one more math quest...

[xoangelxo]

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With one fair die, find the probability of throwing two fours in five attempts?? Thanks! =)

(the answer is supposed to be (10x125)/(36x216)

 

[Streetwolf]

With one fair die, find the probability of throwing two fours in five attempts?? Thanks! =)

(the answer is supposed to be (10x125)/(36x216)

P(throw a 4) = 1/6
P(do not throw a 4) = 5/6

You want two 4s in five attempts. There are (5 choose 2) = 5!/3!2! = 10 ways to select which two rolls of the five are 4s.

Answer = 10*(1/6)^2*(5/6)^3 = (10*125)/(36*216).

You get 10 from the (5 choose 2). You get (1/6)^2 because those are the two rolls that land on 4. You get (5/6)^3 because those are the three rolls that land on any of the other 5 numbers.