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A fair die is cast four times. Calculate
the probability of obtaining exactly two
6's.
Round to the nearest tenth of a
percent.


Sagot :

Answer:

0.1319 or 13.2%

Step-by-step explanation:

You can solve this using the binomial probability formula.

The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.

Then, we can set the equation as follows:

P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k)

n=4, x=2, k=2

when x=2  (4 2)(1/6)^2(5/6)^4-2 = 0.1157

when x=3  (4 3)(1/6)^3(5/6)^4-3 = 0.0154

when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008

Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)

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