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

1 answer:

Mrac [35]2 years ago

6 0

$|\Omega|=6^4=1296\\|A|=1\cdot1\cdot5\cdot5\cdot\dfrac{4!}{2!2!}=25\cdot6=150\\\\P(A)=\dfrac{150}{1296}\approx11.6\%$

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