A die is rolled four times. What is the probability that exactly two multiples of 3 are rolled?
I know that the multiples of 3 in the dice are only 3 and 6. Therefore there would be a for getting three or six would be :
$\frac{2}{6}$$=$$\frac{1}{3}$
Then since there are two die we would have to multiple them together so we get: $\frac{1}{3}$$*$$\frac{1}{3}$$=$$\frac{1}{9}$
This is the part that I don't understand. Would I have to multiply four like this : $4$$*$$\frac{1}{9}$ to get $\frac{4}{9}$ or would I have to do something else.
Help would be appreciated.
Let probability of event A be P(A) & probability of event B be P(B)
Provided that event A is independent of event B,
The probability of BOTH A & B Happening = P(A AND B) = P(A) * P(B)
We write it as P(A $\cap$ B) = P(A) $\times$ P(B)
Hope this helps