mework $$ 12(5.5,5.6,5.7) $$ 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? (b) What is the probability that an individual would win $$ \$ 1 $$ million twice in the second scratch-off game? $$ 86.11 \%, 31 $$ of 36 points Points: 0 of 1

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UpStudy AI · Accepted Answer

(a) The probability of winning \$1 million in the first game is  
$$
P_1=\frac{1}{5\,200\,000}
$$  
and in the second game it is  
$$
P_2=\frac{1}{605\,600}.
$$  
Since the two events are independent, the probability of winning both is  
$$
P_{\text{both}}=P_1\times P_2=\frac{1}{5\,200\,000}\times\frac{1}{605\,600}=\frac{1}{5\,200\,000\times605\,600}.
$$

We can leave the answer in this form or compute the denominator:
$$
5\,200\,000\times605\,600\approx3.14912\times10^{12}.
$$
Thus,
$$
P_{\text{both}}\approx\frac{1}{3.14912\times10^{12}}\approx3.17\times10^{-13}.
$$

(b) For winning \$1 million twice in the second scratch‐off game, the ticket is bought twice, with each win having probability  
$$
P_{\text{single}}=\frac{1}{605\,600}.
$$  
Therefore, the probability of two wins is  
$$
P_{\text{double}}=\frac{1}{605\,600}\times\frac{1}{605\,600}=\frac{1}{605\,600^2}.
$$

Calculating the square gives approximately
$$
605\,600^2\approx3.6683\times10^{11},
$$
so 
$$
P_{\text{double}}\approx\frac{1}{3.6683\times10^{11}}\approx2.73\times10^{-12}.
$$
(a) The probability of winning \$1 million in both games is approximately 3.17 × 10⁻¹³. (b) The probability of winning \$1 million twice in the second scratch-off game is approximately 2.73 × 10⁻¹².

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