Question

$6$ symmetrical dice are thrown $729$ times. The number of times you expect exactly three dice showing a four or five is

• A. $160$
• B. $240$
• C. $12$
• D. $6$

Answer 1

The correct option is A $160$

$P\left(4or5\right)=\frac{1}{3}P\left(not4or5\right)=\frac{2}{3}$

$P(X=3)=6C_3(\frac{1}{3}\ast \frac{1}{3}\ast \frac{1}{3}\ast \frac{2}{3}\ast \frac{2}{3}\ast \frac{2}{3})=\frac{160}{729}$

So, if Dices are rolled $729$ times

$\mu =729\ast \left(P\right(X=3\left)\right)=\frac{160}{729}\ast 729=160$