My Math Forum Basic Stats Problem!

 Probability and Statistics Basic Probability and Statistics Math Forum

 March 12th, 2016, 03:48 PM #1 Newbie   Joined: Mar 2016 From: Oregon Posts: 2 Thanks: 0 Basic Stats Problem! Hey all - Going through an intro to stats class here, and am having an issue with a homework problem that we have not covered in class yet. Here it is: “As a poor college student, I bought four $2 scratchers in hopes to win some extra cash. On the back of the scratcher, it said the odds of winning are 1 out of every 3.8. So with my quick thinking I thought I would for sure win. I scratched all of them and didn't win anything. What is the likelihood of that happening?” I know the reason why he didn't win, but I am looking for some help with the probability numbers itself of him not winning after scratching 4 of them (at a 1 in 3.8 chance) I am sure the answer is trivial and simple, but I am stuck. Really appreciate the help! Last edited by snewman2; March 12th, 2016 at 03:51 PM.  March 12th, 2016, 05:31 PM #2 Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 The probability of him winning on one of the scratchies is$\dfrac{1}{3.8} = \dfrac{5}{19}$. Thus, the probability of him losing on one of the scratchies is$1 - \dfrac{5}{19} = \dfrac{14}{19}$. Since he lost on all four scratchies, the required probability is$\left(\dfrac{14}{19}\right)^4$. Thanks from snewman2 March 12th, 2016, 05:47 PM #3 Newbie Joined: Mar 2016 From: Oregon Posts: 2 Thanks: 0 Quote:  Originally Posted by Azzajazz The probability of him winning on one of the scratchies is$\dfrac{1}{3.8} = \dfrac{5}{19}$. Thus, the probability of him losing on one of the scratchies is$1 - \dfrac{5}{19} = \dfrac{14}{19}$. Since he lost on all four scratchies, the required probability is$\left(\dfrac{14}{19}\right)^4\$.
You are a god sir. Thanks for the quick reply. I completely understand that concept now, and it will really help me going forward.

Appreciate it!

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