The table below describes the smoking habits of a group of asthma sufferers. If one of the 1156 people is randomly selected, find the probability that the person is a man or a heavy smoker. Round to three decimal places as needed.

Submit test 1 is a man or a heavy smoker. Round to three decimal places as needed
$\begin{array}{lccccc}\text { Nonsmoker } & \begin{array}{c}\text { Occasional } \\ \text { Smoker }\end{array} & \begin{array}{c}\text { Regular } \\ \text { Smoker }\end{array} & \begin{array}{c}\text { Heavy } \\ \text { Smoker }\end{array} & \text { Total } \\ \text { Men } & 431 & 50 & 71 & 49 & 601 \\ \text { Women } & 382 & 48 & 86 & 39 & 555 \\ \text { Total } & 813 & 98 & 157 & 88 & 1156\end{array}$
A. 0.557
B. 0.511
c. 0.596
D. 0.554
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Step 1 ：The problem is asking for the probability that a randomly selected person is either a man or a heavy smoker. This can be calculated by adding the probability of the person being a man and the probability of the person being a heavy smoker. However, since there can be overlap (i.e., the person can be both a man and a heavy smoker), we need to subtract the probability of the person being both a man and a heavy smoker to avoid double-counting.

Step 2 ：The probability of an event is calculated by dividing the number of ways the event can occur by the total number of outcomes. In this case, the total number of outcomes is the total number of people, which is 1156.

Step 3 ：The number of ways the event 'the person is a man' can occur is the total number of men, which is 601. So, the probability of the person being a man is \(\frac{601}{1156}\).

Step 4 ：The number of ways the event 'the person is a heavy smoker' can occur is the total number of heavy smokers, which is 88. So, the probability of the person being a heavy smoker is \(\frac{88}{1156}\).

Step 5 ：The number of ways the event 'the person is both a man and a heavy smoker' can occur is the number of men who are heavy smokers, which is 49. So, the probability of the person being both a man and a heavy smoker is \(\frac{49}{1156}\).

Step 6 ：So, the probability of the person being either a man or a heavy smoker is \(\frac{601}{1156} + \frac{88}{1156} - \frac{49}{1156}\).

Step 7 ：Calculating this gives us a probability of approximately 0.554.

Step 8 ：Final Answer: The probability that a randomly selected person is either a man or a heavy smoker is approximately \(\boxed{0.554}\).