Each player rolls two six sided die once each and the sum of the highest roll wins. the first player rolls a 3 and 4 so that his sum is 7, what is the probability that the secon player will win

5/12 or 41.66%

Step-by-step explanation:

When throwing two six-sided dice, you have 36 possible 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} {6,4} {6,5} {6,6}

To find what is the probability the second player will win, we need to see how many of those 36 possibilities have a combined total of 8 or more (to beat the 7 of the first player):

These 15 combinations have a total of 8 or more:

{2,6} {3,5} {3,6} {4,4} {4,5} {4,6} {5,3} {5,4} {5,5} {5,6} {6,2} {6,3} {6,4} {6,5} {6,6}

So, the probability the second player gets 8 or more and wins is:

15/36 or 5/12 or 41.66%

it’s supposed to be -4+45=41

step-by-step explanation: