Solution:
Let X = your earnings

• X = 50 - 1 = 49, when you are winning (Prize money minus cost of playing)

• X = -1, when you are losing (Cost of playing)

Since the order is not important, we will use combination.
Probability of winning, $P(\mathrm{Winning\ \$49}) = \dfrac{1}{({}^{9}C_{3})} = \dfrac{1}{84}$
Probability of losing, $P(\mathrm{Losing\ \$1}) = 1- \dfrac{1}{84} = \dfrac{83}{84}$
Expected Earnings, $E(X) = 49 * \dfrac{1}{84} + (-1)* \dfrac{83}{84} = - \dfrac{34}{84} = -\$0.405$

Correct Answer: D

Solution:
The sample space for the outcome at second rolling is: $S_2 = \{1,2,3,4,5,6\}$ Given that we have already got 1 in the first rolling, \begin{align*} S &= \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)\} \\ &= \{2,3,4,5,6,7\} \end{align*} Each of the above outcome is equally likely, which is $\dfrac{1}{6}^{\mathrm{th}}$ times. \begin{align*} E &= \sum_{i=1}^{n} P_i x_i \\&= \left(\dfrac{1}{6} \times 2\right) + \left(\dfrac{1}{6} \times 3\right) + \left(\dfrac{1}{6} \times 4\right) \\ &+ \left(\dfrac{1}{6} \times 5\right) + \left(\dfrac{1}{6} \times 6\right) + \left(\dfrac{1}{6} \times 7\right)\\ &= \dfrac{1}{6} \left(2+3+4+5+6+7\right)\\ &= \dfrac{1}{6} (27) = 4.5 \end{align*}

Correct Answer: B