Problem Solving

manpsingh87 wrote:

Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.

hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!

manpsingh87 wrote:

Anurag@Gurome wrote:

manpsingh87 wrote:

Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.

hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!

Here, for this problem, your logic is perfectly correct.
But, generally it is advisable to use the formula to avoid errors.

Suppose the question is: what is the probability, that the sum on two faces is greater than 9, given that first dice shows 5?
How will you solve this?

sum greater than 9, i.e. sum can be either 10,11,12, now these sums can be achieved in {(4,6)(6,4)(5,5)(5,6)(6,5)(6,6)} 6 ways therefore the desired probability as per the traditional method suggested by anurag sir above would be

(2/36)/(6/36) = 1/3;

and sir as per my logic it would be 2/6 as the two favorable cases would be (5,6)(5,5) =1/3;

sir still getting the same answer by both the methods...!!!!