My Math Forum Probability
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 June 16th, 2016, 08:31 AM #1 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 Probability Can someone help me and explain how to do this question? Last edited by jamesbrown; June 16th, 2016 at 08:41 AM.
 June 16th, 2016, 08:50 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics Can you show us any work you've done/any previous attempt?
 June 16th, 2016, 09:22 AM #3 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 Yeah i dont even know where to start so I don't have any working?
 June 16th, 2016, 12:28 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If, as you appear to be saying, you have never studied probability, where did you get this problem? If the probability of something happening is 'p' then the probability of it happening twice is $\displaystyle p^2$. Here, since the probability of "heads, twice" is 0.36 so the probability of "heads" is 0.6. The probability of a single tails, then, is 1- 0.6= 0.4. The probability of "tails, twice" is $\displaystyle (0.4)^2= 0.16$. For the second problem you are told that the probability of exactly one head when a coin is tossed twice is 8/25. If the probability of "heads" on one toss is p, then the probability of a tail is 1- p. The probability of one head when tossed twice is $\displaystyle 2p(1- p)= 2p- 2p^2= 8/25$. Solve $\displaystyle 2p^2- 2p+ 8/25= 0$ for p. Then calculate $\displaystyle 25p^2- 25p+ 4$. If there are n red beads out of 10 then the probability the first drawn is red is $\displaystyle \frac{n}{10}$. If Meg draws a red bead and does not replace it, there are now 9 beads left, n-1 red. The probability of getting a red bead on the second draw is $\displaystyle \frac{n-1}{9}$. The probability of drawing two red beads is ]math]\frac{n(n-1)}{90}= \frac{1}{3}[/math]. Solve that for n and calculate $\displaystyle n^2- n- 30$. Let P be the number of people in the club. We are told that 1/3 of the people in the club are men and that is "n". So n= (1/3)P and P= 3n. If there are n men out of P people in the club then the probability one person chosen at random is a man is $\displaystyle \frac{n}{P}$. That leaves n-1 men out of P-1 people so the probability the second person chosen is also man is $\displaystyle \frac{n-1}{P-1}$. So the probability two people chosen are both men is $\displaystyle \frac{n(n-1)}{P(P-1)}= \frac{1}{10}$. But, as above, P= 3n so $\displaystyle \frac{n(n-1)}{3n(3n-1)}= \frac{1}{10}$. Solve that for n and then calculate P. If these questions were for a course and you really "have no idea where to start" you need to talk to your teacher! Last edited by Country Boy; June 16th, 2016 at 01:27 PM.
 June 16th, 2016, 01:24 PM #5 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 $\displaystyle \frac{n(n-1)}{90}= \frac{1}{3}$
 June 17th, 2016, 07:47 PM #6 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 639 Thanks: 85 Multiple n(n - 1). Then cross-multiply the fractions. You will get a quadratic equations that can be factored.

 Tags probability

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rivaaa Advanced Statistics 5 November 5th, 2015 01:51 PM hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.