Question

$6$ symmetrical dice are thrown $1458$ times. The number of times you expect three dice to show a four or five is

• A. $466$
• B. $320$
• C. $480$
• D. $600$

Answer 1

The correct option is A $466$

Probability of getting a $4$ or $5=\frac{2}{6}=\frac{1}{3}$
Hence$p=\frac{1}{3}$ and $q=1-\frac{1}{3}=\frac{2}{3}$, $n=6$ and as at least three dice then say $r=3$
Therefore probability of getting $4$ or $5$ in atleat $3$ dice is $P\left(y\right)=1-\left[P\right(X=0)+P(X=1)+P(X=2\left)\right]$

$P\left(y\right){=}^6{C}_0{q}^6{+}^6{C}_1{q}^{5}p{+}^6{C}_2{q}^4{p}^2$
$P\left(y\right)=1-\left[\right(2/3{)}^6.(1/3{)}^0+6(2/3{)}^5\left(1/3\right)+15\left(2/3{)}^4\right(1/3{)}^2]$

$P\left(y\right)=1-[\frac{64}{729}+\frac{6\times 32}{729}+\frac{15\times 16}{729}]$
$P\left(y\right)=1-\left[\frac{496}{729}\right]$

$P\left(y\right)=\frac{233}{729}$
Therefore,

For $1458$ trials . we can expect $1458\times \frac{233}{729}=466$ times.

Hence, this is the answer.