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You roll a dice $6$ times. What is the probability of rolling at least one $5$ AND at least one $6$?

The answer in the book is $1 - (5/6)^6 - (5/6)^6 + (4/6)^6$. Would someone please explain why that is?

$(1 - (5/6)^6 - (5/6)^6)$ : This is the probability of rolling a $5$ OR $6$ for six rolls of the dice. Correct?

What is $(+ (4/6)^6)$? Isn't that the probability of not rolling a $5$ or $6$? Why do I need to add it.

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The probability of the event is

$P($ at least one $5$ and at least one $6)$

$=P($ at least one $5)+P($ at least one $6)-P($ at least one $5$ & $6)$

$=1-\left(\frac56\right)^6+1-\left(\frac56\right)^6-\{1-\left(\frac46\right)^6\}$

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    @Justin, exactly.2012-12-23