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You are going to throw a 6-sided die. What is the probability that you will throw at least three "one's" in 4 tries?

 Apr 7, 2020
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P ( three 1's)  =  C(4/3) ( 1/6)^3 (5/6)   =   5/324 =  20/1296

 

P(four 1's)  =  (1/6)^4  =  1/1296 

 

So...the total probability =  (20/1296)  + (1/1296)  =  21 / 1296  =  7 / 432

 

 

cool cool cool

 Apr 7, 2020

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