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The probability of success is .15 for each trial. (a) What is the probability of success after 3 trials. (b) How many attempts should realistically be made to ensure success.

(a) I'm thinking p=.15 and q=.85
p(X=3)=(.85)^3 * .15 =.092

(b) E(X) = 1/p = 1/.15 = 7

The numbers don't seem to check out! Help (with explanation) will be appreciated! Thanks!

2006-10-07 10:13:50 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

a) p is 1 - (.85)^3 or ≈ 0.3859 is after 3 trials. 1-(.85)^n is success after n trials.

b) you need to define the word "ensure". the probability is greater than .5 after 5 trials. ≈55.6% = 1-.85^5.
≈80% after 10 trials
≈90% after 14 and so on...

2006-10-07 10:26:04 · answer #1 · answered by Scott R 6 · 1 0

(a) You are almost there.
There are a number of subtle ways that this question can be worded, but I will assume that what is intended is to say, what is the probability of at least one success after 3 trial. In this case you can re-word it to say, what is the probability of all three trials ending in failure. This gives the answer that you have. 0.092.
So the probability of at least one success is 1 - 0.092 = 0.908

(b) Here you need to define "realistically". For instance, how many attempts will give you a 95% chance of success. Well here we are looking at a 5% chance of total failure which is 0.05.
So we have 0.85^x = 0.05.
this gives us x*log0.85 = log 0.05
or x = log0.05/log0.85 = 18.4
Now, 18.4 tries is obviously silly so lets make it 19
And the probability of success after 18 tries is (1 - 0.85^18) = 0.9544 or 95.44%

2006-10-07 17:57:11 · answer #2 · answered by Stewart H 4 · 0 2

For kicks, I wrote a little C# program to "roll the dice" with your parameters. I ran it 1000 times; for each run, I counted how many trials were required to meet your .15 success criteria.

Success after 5 attempts (or less) happened 55.96% of the time.
Success after14 trials happened 92.02% of the time.
Success after 17 trials happened 94.95% of the time

Random myRand = new Random();
int[] sortedRes = new int[30];
int count;

//initialize array
for (int i = 0; i < sortedRes.Length; i++) {
sortedRes[i] = 0;
}

//1000 runs
for (int i = 0; i < 1000; i++) {
count = 0;
do {
count++;
} while (myRand.NextDouble() > .15);
if (count < sortedRes.Length) {
sortedRes[count]++;
}
}

for (int i = 1; i < sortedRes.Length; i++) {
Console.WriteLine(i.ToString() + System.Convert.ToChar(9) + sortedRes[i].ToString());
}

2006-10-07 18:10:25 · answer #3 · answered by Tom D 4 · 0 2

a) The probability of FAILURE is 0.85 for each trial. Success after 3 trials is no failure after three trials

=1 - (0.85 x 0.85x 0.85)

b I dunno

2006-10-07 17:24:02 · answer #4 · answered by Goddess of Grammar 7 · 0 1

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