Quantitative Reasoning Question.. A recent survey determined that two-fifths of practicing dentists

[Hammer Time]

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A recent survey determined that two-fifths of practicing dentists use a water-pik on their own teeth. If 10 dentists are chosen at random, which expression represents the probability that exactly three of them use a water-pik on their own teeth?

So, the answer is C(10,3)*(2/5)^3*(3/5)^7.

Umm...... Could anyone explain how to get there? Also what does the C mean? I tried looking this up to no avail. Thanks in advance 😀

 

[alexis2010]

C stands for combination. Since we have 10 dentists and we are asked the probability that exactly 3 use a water pick, first we need to find how many ways we can select 3 out of 10, then those 3 dentists have to use a water pick hence (2/5)^3 and the remaining 7 don't use a water pick hence (3/5)^7.

 

[Hammer Time]

C stands for combination. Since we have 10 dentists and we are asked the probability that exactly 3 use a water pick, first we need to find how many ways we can select 3 out of 10, then those 3 dentists have to use a water pick hence (2/5)^3 and the remaining 7 don't use a water pick hence (3/5)^7.

Do you know why those numbers are multiplied? It seems odd to me to multiply the chances that it can't happen by the chances it can happen

 

[100%permwinner]

This problem is an example of binomial probability. Hope it helps.
http://www.regentsprep.org/regents/math/algtrig/ats7/blesson.htm
http://math.stackexchange.com/quest...ormula-includes-the-not-happening-probability

 

[Hammer Time]

This problem is an example of binomial probability. Hope it helps.
http://www.regentsprep.org/regents/math/algtrig/ats7/blesson.htm
http://math.stackexchange.com/quest...ormula-includes-the-not-happening-probability

That last website was bomb. Thanks man

 

[Hammer Time]

Ok, so to answer my own question (in case somebody stumbles across this someday). This problem can be thought of as a combinations problem. There are ten dentists, and 3 of them are chosen. (Think of it like out of ten people, 3 are going to the ball game). So its a combinations problem where order doesn't matter.

So, (10x9x8)/(3x2x1). That is the combinations part of the problem. This is represented by C(10,3).

Now, since we were given that 2/5 dentists use the water pick. So now we need to multiply the possible combinations of dentists (the 3 out of the 10 combination) by the probability that the dentists will actually use the water pick.

So, there are two options, either the dentists use the pick or they don't. So out of ten, it is expected that 4 use the pick and 6 do not. But, since they want to know what are the chances that exactly three use the pick we represent that by (2/5)(2/5)(2/5). Now, the chances that the other 7 will not use the pick is (3/5)(3/5)(3/5)(3/5)(3/5)(3/5)(3/5).

Now we need to multiply the possible combinations of dentists by the probability that that three of them will use the pick and that seven will not. So putting it all together:
C(10,3)*(2/5)^3*(3/5)^7.

Or in other words, (10x9x8)/(3x2x1)*(2/5)^3*(3/5)^7.