Question

The probability of throwing at most $2$ sixes in $6$ throws of a single die is:

• A. $\frac{35}{18}{\left(\frac{5}{6}\right)}^3$
• B. $\frac{35}{18}{\left(\frac{5}{6}\right)}^4$
• C. $\frac{5}{18}{\left(\frac{5}{6}\right)}^4$
• D. $\frac{35}{48}{\left(\frac{5}{6}\right)}^3$

Answer 1

The correct option is B $\frac{35}{18}{\left(\frac{5}{6}\right)}^4$

Let us assume $X$ represent the number of times of getting sixes in $6$ throws of the die.
Also, the repeated tossing of a die selection are the Bernoulli trials.
Thus, probability of getting six in a single throw of die, $p=\frac{1}{6}$
Clearly, we have $X$ has the binomial distribution where $n=6$ and $p=\frac{1}{6}$
And, $q=1-p=1-\frac{1}{6}=\frac{5}{6}$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$
${=}^6{C}_x{\left(\frac{5}{6}\right)}^{6-x}.{\left(\frac{1}{6}\right)}^x$
Hence, probability of throwing at most $2$ sixes $=P(X\le 2)$
$=P(X=0)+P(X=1)+P(X=2)$
${=}^6{C}_0{\left(\frac{5}{6}\right)}^6{+}^6{C}_1{\left(\frac{5}{6}\right)}^5.\left(\frac{1}{6}\right){+}^6{C}_2{\left(\frac{5}{6}\right)}^4.{\left(\frac{1}{6}\right)}^2=1.{\left(\frac{5}{6}\right)}^6+6.\frac{1}{6}{\left(\frac{5}{6}\right)}^5+15.\frac{1}{36}.{\left(\frac{5}{6}\right)}^4={\left(\frac{5}{6}\right)}^6+{\left(\frac{5}{6}\right)}^5+\frac{5}{12}.{\left(\frac{5}{6}\right)}^4$
$={\left(\frac{5}{6}\right)}^4[{\left(\frac{5}{6}\right)}^2+\left(\frac{5}{6}\right)+\left(\frac{5}{12}\right)]$
$={\left(\frac{5}{6}\right)}^4.\left[\frac{25+30+15}{36}\right]=\frac{70}{36}.{\left(\frac{5}{6}\right)}^4=\frac{35}{18}.{\left(\frac{5}{6}\right)}^4$