1. ## dice

Throw outcome
1 lose 3.rs
2 lose 2.rs
3 lose 1.rs
4 neither
5 win 1.rs
6 win 5.rs

this table represents a outcome of a single dice when thrown ...now what is probability of having of having at least 5.rs at the end of 2 throws ..

my ans is 1/9 is it correct

2. I get five different pairs each gives at least 5rs.

3. Hello, arunachalam.s!

Edit: Once again, I've made a really stupid blunder . . . embarrassing!

A die is thrown. These are the payoffs.

. . . $\displaystyle \begin{array}{c|c} \text{Throw} & \text{Payoff (\$)} \\ \hline
1 & \text{-}3 \\ 2 & \text{-}2 \\ 3 & \text{-}1 \\ 4 & 0 \\ 5 & +1 \\ 6 & +5 \end{array}$What is the probability of having of having at least$5 at the end of 2 throws?

My answer is $\displaystyle \tfrac{1}{9}$ . Is it correct? . . . . no
There are $\displaystyle 6^2 = 36$ outcomes for two rolls of a die.

To win at least $5, we must: . . Roll a 4 and a 6 in some order: .$\displaystyle (4,6),(6,4)$. . . two ways. . . Roll a 5 and a 6 in some order: .$\displaystyle (5,6), (6,5)$. . . two ways. . . Roll a 6 and a 6: .$\displaystyle {\color{blue}(6,6)}$. . . one way. Hence, there are five ways to have at least$5.

Therefore: .$\displaystyle P(\text{at least \$5}) \:=\:{\color{blue}\frac{5}{36}}\$