homework-Probability-and-statistics

Discussion in 'Math' started by couldi, Feb 28, 2012.

1. couldi Thread Starter New Member

Feb 28, 2012
4
0
I need some help about these 3 questions;

A manufacturing plant makes radios that each contain an integrated circuit (IC) supplied by three sources A, B, and C. The probability that the IC in a radio came from one of the sources is 1/3, the sam efor all sources. ICs are known to be defective with probabilities 0.001, 0.003, and 0.002 for sources A, B, and C, respectively.
(a) What is the probability that any given radio will contain a defective IC?
(b) If a radio contains a defective IC, find the probability it came from source A.

You have two biased coins. Coin A comes up heads with probabaility ¼. Coin B comes up heads with probability ¾. However, you are not sure which is which so you choose a coin randomly and you flip it. If the flip is heads, you guess that the flipped coin is B; othervise you guess that the flipped coin is A. Let events A and B designate which coin was picked. What is the probability P(C) that your guess is correct? (Hint: This is an example of sequential experiments, so you might want to use tree diagrams.)

Assume drivers are independent.
a) If 5% of the drivers fail to stop at a red light, find the probability tahat at least 2 of the next 100 drivers fail to stop.
b) If on the average 3 drivers fail to stop at the red light during each rush hour, what is the probability that at least 2 drivers fail to stop at the red light during tonights rush hour?

2. Georacer Moderator

Nov 25, 2009
5,151
1,266
Since this looks like a homework question, have you done anything so far? Post your progress and we 'll take it from there.

3. couldi Thread Starter New Member

Feb 28, 2012
4
0
second question
use Bayes' theorem:

We know that P(B) = 1/2 because coins A and B are equally likely to be picked. P(head) is also 1/2 because

Therefore,

Additionally, P(A|tail) = 3/4. The answer is *not* 3/4 + 3/4 = 3/2, that's absurd. It is 3/4 because P(head) and P(tail) both have a 1/2 chance of occurring, and we need to multiply by 1/2 to compensate for each cases.

I dont any idea about 1. and 3. questions

4. 1chance Member

Nov 26, 2011
42
184
For #1 & #3 you first need to decide if you are dealing with a binomial distribution, geometric distribution (looking for 1st success), or Poisson distribution. Make sure you consider the Poisson not only for “interval” situation but also for the rare event scenario where n>100 and np< 10. You have 3 choices for the method on each of these—using a table to look up values, using a graphing calculator or applying the “formula”. On #3 you will want to use the complement (that is 1 – computed probability) along with the “cumulative” function on the graphing calculator (or sum using table). I would use a tree diagram on #2 as suggested in the hint.