Question

The chance that a person with two dices, the faces of each being numbered $1$ to $6$, will throw aces exactly $4$ times in $6$ trials is

• A. ${\left(\frac{1}{36}\right)}^4$
• B. ${\left(\frac{1}{36}\right)}^4{\left(\frac{35}{36}\right)}^2$
• C. ${}_4^{6}C{\left(\frac{1}{36}\right)}^4{\left(\frac{35}{36}\right)}^2$
• D. ${}_4^{6}C{\left(\frac{1}{36}\right)}^2{\left(\frac{35}{36}\right)}^4$

Answer 1

The correct option is C ${}_4^{6}C{\left(\frac{1}{36}\right)}^4{\left(\frac{35}{36}\right)}^2$

Total cases for two dice: $6\times 6=36$
Cases when the faces of each being numbered $1$ to $6$ $=1$
Probability when the faces of each being numbered $1$ to $6$ $P=\frac{1}{36}$
Probability for $4$ times in $6$ trials ${=}_4^{6}C{\left(P\right)}^4{\left(\right(1-P\left)\right)}^2$
${=}_4^{6}C{\left(\frac{1}{36}\right)}^4{\left(\frac{35}{36}\right)}^2$