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 May 12th, 2011, 11:35 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability question (hard and confusing) 1. A new skin test is devised for detecting tuberculosis. To evaluate the test before it is put into use, a research indicates tuberculosis in 96% of those who have it and in 2% of those who do not. It is known that 8% of the population have tuberculosis. a) What is the probability of a randomly selected person having tuberculosis given that the skin test is positive? b) What is the probability that a person has tuberculosis given that the test indicates no tuberculosis is present? c) What is the probability of the skin test giving a false positive result? 2. A box contains 6 green balls and 4 red balls. 3 balls are chosen at random. Find the probability of obtaining exactly 2 green balls and 1 red balls are chosen a) with replacement b) without replacement 3. A carton of 12 eggs contains 3 eggs with double yolks. If 4 eggs are selected at random from the carton, without replacement, find the probability that exactly 2 eggs have double yolks. I am quite good in probability, but when comes to these question, I can't solve it and dunno how to start to solve it. I need help and explanation for the question. Thank you.
 May 13th, 2011, 03:09 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 Can you make some attempt (for any of the questions)? Have you been told what the answers should be?
 May 13th, 2011, 06:05 AM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question (hard and confusing) For question 2 a) 2/10 x 1/10 b) 2/10 x 1/9 other I dunno how to do. I need help and explanation.
May 13th, 2011, 11:10 AM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probability question (hard and confusing)

Hello, hoyy1kolko!

Quote:
 2. A box contains 6 green balls and 4 red balls. Three balls are chosen at random. Find the probability of obtaining exactly 2 green balls and 1 red balls. a) with replacement

Since the balls are replaced, the probabilities are independent.

$\begin{Bmatrix}P(G)=&\frac{6}{10}=&\frac{3}{5}\; \\ \\ \\ P(R)=&\frac{4}{10}=&\frac{2}{5}\; \end{Bmatrix}=$

$P(\text{2G,\,1R}) \:=\:{3\choose2,1}\left(\frac{3}{5}\right)^2\left( \frac{2}{5}\right)^1 \:=\:\frac{54}{125}$

Quote:
 b) without replacement

$\text{Drawing 3 of the 10 balls, there are: }\:{10\choose3} \,=\,120\text{ possible outcomes.}$

$\text{We want 2 of the 6 Green balls; there are: }\:{6\choose2} \,=\,15\text{ ways.}$
$\text{We want 1 of the 4 Red balls; there are: }\:{4\choose1} \,=\,4\text{ ways.}$

[color=beige]. . [/color]$\text{Hence, there are: }\:15\,\cdot\,4 \:=\:60\text{ ways to get 2 Green balls and 1 Red ball.}$

$\text{Therefore: }\:P(\text{2 Green, 1 Red}) \:=\:\frac{60}{120} \:=\:\frac{1}{2}$

 May 13th, 2011, 05:20 PM #5 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question (hard and confusing) Thank you sir, I get the concept now. How about question 1 and 3?
 May 15th, 2011, 02:14 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 Are you familiar with the law of total probability and with Bayes's theorem? If so, can you apply these to question 1(a)?
 May 15th, 2011, 04:58 AM #7 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question (hard and confusing) can you explain the concept briefly to me?Thank you.
 May 18th, 2011, 12:03 AM #8 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question (hard and confusing) I need help.
 May 18th, 2011, 06:37 AM #9 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: Probability question (hard and confusing) Hello, hoyy1kolko! I'm sorry, but I must point this out. #2 involves colored balls; #3 involves eggs. [color=beige]. . [/color]Other than that, the problems are the same. If you can't transfer a procedure from one problem to another, [color=beige]. . [/color]you need more help than any of us can provide. I've solved and explained about a dozen probability problems for you. You seem to have learned nothing and keep asking identical questions.
 May 18th, 2011, 08:49 AM #10 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question (hard and confusing) Thank for your advise,sir.I will try to solve the question and post the working for you.

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