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ryzh [129]
3 years ago
13

Consider the following code segment, which is intended to simulate a random process. The code is intended to set the value of th

e variable event to exactly one of the values 1, 2, or 3, depending on the probability of an event occurring. The value of event should be set to 1 if the probability is 70 percent or less. The value of event should be set to 2 if the probability is greater than 70 percent but no more than 80 percent. The value of event should be set to 3 if the probability is greater than 80 percent. The variable randomNumber is used to simulate the probability of the event occurring.
int event = 0;
if (randomNumber <= 0.70)
{
event = 1;
}
if (randomNumber <= 0.80)
{
event = 2;
}
else
{
event = 3;
}
The code does not work as intended. Assume that the variable randomNumber has been properly declared and initialized. Which of the following initializations for randomNumber will demonstrate that the code segment will not work as intended?
A) randomNumber = 0.70;
B) randomNumber = 0.80;
C) randomNumber = 0.85;
D) randomNumber = 0.90;
E) randomNumber = 1.00;
Computers and Technology
1 answer:
lora16 [44]3 years ago
3 0

Answer:

A) randomNumber = 0.70;

Explanation:

The code works perfectly for all values of randomNumber given in the options except option (a)

This is so because:

The first condition

<em>if (randomNumber <= 0.70)  {  event = 1;  }</em>

It checks if randomNumber is less than or equal to 0.70. This is true, so event is set to 1.

However, the next condition after it will also be executed because it is a different conditional to be executed.

So, we have:

if (randomNumber <= 0.80)  {  event = 2;  }

This condition is also true, so event is set to 2.

From the question randomNumber is meant to be event 1. However, this is different from the result of executing the code.

Hence, (a) is true

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