BuilderBoi

The only way Meena loses is if Bob wins the next 2 points. The probability of Bob winning the next 2 points is \(({1 \over 3})^2 = {1 \over 9}\).

So, the probability Meena wins is \(1 - {1 \over 9} = \color{brown}\boxed{8 \over 9}\)

No, I agree with your answer...

There are \(10^4 = 10000\) ways to choose the 4 letters, and there are only 2 ways to choose the word "MATH" (there are 2 T's), so the probability is \(\color{brown}\boxed{1 \over 5000}\)

If the base is odd, the number will never be even, because there are 3 odd numbers. The sum of these numbers will always be odd, because\(\text{odd} \times \text{odd} = \text{odd}\), and \(\text{odd} + \text{odd} + \text{odd} = \text{odd}\) 

However, if the base is even, all the individual digits will yield an even number because \(\text{even} \times \text{odd} = \text{even}\), and \(\text{even} + \text{even} = \text{even}\)

So, how many even digits are there between 7 and 100?

There are \(6 \times 20 = 120 \) total outcomes. 

Of these, \(5 \times 16 = 80\) outcomes have neither Alice nor Bob rolling a multiple of 5

So, there are \(120 - 80 = 40\) successes, so the probability is \({40 \over 120} = \color{brown}\boxed{1 \over 3}\)

There are \(6^3 = 216\) ways to roll the dice

There are only 2 ways to get 16:  \((5, 5, 6)\), \((4, 6, 6)\)

In both cases, there are \(3! \div 2! = 3\) ways to order them

Thus, the probability is \({2 \times 3 \over 216} = {6 \over 216} = \color{brown}\boxed{1 \over 36}\)

https://web2.0calc.com/questions/geometric-probability_8

This was my 1000th answer lol...

We have \(j = 4\), \(a = 11\), and \(z = 16\), meaning \(jaz^2 = 4 \times 11 \times (16^2) = \color{brown}\boxed{11,264}\)

Note that the squared applies to the \(z\) only, not \(j \times a \times z\)