Question

In five throws with a single die what is the chance of throwing (1) three aces exactly, (2) three aces at least?

• A. $\frac{125}{3888}$,$\frac{625}{648}$, Favourable way for event $E$ when one number is selected are $(2,9)$
• B. $\frac{125}{3888}$,$\frac{23}{648}$, Favourable way for event $E$ when one number is selected are $(2,9)$
• C. $\frac{3763}{3888}$,$\frac{23}{648}$, Favourable way for event $E$ when one number is selected are $(2,9)$
• D. $\frac{3763}{3888}$,$\frac{625}{648}$, Favourable way for event $E$ when one number is selected are $(2,9)$

Answer 1

Answer: (B)

$\frac{125}{3888}$,$\frac{23}{648}$, Favourable way for event $E$ when one number is selected are $(2,9)$

Total number of throws i.e. $n=5$
Let $p$ be the probability of throwing an ace and $q$ be of not throwing an ace in a single throw of die.
Probability of getting an ace (1) is $p=\frac{1}{6}$
$\Rightarrow q=1-\frac{1}{6}=\frac{5}{6}$
Probability of exactly $3$ aces is
$P(X=3){=}^5{C}_3{p}^3{q}^2$

$P(X=3)=10\left(\frac{1}{6}{)}^3\right(\frac{5}{6}{)}^2$

$\Rightarrow P(X=3)=\frac{125}{3888}$
Probability of at least $3$ aces is
$P(X=3)+P(X=4)+P(X=5)$

${=}^5{C}_3{p}^3{q}^2{+}^5{C}_4{p}^4{q}^1{+}^5{C}_5{p}^5{q}^0$
$=10\left(\frac{1}{6}{)}^3\right(\frac{5}{6}{)}^2+5\left(\frac{1}{6}{)}^4\right(\frac{5}{6})+(\frac{1}{6}{)}^5$
$=\frac{1}{6^5}[250+25+1]=\frac{276}{7776}=\frac{23}{648}$