two dices are trown. determinde the following conditional probabilities.
a) the sum is 4 given that the first die is 3
b) the first die is 3 given that the sum is 4
c) the events "the sum is 4" and "the first die is 3" are independent?
d)the sum is 4, given taht the first die is even
e) the first die is even, given taht the sum is 4

How can I approach this? I have a class test tomorrow... thanks so much.

2. Originally Posted by 0123
two dices are trown. determinde the following conditional probabilities.
a) the sum is 4 given that the first die is 3
P(A+B=4|A=3) = P(B=1|A=3) = 1/6

b) the first die is 3 given that the sum is 4
Given the sum is 4 the (equally likley) possibiliies are (A=1,B=3) (A=2,B=2)
(A=3,B=1) so P(A=3|A+B=4) = 1/3

c) the events "the sum is 4" and "the first die is 3" are independent?
d)the sum is 4, given taht the first die is even
e) the first die is even, given taht the sum is 4

How can I approach this? I have a class test tomorrow... thanks so much.

3. Hello, 0123!

For two-dice problems, list (or visualize) the 36 outcomes.

. . (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
. . (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
. . (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
. . (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
. . (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
. . (6,1) (6,2) (6,3) (6,5) (6,5) (6,6)

Two dice are thrown.
Determine the following conditional probabilities.

a) the sum is 4, given that the first die is 3
Given: the first die is a 3.
There are six outcomes in the sample space: .(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
. . Among them, exactly one has a sum of 4: .(3,1)
Therefore: .P(sum is 4 | first is 3) .= .1/6

b) the first die is 3, given that the sum is 4
Given: the sum is 4.
There are three outcomes in the sample space: (1,3), (2,2), (3,1)
. . Among them, exactly one has the first die with 3: .(3,1)
Therefore: .P(first is 3 | sum is 4) .= .1/3

c) the events "the sum is 4" and "the first die is 3" are independent?

Events A and B are independent if: .P(A ∩ B) .= .P(A) × P(B)

There are 36 possible outcomes.

"Sum is 4 and first die is 3". .There is one outcome: (3,1)
. . Hence: .P(sum is 4 ∩ first die is 3) .= .1/36

"Sum is 4". .There are three outcomes: (1,3), (2,2), (3,1)
. . Hence: .P(sum is 4) .= .3/36 .= .1/12

"First die is 3". .There are six outcomes: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
. . Hence: .P(first is 3) .= .6/36 .= .1/6

Since 1/36 is not equal to (1/12) × (1/6), the events are not independent.
. . They are dependent events.

d) the sum is 4, given that the first die is even.
Given: the first die is even.
There are eighteen outcomes in the sample space:
. . (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
. . (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
. . (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Among them, exactly one has a sum of 4: .(2,2)
Therefore: .P(sum is 4 | first is even) .= .1/18

e) the first die is even, given that the sum is 4
Given: the sum is 4. .There are three outcomes: .(1,3), (2,2), (3,1)
. . Among them, exactly one has the first die even: .(2,2)
Therefore: .P(first is even | sum is 4) .= .1/3

4. Soroban you're great. Now I have really understood. Thanks.