Destroyer QR #65 help

[richard86]

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I am having trouble understanding QR destroyer #64 (not 65, sorry I made a typo on the thread title and it won't let me edit title after submit). The solution says to use this formula P=(nCr)(p)^n (1-p)^r, where P is the desired outcome probability, p is the probability of a desired event occuring and n=# of trials, r=# of trials desired.. I don't understand why the solutions page says to plug in 2 as n and 1 as r.. isn't n=3, and r=2?? im not sure if i am allowed to post the actual question, so if anyone has the actual destroyer and has completed this question please help me? Thank you guys!

 

[Streetwolf]

I am having trouble understanding QR destroyer #64 (not 65, sorry I made a typo on the thread title and it won't let me edit title after submit). The solution says to use this formula P=(nCr)(p)^n (1-p)^r, where P is the desired outcome probability, p is the probability of a desired event occuring and n=# of trials, r=# of trials desired.. I don't understand why the solutions page says to plug in 2 as n and 1 as r.. isn't n=3, and r=2?? im not sure if i am allowed to post the actual question, so if anyone has the actual destroyer and has completed this question please help me? Thank you guys!

Is that the probability of 1 son out of 3 children or something? I can't help much without the actual question.

The formula normally reads:

P = (nCr) * (p)^r * (1-p)^(n-r)

n = number of events
r = number of positive outcomes
p = probability of a positive outcome

 

[simplyome]

If this is the destroyer question I think you are talking about with rain days. The solutions tells you the wrong formula. The formula that streetwolf wrote is the actual formula - which I checked with my statistics textbook.