Question

The probability of getting a total of $9$ exactly twice in $6$ tosses of a pair of dice is

• A. ${}_2^{6}C{\left(\frac{8}{9}\right)}^4{\left(\frac{1}{9}\right)}^2{+}_3^{6}C{\left(\frac{8}{9}\right)}^3{\left(\frac{1}{9}\right)}^3$
• B. ${}_2^{6}C{\left(\frac{8}{9}\right)}^4{\left(\frac{1}{9}\right)}^2$
• C. $1{-}_2^{6}C{\left(\frac{8}{9}\right)}^4{\left(\frac{1}{9}\right)}^2$
• D. ${\left(\frac{8}{9}\right)}^4{\left(\frac{1}{9}\right)}^2$

Answer 1

Answer: (B)

${}_2^{6}C{\left(\frac{8}{9}\right)}^4{\left(\frac{1}{9}\right)}^2$

We get a sum of $9$ for $(5,4),(4,5)$ and $(6,3),(3,6)$
Hence probability of getting a sum $9$, $=\frac{4}{36}$$=\frac{1}{9}$
Now for $6$ tosses, getting a sum of $9$ exactly twice is
${}^6{C}_2\left(\frac{1}{9}{)}^2\right(1-\frac{1}{9}{)}^{6-2}$

$={}^6{C}_2\left(\frac{1}{9}{)}^2\right(\frac{8}{9}{)}^4$