A pair of 7 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 4?

40/49

Step-by-step explanation:

One way to answer this is to ask the opposite question.

What is the probability that neither will show a number 4 or more?

That would mean BOTH have to each be a 3 or less.

For the first die to show 3 or less       -->  probability = 3/7

For the second die to show 3 or less -->  probability = 3/7

For them BOTH to show 3 or less       -->  probability = 3/7 * 3/7 = 9/49

Probability a thing happens + probability it doesn't happen = 1. Always.

Prob(both dice 3 or less) + Prob(at least one die 4 or more) = 1

So, Prob(at least one die 4 or more) = 1 - Prob(both dice 3 or less)

                                                            = 1 - 9/49

                                                            = 49/49 - 9/49

                                                             = 40/49

the one on the bottom left

step-by-step explanation:

all of the other ones have a specific pattern

the answer is 5636.7