When two balanced dice are rolled, there are 36 possible outcomes. Find the probability that the first die is 3 or that doubles are rolled.

Step-by-step explanation:

The space comprising all possible outcomes (sample space) is represented by the 6×66×6 matrix formed by the possible pairs representing the outcome of dice:

From the above sample space, there are six events wherein the first die is 44 (all elements of the fourth row in the matrix). That is, n(F)=6n(F)=6. Moreover, there are six events wherein doubles are rolled (all diagonal elements in the matrix). That is, n(D)=6n(D)=6. The result (4,4)(4,4) is the intersection of both events. That is, n(D∩F)=1n(D∩F)=1.

Therefore, the possibility/probability that the first die is 4 or that doubles are rolled is computed as follows:

P(D∪F)=P(D)+P(F)−P(D∩F)=n(D)n(S)+n(F)n(S)−n(D∩F)n(S)=636+636−136=1136P(D∪F)=P(D)+P(F)−P(D∩F)=n(D)n(S)+n(F)n(S)−n(D∩F)n(S)=636+636−136=1136

there is no question content here, but you for the free points! : d