What is the probability of getting either a sum of 8 or at least one 4 in the roll of a pair of dice?

Question

Abigail Allen · Accepted Answer

The probability of either of the incidents happening is #5/12#.

Here we consider two events: A - (finding a sum of #8#) & B (getting at least one #4#) A : Probability of A is #5/36# because there are #7# such events out of #6xx6=36# events viz. #(2,6),(3,5),(4,4),(6,2),(5,3)# B Probability of B is #11/36# because there are #11# such events out of #36# events viz. #(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)#. Now it is observed that the two incidents are not mutually exclusive as the incident resulting in #(4,4)# is occurs in both. Now, the probability of either of the incidents happening is #P(AUB)=P(A)+P(B)-P(AnnB)# i.e. #=5/36 +11/36 -1/36=15/36=5/12#

Abigail Allen · Answer

undefined To find the probability of getting either a sum of 8 or at least one 4 in the roll of a pair of dice, we can calculate the probabilities of each event separately and then add them together.

1. Probability of getting a sum of 8:
   There are 5 combinations of numbers that sum up to 8: (2,6), (3,5), (4,4), (5,3), and (6,2). Each combination has a probability of $ \frac{1}{36} $ since there are 36 possible outcomes when rolling two dice.

2. Probability of getting at least one 4:
   There are 11 combinations that include at least one 4: (1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,4), and (6,4). Each of these combinations has a probability of $ \frac{1}{36} $.

To find the overall probability, we add the probabilities of getting a sum of 8 and getting at least one 4:

$$ P(\text{sum of 8 or at least one 4}) = P(\text{sum of 8}) + P(\text{at least one 4}) $$

$$ P(\text{sum of 8 or at least one 4}) = \frac{5}{36} + \frac{11}{36} = \frac{16}{36} = \frac{4}{9} $$

So, the probability of getting either a sum of 8 or at least one 4 in the roll of a pair of dice is $ \frac{4}{9} $.