3
$\begingroup$

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.

1 Answers 1

2

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 $5$ & $6)$

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

  • 0
    Thanks for your reply! What is the difference between P(at least one 5 and at least one 6) and P(at least one 5&6) ?2012-12-23
  • 2
    P(at least one 5 & 6)= 1-probability of rolling any one number among $1,2,3,4$.2012-12-23
  • 1
    So let me get this straight because I'm a bit confused. You need to subtract the probability of getting a 5 or 6 because there is some overlap and the events are not mutually exclusive?2012-12-23
  • 0
    @Justin, exactly.2012-12-23