Question

assume that 30% of all plants in a field are infested with aphids. suppose that you pick 2 plants at random. what is the probability that exactly one of them carried aphids?

Answer 1

It depends on how many plants are in the field

Answer 2

that's the question. i got 0.49

Answer 3

0.3= infested and 0.7=not infested

Answer 4

so i did 1-0.7^2=1-0.51=0.49

Answer 5

So how many plants are there in total in the field. This is needed to calculate the probability

Answer 6

it doesn't say how many plants are in the field

Answer 7

it gives u 30% of the plants are infested
which means that 70% of the plants are not infested

Answer 8

so in that field which an unknown amount of plants it's asking for the probability that exactly one of them carried aphids.

Answer 9

Is this out of a text book?

Answer 10

yes

Answer 11

it's an even number question and has no answers so i just wanted to be sure that i'm doing it right

Answer 12

I'm writing a program to test different plant numbers and get the probability...

Answer 13

ok. thank you :)

Answer 14

Definitely needs plant numbers:

In a field with 10 plants the probability is 0.4666666666666667
with 21/45 combinations carrying 1 aphid infested plant

In a field with 100 plants the probability is 0.42424242424242425
with 2100/4950 combinations carrying 1 aphid infested plant

In a field with 100 plants the probability is 0.42042042042042044
with 210000/499500 combinations carrying 1 aphid infested plant

So it depends. Would you like to verify the code (if you understand it)

Answer 15

sure

Answer 16

thanks

Answer 17

Sure you would like to verify it?

OK:
public class PlantsInField {
public static void main(String[] args) {
int limit = 1000;
double prob = 0;
double total = calcTotal(limit);
for (int A = 1; A <= limit; A++) {
for (int B = A + 1; B <= limit; B++) {
int x = 0;
int y = 0;
if (A <= limit * 0.3) x = 1;
if (B <= limit * 0.3) y = 1;
if ((x^y) == 1) {
prob += 1;
}
}
}
System.out.println(prob+"/"+total);
prob /= total;
System.out.println(prob);
}

static int calcTotal(int N) {
int n = N;
int product = 1;
for (; n >= N - 1; n--) {
product *= n;
}
return product / 2;
}
}

Answer 18

that's fine. i don't understand programming. Thank you though :)

Answer 19

Sure thing