways, 12 only one way

Possible outcomes, given >6, are 6+5+4+3+2+1= 21 ways for greater than 6

Of those outcomes even outcomes happen 5+3+1 ways = 9

9/21 = 3/7 = P(A/B)