At 95% confidence interval, the z score is 1.96. Using the
formula:
Confidence Interval = x ± z s / sqrt (n)
where x is the average time served, s is standard deviation
and n is number of samples
Confidence interval = 7.5 ± (1.96 * 3) / sqrt (400)
Confidence interval = 7.5 ± 0.294
<span>Confidence interval = 7.206 to 7.794</span>
Multiples of 11 between 1 and 30:
11, 22
So there are 2 numbers that are multiples of 11 in the bin. There are a total of 30 cards, so the probability is written as 2/30. Or we can simplify it to 1/15.
For the next question:
There are a total of 3 + 8 = 11 balls in the bag.
The probability of choosing a red ball is 3/11.
The probability of choosing a green ball is 8/11.
Multiply the three fractions:
3/11 * 3/11 * 8/11 = 72/1331
So the probability is 72/1331.
For the last question:
A standard deck of cards has 52 cards.
There are 4 queens and 4 kings in the deck.
Probability of choosing a queen is 4/52, and the probability of choosing a king AFTER you already chose a queen is 4/51.
Multiply the two fractions:
4/52 * 4/51 = 16/2652
So the probability is 16/2652 or 4/663
First you would take 312 and divide it by 13. Using that answer and put it over 1. Then times the answer that you just got by twelve. Then put that in a mixed number in its simplest form.
Answer:
0.00002 = 0.002% probability of actually having the disease
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Positive test
Event B: Having the disease
Probability of having a positive test:
0.05 of 1 - 0.000001(false positive)
0.99 of 0.000001 positive. So
Probability of a positive test and having the disease:
0.99 of 0.000001. So
What is the probability of actually having the disease
0.00002 = 0.002% probability of actually having the disease
