Answer:
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 probability of having atleast one dice with a value greater than or equal to 4 is 0.8163
Using a sample space which is attached below :
Recall :
Probability = required outcome / Total possible outcomes
Required outcomes = 40
Total possible outcomes = 7² = 49
The probability of having atleast one dice with a value greater than or equal to 4 will be :
Probability = 40 / 49 = 0.8163
Learn more : https://brainly.com/question/18153040
