Question

If getting a number greater than $4$ is a success in a throw of a fair die, then the probability of at least $2$ successes in six throws of a fair die is

• A. $0.649$
• B. $0.351$
• C. $0.267$
• D. $0.667$

Answer 1

Answer: (A)

$0.649$

Answer 2

p = P (success) = P (getting number greater than 4)
$=P(5,\,6)$
$=\frac{2}{6}=\frac{1}{3}$
$\therefore$ $q=P(failure)=1-P$
$=1-\frac{1}{3}=\frac{2}{3}$
By Binomial distribution $P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$ P (probability of atleast 2 successes)
$=1-P(X=0,1)$
$=1-\\{(P(0)+P(1)\\}$
$=1-{{\\{}^{6}}{{C}_{0}}{{p}^{0}}{{q}^{6}}{{+}^{6}}{{C}_{1}}{{p}^{1}}{{q}^{5}}\\}$
=1-\left\\{ 1{{\left( \frac{1}{3} \right)}^{0}}{{\left( \frac{2}{3} \right)}^{6}}+6\times \left( \frac{1}{3} \right).{{\left( \frac{2}{3} \right)}^{5}} \right\\}
$=1-\left[ {{\left( \frac{1}{3} \right)}^{0}}\times \frac{{{2}^{6}}}{{{3}^{6}}}+2\times \frac{{{2}^{5}}}{{{3}^{5}}} \right]$
$=1-\left[ \frac{64}{729}+\frac{2\times 32}{243} \right]=1-\left[ \frac{192+64}{729} \right]$
$=1-\frac{256}{729}=1-0.35=0.649$