$$ 12(5.5,5.6,5.7) $$ 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 5.2 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? 5.7 .25 (b) What is the probability that an individual would win $$ \$ 1 $$ million twice in the second scratch-off game? Part 1 of 2 (a) The probability that an individual would win $$ \$ 1 $$ million in both games if they bought one scratch-off ticket from each game is $$ \mathbf{3 1} $$. (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) Winning in both games**

1. Let the probability of winning in the first game be
   $$
   p_1=\frac{1}{5.2\times10^6}.
   $$
2. Let the probability of winning in the second game be
   $$
   p_2=\frac{1}{605600}.
   $$
3. The probability of winning both games with one ticket each is obtained by multiplying the two probabilities:
   $$
   p=p_1\times p_2=\frac{1}{5.2\times10^6}\times\frac{1}{605600}.
   $$
4. Multiply the denominators:
   $$
   5.2\times10^6\times605600=5.2\times605600\times10^6.
   $$
5. Compute $$5.2\times605600$$:
   $$
   5.2\times605600=5.2\times600000+5.2\times5600=3120000+29120=3149120.
   $$
6. Thus, the combined denominator is
   $$
   3149120\times10^6=3.14912\times10^{12}.
   $$
7. The probability for part (a) is therefore
   $$
   p=\frac{1}{3.14912\times10^{12}}\approx3.2\times10^{-13}.
   $$

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**(b) Winning twice in the second game**

1. The probability of winning once in the second game is
   $$
   p_2=\frac{1}{605600}.
   $$
2. Assuming two independent tickets are purchased, the probability of winning twice is:
   $$
   p=\left(\frac{1}{605600}\right)^2=\frac{1}{(605600)^2}.
   $$
3. Express $$605600$$ in scientific notation: $$605600=6.056\times10^5$$. Then
   $$
   (605600)^2=(6.056\times10^5)^2=6.056^2\times10^{10}.
   $$
4. Calculate $$6.056^2$$:
   $$
   6.056^2\approx36.67.
   $$
5. Therefore,
   $$
   (605600)^2\approx36.67\times10^{10}=3.667\times10^{11}.
   $$
6. The probability of winning twice in the second game is then
   $$
   p\approx\frac{1}{3.667\times10^{11}}\approx2.7\times10^{-12}.
   $$
**(a) Probability of winning both games:** The probability is approximately $$3.2 \times 10^{-13}$$. **(b) Probability of winning twice in the second game:** The probability is approximately $$2.7 \times 10^{-12}$$.

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