# Probabilty Question

This question seems easy, but for some reason I’m having a hard time grasping the concept. Is there an easy way to remember how to do these? It’s from the Schweser Q Bank A casual laborer has a 70 percent chance of finding work on each day that she reports to the day labor marketplace. What is the probability that she will work three days out of five? A) 0.3192. B) 0.5165. C) 0.6045. D) 0.3087. Your answer: B was incorrect. The correct answer was D) 0.3087. P(3) = 5! / [(5 ¨C 3)! ¡Á 3!] ¡Á (0.73) ¡Á (0.32) = 0.3087 = 5 ¡ú2nd¡ú nCr ¡ú 3 ¡Á 0.343 ¡Á 0.09

number of combinations to choose three days out of five days is 5!/(3!*(5-3)!)=5*4/2=10 then probability that she finds work on each of those three days is .7, on all three days (.7)^3. probability that she doesn’t find work on each of the other two days is 1-.7 = .3, on all two other days (0.3)^2 Answer is 10*(0.7)^3*(0.3)^2 = 0.3087

maratikus, thanks also, I assume this question is testing our knowledge on binomial random variables?

I get it until 5!/(3!*(5-3)!)=5*4/2=10 . From then, how do u get 0.3087? maratikus, could u pls elaborate? tks in advance.

Let’s say you toss a coin once, the chance of you getting a tail is 1/2 (two scenarios - either head with prob 1/2 or tails with prob 1/2). The chance of getting two tails in a row in 1/2*1/2 (because there are four scenarios head, head; head, tail; tail, head; tail, tail - each has probability of 1/4 = 1/2*1/2). Now if there are three tails chances are 1/2*1/2*1/2 = 1/8. Now let’s assume the coin is not symmetrical and probability of tails is 0.7, then chances of three tails in a row is (0.7)*(0.7)*(0.7) = (0.7)^3.

rjs157 Wrote: ------------------------------------------------------- > maratikus, thanks > > also, I assume this question is testing our > knowledge on binomial random variables? yes, It’s testing the knowledge of B-RV’s ---------------------------------------------------------------------------------------- P(X = x) = (number of ways to choose x from n) * §^x * (1- p)^(n-x) ---------------------------------------------------------------------------------------- So for the above problem n=5, x=3 & “Probability of finding work” is what I’ll label as SUCCESS, therefore p = 0.70 P(3) = 5!/(2! * 3!) * (0.70)^3 * (1-0.70)^(5-3) = 5!/(2! * 3!) * (0.70)^3 * (0.3)^(2) = 0.3087 - Dinesh S

Dinesh, thanks It’s becoming more clear the more I look at it. I just wish it could come to me intuitively, rather than having to memorize the formula

For me, to intuitively understand it, i break the problems answer into two parts: first is that one way it can be done is find work the first three days and not the last two…which is .7x.7x.7x.3x.3…and then i multiply that answer by the number of combinations of days finding work and days not finding…which has already been shown to be ten. Its easy to memorize the formulas, but for me, remembering how and why to apply them makes it easier. so: what is the probablility it can be done one way, and how many different ways can it be done. i hope that helps.

petetini, great explanation i like your approach to breaking down the problem into parts, then putting it all together. It helps thanks rjs157

my pleasure, i found this forum today…up until now i have been studying in a box, its so nice to not be alone.

I agree. This forum is great. There is a good amount of people on this board that know what they’re doing. I plan to spend more time here between now and the exam. It’s a great way to get different perspectives on approaching problems.