$$ 12(5.5,5.6,5.7) $$ Once a woman won $$ \$ 1 $$ million in a scratch-off game from a lottery. Some years later, she won $$ \$ 1 $$ million in another scratch-off game. In the first game, she beat odds of 1 in 4.2 million to win. In the second, she beat odds of 1 in 605,600 . (a) What is the probability that an individual would win $$ \$ 1 $$ million in both games if they bought one scratch-off ticket from each game? (b) What is the probability that an individual would win $$ \$ 1 $$ million twice in the second scratch-off game? (a) The probability that an individual would win $$ \$ 1 $$ million in both games if they bought one scratch-off ticket from each game is (Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to the nearest tenth as needed.)

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**(a) Calculation for winning in both games:**

1. The probability of winning in the first game is  
   $$
   p_1=\frac{1}{4.2\times10^6}.
   $$

2. The probability of winning in the second game is  
   $$
   p_2=\frac{1}{605600}.
   $$

3. Since the events are independent, the probability of winning both is  
   $$
   p=p_1\times p_2=\frac{1}{4.2\times10^6}\times\frac{1}{605600}.
   $$

4. Multiply the denominators:  
   $$
   (4.2\times10^6)(605600)=4.2\times605600\times10^6.
   $$
   First, compute  
   $$
   4.2\times605600=2,522,352.
   $$
   Thus,  
   $$
   (4.2\times10^6)(605600)=2,522,352\times10^6=2.522352\times10^{12}.
   $$

5. Therefore,  
   $$
   p=\frac{1}{2.522352\times10^{12}}.
   $$

6. Converting to scientific notation and rounding to the nearest tenth gives approximately  
   $$
   p\approx3.9\times10^{-13}.
   $$

**(b) Calculation for winning twice in the second game:**

1. The probability of winning in a single ticket of the second game is  
   $$
   p_2=\frac{1}{605600}.
   $$

2. For two independent attempts (buying two tickets) in the same game, the probability is  
   $$
   p=(p_2)^2=\left(\frac{1}{605600}\right)^2.
   $$

3. Calculate the denominator:  
   $$
   (605600)^2\approx3.6675\times10^{11}.
   $$

4. Thus,  
   $$
   p\approx\frac{1}{3.6675\times10^{11}}\approx2.7\times10^{-12}.
   $$

**Answers:**

(a) $$3.9\times10^{-13}$$

(b) $$2.7\times10^{-12}$$
**(a) Probability of winning both games:** The chance of winning both scratch-off games is approximately $$3.9 \times 10^{-13}$$. **(b) Probability of winning twice in the second game:** The chance of winning the second scratch-off game twice is approximately $$2.7 \times 10^{-12}$$.

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